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Unformatted text preview: Calculations In Chemistry
Modules 19 and above have been renumbered.
The former Module 28 on Weak Acids is now Module 30
If you are looking for Weak Acid topics, check Module 30
At www.ChemReview.Net ***** Module 28: Equilibrium
Module 28 – Equilibrium.............................................................................................. 817
Lesson 28A:
Lesson 28B:
Lesson 28C:
Lesson 28D:
Lesson 28E:
Lesson 28F:
Lesson 28G:
Lesson 28H:
Lesson 28I:
Lesson 28J: Le Châtelier’s Principle..................................................................................... 818
Powers and Roots of Exponential Notation................................................... 830
Equilibrium Constants...................................................................................... 840
K Values ............................................................................................................. 847
Kp Calculations .................................................................................................. 850
K and Rice Moles Tables .................................................................................... 856
K Calculations From Initial Concentrations .................................................. 863
Q: The Reaction Quotient ................................................................................. 869
Calculations Using K and Q ............................................................................. 872
Solving Quadratic Equations ........................................................................... 879
For additional modules, visit www.ChemReview.Net ©2010 ChemReview.net v. 3d Page i Module 28 — Equilibrium Module 28 — Equilibrium
Pretests: In this module, if you have had a prior chemistry course, the initial lessons may
be review. If you think you are familiar with the topic of a lesson, try a few problems at the
end of the last problem set in the lesson. If you get those right, move to the next lesson.
***** Introduction
Chemical reactions can be divided into three types.
1. Reactions that go nearly 100% to completion. Burning paper is one such reaction.
Once the reaction begins, the reaction goes until one of the reactants (the paper or
oxygen) is essentially used up.
2. Reactions that don’t go. Trying to convert carbon dioxide and water into paper is very
difficult to do in a chemist’s laboratory (though plants are able to accomplish most steps
in this reaction by the remarkable process of photosynthesis).
3. Reactions that are reversible and go partially to completion. In reversible reactions, as
the reaction proceeds, the reactants are gradually used up. As a result, the forward
reaction slows down. As product concentrations increase, they more frequently collide
and react to reform the reactants. Finally, both the forward and reverse reactions are
going at the same rate. As long as no substances or energy are added to or removed
from the reaction system, the two rates will remain equal and no further reaction seems
to take place. The system is said to be at equilibrium.
For equilibrium to exist,
• all reactants and products must be present in at least small quantities, and • the reaction must be in a closed system: no particles or energy can be entering or
leaving the reaction vessel. At equilibrium, no reaction seems to be occurring, but this appearance is deceiving.
Equilibrium is dynamic: the forward and reverse reactions continue. However, because
the rates of reaction of the forward and reverse reactions are the same, there is no net
change.
In theory, all reactions that occur are reversible, and all reactions go to equilibrium. In
practice, many equilibria favor the products so much that nearly all of the limiting reactant
is used up, and the reaction is considered to go “to completion.” For those reactions, if the
limiting reactant is known, calculation of the amounts of reactants used up and products
formed can be done using conversion stoichiometry.
When reactions go only partially to completion, there is no limiting reactant completely
used up that decides how much of the products form. Reactions that go to equilibrium
require an accounting system to track the particles used up and formed. ©2010 ChemReview.net v. 3d Page 817 Module 28 — Equilibrium Lesson 28A: Le Châtelier’s Principle
Timing: Depending on the sequence in your course, problems based on Le Châtelier’s
Principle may be assigned either at the beginning or end of a unit on Equilibrium. This
lesson may be completed at either point. When Le Châtelier’s Principle is assigned in your
class, read the introduction above and then complete this lesson.
Pretest: If you are familiar with Le Châtelier’s Principle, try Problems 3 and 4 in Practice D
at the end of this lesson. If you can do those problems correctly, you may skip this lesson.
***** Shifts In Equilibrium
Equilibrium is important because many reactions, including those in biological systems, are
in practice reversible. For reversible reactions, we want to
• predict what happens when a system at equilibrium is disrupted, and • shift an equilibrium to make as much of a wanted substance as possible. Shifts in an equilibrium can be predicted by
Le Châtelier’s Principle
If a system at equilibrium is subjected to a change, processes occur which tend to
counteract that change.
Le Châtelier’s Principle predicts the direction that a reversible reaction will shift when a
reaction mixture at equilibrium is subjected to changes in concentration, temperature, and
pressure. Changes in Concentration
To predict shifts in equilibrium due to changes in concentration, it is helpful to restate Le
Châtelier’s Principle in the following manner. For reversible reactions at equilibrium,
To Predict Shifts Due to Concentration Changes:
Write the balanced reaction equation using a twoway (
rules. ) arrow, then apply these a. Increasing a [substance] which appears on one side of an equilibrium equation
shifts an equilibrium to the other side. The other substance concentrations on the
same side as the [increased substance] are decreased, and the substance
concentrations on the other side are increased.
b. Decreasing a [substance] which appears on one side of an equilibrium equation
shifts the equilibrium toward that side. The other [substances] on the same side are
increased, and the [substances] on the other side are decreased.
Memorize the rules in the two boxes above, then apply the rules to the following problem. ©2010 ChemReview.net v. 3d Page 818 Module 28 — Equilibrium Q. Chromate ions react with acids to form dichromate ions in this reversible reaction.
2 CrO42―
(aq)
(chromate ion yellow in solution) + 2 H+ (aq) (acid) +
Cr2O72―
(aq) H2O (l) (dichromate ion orange in solution) Below, cover below the * * * * * line, and answer the questions above the line.
1. If acid (H+) is added to a yellow chromate ion solution at equilibrium,
a. Which direction will the equilibrium shift (left or right)? _________________
b. The [CrO42―] will (increase or decrease?) ____________________.
c. The [dichromate ion] will (increase or decrease?) ____________________.
d. What color change will tend to occur? ___________________________________
*****
Answers
a. The equilibrium will shift to the right. Increasing the concentration of the
H+ found on the left side shifts the equilibrium toward the right side.
b. The [CrO42―] will decrease. Increasing the concentration of a substance that
appears on one side decreases the concentration of the other substances on
that side.
c. The [dichromate ion] will increase. Increasing the [H+] that appears on the
left increases the concentration of the substances on the right.
d. Since adding acid decreases the [chromate] and increases the [dichromate],
the solution color shifts from yellow toward orange.
These shifts are consistent with what we know about chemical reactions.
• When [H+] increases, there will be more collisions between the H+ and the
chromate ions. Though the percentage of collisions that result in a reaction stays
the same if the temperature remains constant, more collisions means more
forward reaction. Increasing the [H+] means that the rates of the forward and
reverse reaction, equal at equilibrium, are thrown out of balance. The increased
forward reaction uses up chromate and forms more dichromate.
In terms of Le Châtelier’s Principle, when H+ is added, the system response is to
decrease the [H+] by increasing the rate at which H+ is used up.
• As more dichromate forms, its collisions with water increase, and the speed of
the reverse reaction increases. A new balance is reached, but only after some of
the yellow chromate has been used up and more orange dichromate has formed. *****
2. If a base is added to the orange solution that results after adding acid,
a. What will happen to the acid concentration? _________________
***** ©2010 ChemReview.net v. 3d Page 819 Module 28 — Equilibrium a. Bases neutralize acid (see Lesson 14A). This lowers the [H+].
b. Which direction will the equilibrium shift (left or right)? _________________
c. The [CrO42―] will (increase or decrease) ____________________.
d. The [Cr2O72―] will (increase or decrease) ____________________.
e. What color change will occur? ______________________________________
*****
b. Decreasing the concentration of H+, which is in the equation on the left side,
will shift the equilibrium toward that side.
c. Shifting to the left means that the [CrO42―] will increase.
d. Decreasing the concentration of a term on the left shifts the equilibrium to
the left and decreases the concentration of the terms on the right. The
[Cr2O72―] will decrease.
e. Adding base decreases the [acid]. The equilibrium shifts toward the side
with the acid term, increasing the [yellow chromate] and decreasing the
[orange dichromate]. The color changes from orange toward yellow.
In terms of what we know about chemical reactions, these shifts are logical.
• When [H+] decreases, the balanced rates at equilibrium are upset. There will be
fewer collisions between the acid and the chromate ions. Fewer collisions mean that
the reverse reaction is now faster than the slowed forward reaction. The reverse
reaction uses up orange dichromate and forms yellow chromate until a new
balanced equilibrium is reached. Driving a Reversible Reaction
Given a reversible reaction at equilibrium, if the system is opened and a reactant or product
is allowed to escape (such as letting a gas product escape from a solution reaction), the
reversible reaction is no longer in a closed system at equilibrium because a particle is
escaping. In this case, the reaction and the mixture composition will shift toward the side
that contains the escaping particle. This type of shift can be used to drive a reversible
reaction toward a side with products that are wanted. Practice A: First learn the rules, then do the problems applying the rules from memory.
Check your answers after Part a.
1. For the Haber process reaction: N2 (g) + 3 H2 (g) 2 NH 3 (g) a. If the [H2] is increased,
1. Equilibrium will shift to the (left or right?) _____________________.
2. [N2] will (increase or decrease?) _____________________.
3. [NH3] will (increase or decrease?) _____________________. ©2010 ChemReview.net v. 3d Page 820 Module 28 — Equilibrium b. If the [N2] is decreased,
1. Equilibrium will shift to the (left or right?) _____________________.
2. [H2] will (increase or decrease?) ________________________.
3. [NH3] will _________________________. Changes in Temperature
For a reaction at equilibrium, changes in energy follow Le Châtelier’s Principle in a manner
similar to changes in concentration.
• Adding energy shifts the equilibrium away from the side with the energy term, and • removing energy shifts the equilibrium toward the side with the energy term. Energy can be added to a system by increasing its temperature. Energy can be removed by
cooling a system. When energy is produced by a shift in equilibrium, the temperature of
the system goes up. When energy is used up by a reaction, the system’s temperature goes
down.
Apply those rules to the following problem. Check Part 1 answers before doing Part 2.
Q. For the reaction: N2 (g) + O2
+ 90.0 kilojoules
(g) 2 NO (g) 1. If the temperature of the reaction vessel is increased,
a. The equilibrium will shift to the (left or right?) _________________
b. The [O2] will (increase or decrease?) ____________________.
c. The [NO] will _____________________________.
2. If the [O2] is decreased,
a. The equilibrium will shift to the ______________________
b. The [N2] will _________________________.
c. The temperature will _________________________.
*****
Answers
1. a. The equilibrium shifts to the right. Energy is a term on the left, so if energy
is added, the system uses up energy by shifting the equilibrium to the right.
b. The [O2] decreases. If a substance is added that appears on one side, the
added substance reacts additionally with and uses up other substances on
that side.
c. The [NO] increases. Increasing energy that is a term on the left causes more
of substances on the right to form.
2. a. The equilibrium shifts to the left. ©2010 ChemReview.net v. 3d b. The [N2] increases. Page 821 Module 28 — Equilibrium c. The temperature of the mixture increases. Shifting the equilibrium to the
left makes more of the terms shown on the left, including more energy.
*****
When working with Le Châtelier’s Principle, if an energy term is expressed in ΔH notation,
it helps to rewrite the equation with the energy term written inside the equation with a
positive sign.
Examples: To apply Le Châtelier’s Principle for energy Convert a negative ΔH to a positive term on the right.
N2
Rewrite as: N2 (g) + 3 H2 (g) 2 NH 3 (g) + 3 H2 (g) 2 NH 3 ΔH = ―92.4 kJ (g) (g) + 92.4 kJ Convert a positive ΔH to a positive term on the left.
H2O
H2O Rewrite as: Practice B H2O (l) (l) ΔH = + 44 kJ (g) + 44 kJ H2O (g) First learn the rules, then do the problems. 1. Convert these reactions to notation where the energy term is written as a positive
quantity on the left or right. (For additional practice, see Lesson 22B.)
a. N2
b. C (g) (g) + 2 O2 + O2 2 NO2 (g) C O2 (g) ΔH = +68 kJ (g) ΔH = ―394 kJ (g) 2. In the equilibrium for this synthesis of methyl alcohol,
CO (g) + 2 H2 (g) + energy
CH3OH
(g) a. If the temperature is decreased,
1. equilibrium will shift to the (left or right?) _____________________.
2. [CH3OH] will (increase or decrease?) _____________________.
b. If the [H2] is increased,
1. equilibrium will shift to the _____________________.
2. [CH3OH] will _____________________.
3. The temperature in the vessel will ____________________.
3. In a closed system, for the reaction: PCl3 (g) + Cl2 (g) PCl5 (g) ΔH = ―93 kJ a. Write the reaction with energy as a positive term in the equation.
b. If the temperature is increased,
1. equilibrium will shift to the ____________. 2. [PCl5] will __________________. ©2010 ChemReview.net v. 3d Page 822 Module 28 — Equilibrium c. If the [Cl2] is increased,
1. equilibrium will shift to the _______________. 2. [PCl3] will ________________.
3. The temperature in the vessel will (inc. or dec.?) ____________________. Changes in Pressure
Pressure will affect an equilibrium which is composed of reacting gases, but only if one side
of the equation has more moles of gas than the other. For a reversible reaction at
equilibrium that includes gases that are reacting,
• Increasing the pressure on the system will shift the equilibrium to create fewer total
moles of gas. • Decreasing the pressure on the system will shift the equilibrium to create more
moles of gas. • Changing the pressure does not shift the equilibrium if the total of the gas
coefficients is the same on both sides. To find which side has more moles of gas, simply add the coefficients of the gas molecules on
each side of the balanced equation.
If the volume of a reaction vessel containing gases is increased or decreased by using a
tightly sealed but moveable piston attached to the reaction vessel, pushing in the piston to
decrease the volume of the container increases the pressure on the gases and the balancing
pressure exerted by the gases. Pulling the piston out a distance increases the container
volume and decreases the pressure on and of the gases.
For a system of reacting gases at equilibrium (which must be a closed system),
• Decreasing the volume increases the pressure, shifting the equilibrium to create
fewer gas molecules. • Increasing the volume decreases the pressure, shifting the equilibrium to create
more gas molecules. However, adding a nonreacting gas to a reaction mixture at constant volume will not shift
the equilibrium. Such a change will not change the moles per liter of the reacting gases or
the partial pressures they exert, and as a result the equilibrium will not shift.
*****
Write and recite the rules above until they are committed to memory. Then apply the rules
to the following problem. Check your answers after each numbered part.
Q. For the reaction at equilibrium: 2 NO2 (g) N2O4 (g) + energy 1. If the pressure on the gases in the reaction vessel is decreased,
a. The equilibrium will shift to the (left or right?) _________________
b. The moles of NO2 will (increase or decrease?) ____________________. ©2010 ChemReview.net v. 3d Page 823 Module 28 — Equilibrium c. The moles of N2O4 will (increase or decrease?) ____________________.
d. The temperature in the vessel will (increase or decrease?) __________________.
2. If the volume of the reaction container is decreased by a piston,
a. The equilibrium will shift to the (left or right?) _________________
b. The moles of N2O4 will (increase or decrease?) ____________________.
c. The temperature in the vessel will (increase or decrease?) __________________.
3. If helium gas is added to the vessel, equilibrium shifts (left or right?) ____________
*****
1a. Equilibrium shifts left. Decreasing pressure favors the side with more gas
molecules.
1b. Moles of NO2 increase as equilibrium shifts left. 1c. Moles of N2O4 decrease. 1d. The temperature in the vessel will decrease. As a term on the right side, energy
must be used up as the equilibrium shifts to the left.
2a. Decreasing the volume increases the pressure. The equilibrium will shift to the
right. The left side has two moles of gas and the right has only one. When the
pressure is increased, the system’s response is to oppose the change and decrease
the pressure. A way it can do so is to make fewer gas molecules.
2b. Moles of N2O4 increase as the equilibrium shifts to the side with fewer gas moles.
2c. The temperature in the vessel will increase. As a term on the right side, energy
must be formed as the equilibrium shifts to the right.
3. Helium, a noble gas, does not react with other substances. Adding nonreacting
gases does not shift an equilibrium with reacting gases. Practice C
1. In the equilibrium for the formation of nitrogen dioxide:
1/2 N2 (g) + O2
+ 34 kilojoules
(g) NO2 (g) a. Which side has fewer gas molecules?
b. If the pressure on the gas mixture is increased,
1. equilibrium will shift to the _____________________.
2. Moles of NO2 will _____________________.
3. The temperature in the vessel will ____________________.
c. If the volume of the reaction vessel is increased,
1. Equilibrium will shift to the _____________________.
2. Moles of O2 will (increase or decrease?) _____________________.
3. The temperature in the vessel will ____________________. ©2010 ChemReview.net v. 3d Page 824 Module 28 — Equilibrium 2. In the equilibrium: C6H12O6 (s) + 6 O2 (g) 6 C O2
+ 6 H2O
(l)
(g) If the pressure on the reaction vessel is decreased, the equilibrium will shift to the
(left or right?) _________________
3. If, for an equilibrium that is found in sealed carbonated beverages,
CO2 (g) + H2O (l) H+ (aq) , the cap is left off the bottle,
+ HCO3―
(aq) a. in which direction will the reaction shift? _____________________
b. What will happen to the acidity of the beverage? ________________________. Concentration Changes: Special Cases
There are two special rules for concentration changes.
1. Adding or removing a solid or a liquid from a reaction at equilibrium does not shift the
equilibrium.
Solids have a constant concentration determined by their density. Adding or removing
solid does not change the solid’s concentration and does not shift an equilibrium.
Example. If solid table salt (NaCl) is added to a glass of water, initially all of the salt
dissolves. However, if enough salt is added and stirred, the solution becomes
saturated: at a given temperature it reaches the limit of dissolved salt it can hold.
This result is in agreement with Le Châtelier’s Principle. The equilibrium for this
reaction: NaCl(s)
NaCl(aq) cannot not be reached until some solid remains
after stirring, since equilibrium requires that all reactants and products be present.
However, once equilibrium is reached, adding more salt crystals increases the
amount of solid salt on the bottom, but it does not increase the concentration of the
solid salt or the dissolved salt. Adding a solid does not shift equilibrium
concentrations.
In a solid substance, the particles are tightly packed: about as concentrated as they
can be. If you hit a solid substance with a hammer, its particles may separate into
pieces, but they will not permanently compress. Changing a concentration shifts an
equilibrium, but a concentration of a solid substance cannot substantially change.
Pure liquids also have a constant concentration at a given temperature. Adding or
removing a pure liquid from a system at equilibrium will not change the concentration
of the liquid, and therefore will not shift the equilibrium.
2. If a solvent for a reaction is a term in the equation, adding more solvent, or using up
or forming solvent in the reaction, does not shift the equilibrium, because the high
concentration of a solvent is difficult to change.
By definition, a solvent is a substance present in very high concentration compared to
the other substances present in a solution. This means that in reasonably dilute ©2010 ChemReview.net v. 3d Page 825 Module 28 — Equilibrium solutions, the solvent concentration remains very close to constant even if it is used up
or formed in a reaction occurring in the solvent.
The most common solvent encountered in chemistry and in living systems is liquid
water. Adding more water to an aqueous solution at equilibrium will not substantially
shift the equilibrium. Even reactions that use up or form water will not significantly
change the concentration of water if water is the solvent for the reaction. Catalysts: No Shift
Adding a catalyst to a mixture at equilibrium will not shift the equilibrium. However,
adding a catalyst will cause the reaction equilibrium to be reached more quickly. Practice D. Commit the rules for Le Châtelier’s Principle to memory, then work these
problems. Check your answers after each lettered section. Save the last problem as a
review for your next study session.
1. For the reaction at equilibrium at high temperature
C (s) + 2 Cl2 (g) CCl4 (g) + energy a. If the [Cl2] is decreased,
1. Equilibrium will shift to the (left or right?) _____________________.
2. [CCl4] will (increase or decrease?) _____________________.
3. [C] will _____________________.
4. The amount of carbon will _____________________
5. The temperature in the reaction vessel will ___ ___________________.
b. If the pressure in the reaction vessel is decreased,
1. Equilibrium will shift to the _____________________.
2. Moles of Cl2 will ________________________.
3. The temperature in the reaction vessel will ___________________.
c. If a catalyst is added, equilibrium will shift to the _____________________.
2. Consider this reaction at equilibrium.
CH3COO―
+ H2O
(l)
(aq) CH3COOH
+ OH―
(aq)
(aq) a. If liquid water is added, equilibrium will shift to the (left or right?) ______________.
b. If the [OH―] is increased,
1. equilibrium will shift to the _____________________.
2. [CH3COOH] will _________________. 3. [H2O] will _____________________. ©2010 ChemReview.net v. 3d Page 826 Module 28 — Equilibrium c. If H+ is added,
1. equilibrium will shift to the ____________. 2. [CH3COO―] will _____________.
3. For the reaction at equilibrium,
2 H2O (g) + 2 Cl2 (g) + energy 4 HCl
+ O2
(g)
(g) a. if the [O2] is increased,
1. equilibrium will shift to the ______________. 2. [HCl] will ________________.
b. If the [HCl] is decreased,
1. equilibrium will shift to the _____________________.
2. [O2] will _____________________. 3. [Cl2] will _______________________.
4. The temperature in the reaction vessel will ___________________.
c. If the temperature in the reaction vessel is decreased,
1. equilibrium will shift to the _____________________.
2. [O2] will ______________________. 3. [H2O] will _________________________.
d. If the total pressure on the gases in the vessel is increased,
1. equilibrium will shift to the (left or right?) _____________________.
2. Moles of O2 will________________________.
3. Moles of Cl2 will _______________________.
4. The temperature in the reaction vessel will ___________________.
e. If a piston is used to increase the volume of the vessel,
1. equilibrium will shift to the _____________________.
2. Moles of O2 will ______________________.
3. Moles of H2O will _______________________.
f. If a catalyst is added to the reaction mixture,
1. equilibrium will shift to the ________________. 2. [O2] will ________________. 4. For the reaction at equilibrium:
4 PCl5 (g) + energy P4 (s) + 10 Cl2 (g) a. If the temperature in the reaction vessel is decreased,
1. equilibrium will shift to the ____________. 2. [Cl2] will _________________.
3. [PCl5] will ______________________. 4. [P4] will ________________________. ©2010 ChemReview.net v. 3d Page 827 Module 28 — Equilibrium b. If a piston is used to decrease the volume of the vessel,
1. equilibrium will shift to the _____________________.
2. Moles of Cl2 will _______________. 3. Moles of PCl5 will __________________.
4. The number of gas molecules in the vessel will _________________________. ANSWERS
Practice A
1. For N2 + 3 H2
2 NH3
(g)
(g)
(g)
a. If [H2] is increased: 1. Equilibrium shifts right. 2. [N2] will decrease. 3. [NH3] will increase.
b. If [N2] is decreased: 1. Equilibrium shifts left. 2. [H2] will increase. 3. [NH3] will decrease. Practice B
1a. N2 + 2 O2 + 68 kJ
2 NO2
1b. C + O2
CO2 + 394 kJ
(g)
(g)
(g)
(g)
(g)
(g)
2. For CO + 2 H2
+ energy
CH3OH
(g)
(g)
(g)
a. If the temperature is decreased: 1. Equilibrium shifts right. 2. [CH3OH] will increase.
b. If [H2] increases: 1. Equilibrium shifts right. 2. [CH3OH] increases. 3. Energy is on the right, so the
temperature increases.
3. a: PCl3 + Cl2
PCl5 + 93 kJ
(g)
(g)
(g)
b. If temperature is increased: 1. Equilibrium shifts left. 2. [PCl5] will decrease.
c. If [Cl2] is increased: 1. Equilibrium shifts right. 2. [PCl3] will decrease. 3. Temp. increases. Practice C
1. For 1/2 N2 + O2
+ 34 kilojoules
NO 2
(g)
(g)
(g)
a. Which side has fewer gas molecules? Left has 1.5 moles, right has 1 mole.
b. If pressure is increased: 1. Equilibrium shifts right. 2. Moles of NO2 increase. 3. Temp. decreases.
c1. If volume is increased, pressure is decreased; equilibrium shifts to left side which has more gas moles.
c2. Moles of O2 will increase. c3. The temperature will increase.
2. For the equilibrium: C6H12O6 + 6 O2
(s)
(g) 6 CO2
+ 6 H2O
(l)
(g) Pressure will not shift the equilibrium, since both sides have the same 6 moles of gas. 3. For CO2(g) + H2O(l) H+ (aq) + HCO3― (aq) if the cap is left off the bottle, a. As the CO2 escapes, the forward rate slows, and the net reaction shifts left.
b. As the CO2 escapes, the [H+], which determines the level of acidity, will decrease. ©2010 ChemReview.net v. 3d Page 828 Module 28 — Equilibrium Practice D
1. For C (s) + 2 Cl2 (g) CCl4 (g) + energy a.. If the [Cl2] is decreased: 1. Equilibrium shifts left. 2. [CCl4] will decrease. 3. [C] will not change,
because C is solid. 4. Amount of carbon increases. 5. Temp. decreases.
b. If pressure is decreased: 1. Equilbrium shifts left because left has 2 gas moles and right has one.
Decreasing pressure favors side with more gas moles. 2. Moles of Cl2 increase. 3. Temp.
decreases.
c. Catalysts do not shift the position of an equilibrium.
2. For CH3COO―
+ H2O
(l)
(aq) CH3COOH
+ OH―
(aq)
(aq) a. If water is added, the equilibrium does not shift. In an aqueous reaction, water is the solvent, and
adding solvent does not shift an equilibrium.
b. If [OH―] is increased: 1. Equilibrium shifts left. 2. [CH COOH] will decrease.
3 3. [H2O] will not change. Adding solvent does not shift an equilibrium.
c. If H+ is added: 1. H+ neutralizes OH―, lowering [OH―], so equilibrium shifts right.
2. [CH COO―] decreases.
3 3. For 2 H2O + 2 Cl2
+ energy
4 HCl
+ O2
(g)
(g)
(g)
(g)
a. If [O2] is increased: 1. Equilibrium shifts left. 2. [HCl] will decrease. b. If [HCl] is decreased: 1. Equilib. shifts right. 2. [O2] will increase. 3. [Cl2] will decrease.
4. Energy is used up when the reaction goes left, so temperature will decrease.
c. If temperature is decreased: 1. Equilibrium shifts left. 2. [O2] will decrease.
3. [H2O] will increase. If water is in the gas state as steam, its concentration can change.
d. If the total P is increased: 1. The right has 5 gas moles, the left has 4; lowering pressure favors the
left side. 2. Moles of O2 decrease. 3. Moles of Cl2 increases. 4. Temperature increases.
e. If V increases: 1. P decreases; equilib. shifts right. 2. Moles O2 increase. 3. Moles H2O decrease. f. If catalyst is added: 1. Equilibrium will not shift 4. For: 2. [O2] will not change. 4 PCl5 + energy
P4 + 10 Cl2
(g)
(s)
(g)
a. If temp. is decreased: 1. Equilib. shifts left. 2. [Cl2] will decrease. 3. [PCl5] gas increases.
4. [P4] , since it is a solid, will not change. b. If V decreases: 1. P increases, and equilibrium shifts to left side with fewer moles of gas.
2. Moles of Cl2 decreases. 3. Moles of PCl5 gas increases. 4. Number of gas molecules decreases.
***** ©2010 ChemReview.net v. 3d Page 829 Module 28 — Equilibrium Lesson 28B: Powers and Roots of Exponential Notation
Timing: Complete this lesson before calculations involving K (the equilibrium constant).
Prerequisites: Review Module 1 on Scientific Notation if needed.
*****
Pretest: If you can solve these problems correctly, skip to Lesson 28C. Express your final
answers in scientific notation. Answers are at the end of this lesson.
1. Do this problem without a calculator. (3.0 x 103)3 =
2. Use a calculator. (4.5 x 10─4)2 =
3. Do not use a calculator. Write the cube root of 8.0 x 1024.
4. Use a calculator for all or part of this problem. (2.5 x 105)1/3 =
***** Taking Numbers to a Power
Many calculators have an x2 key. To calculate both squares and higher powers, most
calculators also have power functions labeled xy or yx or (y)^x or ^ .
To learn to use the power keys, you may check your calculator manual, or experiment
using simple examples for which you know the answer, or (preferably) do both.
Cover the answer below the * * * * * , then do this calculation in your head.
23 = _____
*****
23 = 2 x 2 x 2 = 8 .
Now, by entering 2 and using the keys noted above, find a key sequence that will give the
same answer on your calculator.
Write your key sequence: _______________________________________________.
To test your key sequence, first solve in your head, then apply your sequence on the
following problems. Make sure that you get the same answers in your head and on the
calculator.
a. 24 = _____
*****
Answers: a. 16 b. 34 = _____
b. 81 c. Do two ways on keys: 122 = _____ . c. 12 x 12 = 144 On each problem below, use your powerkey sequence to do the calculation. Write the
answer. Then use the calculator to multiply each number by the number of times indicated
by the power (as in 23 = 2 x 2 x 2 ) and compare your answers.
d. 163 = _________ e. 2.54 = _______ f. Do two ways on keys: 0.502 = _______ ***** ©2010 ChemReview.net v. 3d Page 830 Module 28 — Equilibrium Answers: d. 4096 e. 39.0625 f. 0.25 Note that all of the calculations above can be done in at least two ways. Complex calculator
operations should always be done two different ways as a check on your work. Taking Exponential Notation to a Power Without a Calculator
1. Memorize this rule: To take exponential terms to a power, multiply the exponents. (As
used here, power and exponent have the same meaning.)
Examples: (103)2 = 106 (105)―2 = 10―10 (10―3)―4 = 10+12 (Recite: “a minus times a minus is a plus.”)
Without a calculator, write answers to these, then check below.
a. (106)2 = c. (10―12)―3 = b. (105)―5 = *****
Answers: a. 1012 b. 10―25 c. 1036 2. When taking exponential notation to a power, the fundamental rules apply:
Do numbers by number rules and exponents by exponential rules.
Example: (2.0 x 104)3 = 8.0 x 1012
Treat numbers as numbers. 2 cubed is 8.
Treat exponents as exponents. (104)3 is 1012 . Without a calculator, write answers to these, then check below.
a. (3 x 103)2 = b. (2 x 10―5)3 = *****
Answers: a. 9 x 106 b. 8 x 10―15 On the two problems below, without a calculator, write answers, then rewrite the
answers converted to scientific notation.
c. (5 x 104)2 = d. (2 x 10―3)4 = *****
Answers: c. 25 x 108 = 2.5 x 109 d. 16 x 10―12 = 1.6 x 10―11 Taking Exponential Notation to a Power With a Calculator
To take exponential notation to a power, most calculators use the same keys that you used
above to take numbers to a power. However, you should also know how to take
exponential notation to a power
• without entering the powers of 10, and • by estimating an answer without using a calculator at all. ©2010 ChemReview.net v. 3d Page 831 Module 28 — Equilibrium Let’s learn by example. For this calculation: (3.5 x 10─4)3
a. Using a calculator as needed, write an answer (3.5 x 10─4)3 = _________________
Now let’s check the answer. Write your answers to parts b and c below without converting
to scientific notation at the end and without a calculator.
b. (3 x 10─4)3 = _____________________. c. (4 x 10─4)3 = _______________________. *****
For both: (10─4)3 = 10─12; (b) 3 x 3 x 3 = 27 x 10─12 (c) 4 x 4 x 4 = 64 x 10─12 d. Based on the (b) and (c) answers,
write an estimate of what (3.5 x 10─4)3 should be: _________________________.
*****
3.5 is half way between 3 and 4, so you might estimate that half way between 33 =
27 and 43 = 64 … is about 45 x 10─12.
e. Calculate (3.5)3 using a
f. yx type calculator function: _______________________ Calculate (3.5)3 on the calculator without the yx or ^ type keys. *****
Multiply 3.5 x 3.5 x 3.5 = ________ . Does this match your part (e) answer ?
g. Based on (e) and (f), write a precise answer to (3.5 x 10─4)3. Compare your answer
to your estimate in (d) above. Then convert your answer to scientific notation
and apply sf : __________________________
*****
42.9 x 10─12 = 4.3 x 10─11 in scientific notation. h. Now try (3.5 x 10─4)3 by plugging everything into the calculator. You will probably
need keys labeled xy or yx or ^ or (y)^x .
• A “standard TItype” calculator may use 3.5 E or EE 4 +/ • A graphing calculator might use • On an RPN calculator, try yx 3 = 3.5 EE () 4 enter ^ 3 enter 3 .5 E or EE or EXP 4 +/ enter 3 yx An online search with your calculator name and model number and “exponential
notation” may offer a better approach. Try to work through the logic for key
sequences. Then practice the calculation until you can repeat it without looking at
hints or directions.
Write the calculator answer, rounded to proper sf :. _________________________.
i. Compare answers in steps (g) and (h). They should agree. They should also be
close to the value of your estimate in step (d). ©2010 ChemReview.net v. 3d Page 832 Module 28 — Equilibrium Which method is easier: Numbers on the calculator but exponents in your head, or all on
the calculator? Which method is easier to remember?
You may do calculations using any method you choose, but doing numbers on the
calculator and exponents by mental arithmetic can speed and simplify your work.
In addition, every calculation should be done two different ways as a check on your
calculator use. Estimating the number math “in your head” is one way to check a
calculator answer. Practice A: Convert final answers to scientific notation. Do not use a calculator on these first four.
1. (10─3)2 = 2. (10─5)─2 = 3. (2.0 x 104)4 = 4. (3.0 x 10─1)3 = On 58 below,
• first write an estimated answer, then rewrite it in scientific notation; • then use the calculator for whatever parts you wish and write a final answer in
scientific notation. Try any two. Need more practice? Do more. Check answers as you go.
5. (2.1 x 106)2 Est = ______________ Sci Note = ______________ Calc =_________________
6. (3.9 x 10─2)3 Est = _____________ Sci Note = ______________ Calc =_________________
7. (7.7 x 104)4 Est = ______________ Sci Note = ______________ Calc =_________________
8. (5.5 x 10─2)3 Est = ______________ Sci Note = ______________ Calc =_________________ Roots
To calculate a square root, some calculators have a square root button: √x or x1/2 .
Other calculators use this twokey sequence: 2nd or INV x2 .
However, to take a cube root or higher, you will need to use a different key sequence. You
will also need to know multiple ways to take roots in order to check your calculations. One
way to do this is to treat taking a root of a term as taking the term to a fractional power. Roots as Exponents
“Taking the root” of a quantity is the same is assigning the quantity the fractional
exponent (or reciprocal exponent) of the root.
a. The cube root of x can be written as x1/3 . 161/4 means the fourth root of 16.
y
b. In general, x = x1/y . ©2010 ChemReview.net v. 3d Page 833 Module 28 — Equilibrium Taking Higher Roots
On many calculators, both square and higher roots may be calculated using the 2nd or INV
key followed by the keys used to calculate powers, such as yx or (y)^x or ^ .
It may also be possible to calculate roots by entering fractional exponents as powers using
the reciprocal key 1/x or x―1 or the division operation (1/x = 1 ÷ x ) .
Knowing at least two ways to calculate roots is necessary in order to check your calculator
answers. A web search with the name and model number of your calculator, plus the word
root, may help in learning ways to calculate a root.
Once you determine two key sequences that work, it is important to practice and test those
sequences by entering sample calculations that are easy to check. Do the following
calculation in your head.
The cube root of 8 = 81/3 = _____
*****
8 = 2 x 2 x 2 , so 81/3 = 2
Now, by entering 8 and using the inverse, or reciprocal, or division, and/or power keys,
see what key sequences give the same answer for the root on your calculator.
*****
One or more of these sequences may work. Others may work as well.
• On a standard TItype, try 8 2nd or INV yx 3 = and/or try
and/or try
• 8 yx 8 yx 0.33333333 = On a graphing calculator (if allowed), try
and/or try • (1÷3)= On an RPN calculator, try
and/or try 8 ^ ( 1 ÷ 3 ) enter 8 ^ 3 1/x or x―1 enter
8 enter 3 1/x yx 8 enter 0.33333333 yx Write one and if possible two key sequences that work and make sense to you.
To take a root: ____________________________________________________
_____________________________________________________ Then, on the problems below, check your key sequence. First try the problem “in your
head” and write your answer. Then solve two different ways using the calculator.
a. 161/4 = b. 1251/3 = c. (0.001)1/3 = d. (0.008)1/3 = ***** ©2010 ChemReview.net v. 3d Page 834 Module 28 — Equilibrium Answers: a. 2 b. 3 c. 0.1 d. 0.2 One way to check a root calculation is to reverse the process: Take the answer to the power of
the root. This should result in the original number. For example,
• If the cube root of 125 = 1251/3 = 5, then 53 should equal _________. • For problems c and d above, start from the answer, reverse the process using your
power key sequences, and see if the result is the (original number). Roots of Divisible Powers of 10
When estimating roots for numbers in exponential notation, it is important to be able to
take a root of 10x without a calculator. In some cases, this is easy.
•
• Roots of exponential terms can be taken without a calculator if the power of 10,
when multiplied by the fractional exponent, results in a whole number.
Another way to say this: the root of 10x can be found without the calculator if x is
evenly divisible by 2 to find a square root, and by 3 to find a cube root, etc. To calculate the root of an evenly divisible exponential term, use these steps.
1. Write the root as a fractional exponent.
Write the square root of 10x as (10x)1/2, and the fourth root of 10x as (10x)1/4.
2. Apply the rule: to take an exponential term to a power, multiply the exponents.
Examples: The square root of 104 = (104)1/2 = 102
The cube root of 10―9 = (10―9)1/3 = 10―3 Practice B:
1. Take these roots by entering the numbers into a calculator. Try each two ways.
a. The square root of 9,025 = b. (0.004096)1/3 = 2. Do not use a calculator. Write answers to these as powers of 10.
a. The square root of 1012 = b. (106)1/2 = c. The cube root of 10―6 = d. (10―12)1/4 = 3. Which root is this equivalent to: x0.125 ? 4. Without a calculator: 810.25 = Roots of Exponential Notation
To find a root of numbers written in exponential notation, apply the fundamental rule:
Treat numbers as numbers, and exponents as exponents.
Cover the answer below and, without a calculator, try: (8.0 x 1015)1/3 =
***** ©2010 ChemReview.net v. 3d Page 835 Module 28 — Equilibrium Answer: Treat numbers as numbers. The cube root of 8 is 2.
Treat exponentials as exponentials. (1015)1/3 = 105
(8.0 x 1015)1/3 = 2.0 x 105 For roots that cannot be solved by inspection, use these steps.
1. If the exponential term is not evenly divisible by the root, make the exponent smaller
until the exponent times the fractional power results in a whole number.
Try that step on
***** (8.04 x 10―5)1/3 = (8.04 x 10―5)1/3 = (80.4 x 10―6)1/3
To make the exponent divisible by 3, it is lowered from 10―5 to 10―6. When you
make the exponent smaller, make the significand larger.
2. Without a calculator, write a rough estimate of the root of the significand in front of the
evenly divisible exponent. Find the exact root of the exponential term. Combine these
two parts and write the estimate for the root.
Apply step 2 to the step 1 answer: (80.4 x 10―6)1/3 ≈ __________________
*****
(80.4 x 10―6)1/3 = (80.4)1/3 x (10―6)1/3
To estimate a cube root of 80, since 4 x 4 x 4 = 64, and 80 is a little higher than 64,
guess ≈ 4.2. ( ≈ means approximately equals.)
Handle exponents as exponents. (10―6)1/3 = 10―2
Combine the two parts. Estimate ≈ 4.2 x 10―2
3. To get a precise answer,
• Write the value that has the evenly divisible exponent. • Find the precise root of the significand on the calculator. • Take the root of the exponential term without the calculator. Try those steps on the above problem.
*****
(8.04 x 10―5)1/3 = (80.4 x 10―6)1/3
(10―6)1/3 is 10―2. Answer = The calculator cube root of 80.4 is 4.32 . 4.32 x 10―2 4. Compare the step 3 calculator answer to the estimate. They should be close.
5. Now take the root by entering the original number in the problem into the calculator.
** *** ©2010 ChemReview.net v. 3d Page 836 Module 28 — Equilibrium One or more of these key sequences (and others) may work.
• On a standard TItype, try
and/or try 8.04 EE 5 +/ 2nd or INV 8.04 E or EE or EXP 5 +/ • On a graphing calculator, try yx yx 3 = (1÷3)= 8.04 EE () 5 enter ^ ( 1 ÷ 3 ) enter • On an RPN calculator, try 8.04 E or EE or EXP 5 +/ enter 3 1/x yx The calculator answer should match the step 3 answer.
Circle or write one or two key sequences that work: _____________________________
Whatever sequence you use, work through the logic of why it works. Without the why, it
will be difficult to remember the correct sequence.
Once you have debugged and practiced a key sequence to calculate roots, doing the entire
calculation on the calculator may be faster than converting to an evenly divisible root.
However, using that conversion to estimate the root is a good way to check the calculator
result. Complex calculations should be done in two ways. Roots of NonDivisible Powers of 10
Changing an exponent to make it divisible by the root can put a number in front of the
exponential term that was not there before. Try this example.
First, estimate this answer without a calculator: The cube root of 1016 = ____________
*****
To find the root without a calculator, the power of 10 must be divisible by 3. Try
changing the exponent to 15, to make it smaller and divisible. This puts the number 10
in front of the exponential term.
*****
Answer: (1016)1/3 = (10 x 1015)1/3 = (101/3 x 105) ≈ 2.2 x 105
Now try taking the cube root of 1016 on the calculator.
*****
You may need to enter 1 x 1016 to take the root on the calculator.
*****
2.15 x 105
***** Summary: Roots and Powers
1. If you are not certain that you are using calculator keys correctly, do a simple similar
calculation, first on paper and then on the calculator.
2. For complex operations on a calculator, do each calculation a second time. Use estimates
on paper with rounded numbers or use different steps or keys.
3. On complex calculations, it is often easier to do the numbers on the calculator but the
exponents on paper. ©2010 ChemReview.net v. 3d Page 837 Module 28 — Equilibrium 4. In calculations using exponential notation, handle numbers and exponential terms
separately. Do numbers by number rules and exponents by exponential rules.
• When you multiply exponentials, you add the exponents. • When you divide exponentials, you subtract the exponents. • To take an exponential to a power, multiply the exponentials. 5. To take roots of exponential notation without a calculator,
• convert roots to fractional exponents, then • adjust the significand and exponential to make the exponent evenly divisible by the
root. Practice C
1. Fill in the blanks.
a. 4 x 4 x 4 = _______ , so (_______)1/3 = 4.
b. Using your calculator, find the cube root of 64: _______
c. 64 = 6.4 x 101, so (6.4 x 101)1/3 should equal ________.
Find this root using the calculator: (6.4 x 101)1/3 = __________________ d. 0.4 x 0.4 x 0.4 = 0.064 = 6.4 x 10―2, so (6.4 x 10―2)1/3 in your head = _________
See if you get the same answer to (6.4 x 10―2)1/3 on the calculator.
2. Complete the problems below using the following steps. Do as many as you need to
feel confident. Check your answer after each part.
A. First convert to an exponent with a divisible root.
B. Write an estimated answer for the root.
C. Starting from the divisible root, use the calculator for the root of the number,
take the root of the exponential in your head, write the answer, then convert the
answer to scientific notation. Round the significand to two digits.
D. Take the root of the original exponential notation on the calculator.
E. Compare your answers in steps B, C and D.
a. (6.0 x 1023)1/3 Divisible = ___________________ RootEst. = __________________
Estimate in SciNote = ___________________ Calculator =____________________
b. (1015)1/4 Divisible = ___________________ RootEst = __________________
Estimate in SciNote = ___________________ Calculator =____________________
c. The cube root of 1.25 x 10―7 = Divisible = ______________ Est = _________________
Estimate in SciNote = ___________________ Calculator =____________________ ©2010 ChemReview.net v. 3d Page 838 Module 28 — Equilibrium d. (1.6 x 10―11)1/4 Divisible = ___________________ RootEst = __________________ Estimate in SciNote = ___________________ Calculator =____________________ Practice D : Use a calculator as needed. 1. 24 = 2. (0.25)4 = 3. (4.5 x 103)5 = 4. (2.0 x 105)6 = 5. (3.3 x 10―3)8 = 6. (4.7 x 10―4)4 = 7. (81)1/4 = 8. 5 0.01024 = 9. (6.20 x 104)1/8 = 10. The sixth root of 9.5 x 1015 = 11. (3.3 x 10―3)1/9 = 12. (6.5 x 10―3)7 = ANSWERS
2. 2.0 x 10─7 Pretest: 1. 2.7 x 1010 3. 2.0 x 108 4. 6.3 x 101 Practice A
1. (10―3)2 = 10―6 2. (10―5)―2 = 10+10 3. (2.0 x 104)4 = 16 x 1016 = 1.6 x 1017 4. (3.0 x 10―1)3 = 27 x 10―3 = 2.7 x 10―2 5. (2.1 x 106)2 = 4.4 x 1012 6. (3.9 x 10―2)3 = 59 x 10―6 = 5.9 x 10―5 7. (7.7 x 104)4 = 3,515 x 1016 = 3.5 x 1019 8. (5.5 x 10―2)3 = 166 x 10―6 = 1.7 x 10―4 Practice B
1a. 95
3. 2a. 106 1b. 0.16 2b. 103 x0.125 = x1/8 = the eighth root 4. 2c. 10―2 2d. 10―3 810.25 = 811/4 = ( 811/2 )1/2 = ( 9 )1/2 = 3.0 Practice C
1a. 64 1b. 4 1d. 0.40 or 4.0 x 10―1 1c. 4 2a. (6.0 x 1023)1/3 = (600 x 1021)1/3 = (6001/3 x 107) = 8.4 x 107
2b. (1015)1/4 = (103 x 1012)1/4 = (1000 x 1012)1/4 = (10001/4 x 103) = 5.6 x 103
2c. (1.25 x 10―7)1/3 = (125 x 10―9)1/3 = (1251/3 x 10―3) = 5.00 x 10―3
2d. (1.6 x 10―11)1/4 = (16 x 10―12)1/4 = (161/4 ) x (10―12)1/4 = 2.0 x 10―3 Practice D
1. 16 2. 3.9 x 10―3 7. 3.0 8. 0.4000 3. 1.8 x 1018
9. 3.97 4. 6.4 x 1031 10. 4.60 x 102 5. 1.4 x 10―20 11. 0.53 6. 3.7 x 10―14
12. 4.9 x 10―16 ***** ©2010 ChemReview.net v. 3d Page 839 Module 28 — Equilibrium Lesson 28C: Equilibrium Constants
Prerequisites: Read the introduction to Module 28 and complete Lesson 28B before
starting this lesson.
Pretest: If you are familiar with equilibrium constants, try Problems 2 and 3 in Practice C at
the end of this lesson. If you can do those problems correctly, you may skip this lesson.
***** The Law of Mass Action
At a fixed temperature, for the general reversible reaction
aA + bB cC + dD the Law of Mass Action states that for a reaction at equilibrium, the ratio
[C]c[D]d =
[A]a[B]b product of the [Products] will be constant. product of the [Reactants] This ratio is called the equilibrium constant, symbol K. In an equilibrium constant,
• The concentrations of the substances on the right side of the reaction equation are
multiplied on top, • The concentrations of substances on the left side are multiplied on the bottom. • The coefficient for each substance becomes the power of its concentration. An equilibrium constant equation has two parts:
• The equilibrium constant expression is the ratio which shows the symbols for the
substance concentrations and their powers. • The equilibrium constant value is the number that is the value of the ratio. • An equilibrium constant equation is the combination of the expression and value. Example:
For the reaction S2 (g) The equilibrium constant expression is + 2 H2
K= 2 H2S (g) (g) [H2S]2
[S2][H2]2 and the equilibrium constant value is 1.1 x 107 at 973 K.
(We abbreviate kelvins as K and the equilibrium constant as an italicized K.)
The equilibrium constant equation at 973 K is K= [H2S]2 = 1.1 x 107 [S2][H2]2
As temperature changes, K values change, but the K expression stays the same. ©2010 ChemReview.net v. 3d Page 840 Module 28 — Equilibrium Why K Values Do Not Have Units
K values are numbers without units. This is in part because the K expression is a “shortcut”
equation: a convenient simplification of relationships derived from complex equations in
thermodynamics that are based on a unitless quantity called activity.
Concentration and gas partial pressures are related to activity but are easier to measure
directly, so we use their values in most calculations involving K. For “real” relationships in
science, units must cancel properly, but in “shortcut” simplified equations, units may not.
For now, it is important to remember the following.
When solving calculations based on a K equation,
• units of concentration or pressure data must be converted if needed to be consistent.
• K values are not assigned units.
• When solving for unknown concentration or pressure using K values, the solved
quantity must be assigned a unit consistent with the other units. Equilibrium Constant Expressions
To write a K expression, all that is needed is a balanced equation for a reaction. Cover
below the * * * * * line below and then try this problem.
Q. For the Haber process reaction: N2 (g) + 3 H2 (g) 2 NH3 (g) write the equilibrium constant expression.
*****
K= [NH3]2 with all concentrations measured at equilibrium. [N2][H2]3
This equation means that at a given temperature, if a reaction is run until equilibrium is
reached,
• whether starting with all of the substances written on the left side of the equation, or
all substances on the right, or any mixture of those substances, • no matter how much of each substance is present at the beginning, nor how much
an equilibrium is subsequently shifted by adding or removing substances, when equilibrium is reached, the concentrations of the substances can differ, but the ratio
calculated by this equilibrium constant will always have the same numeric value. Practice A: Write the equilibrium constant expression for these reactions. 1. CH4 2. 2 C4H10
+ 13 O2
(g)
(g) (g) + 2 O2 (g) ©2010 ChemReview.net v. 3d C O2 (g) + 2 H2O (g) 8 CO2
+ 10 H2O
(g)
(g) Page 841 Module 28 — Equilibrium Concentrations That Are Constant
The correlation between a balanced equation and its equilibrium constant expression is
simple, but there is one important exception.
Only concentrations that can change are included in K expressions.
• Solids, pure liquids, and solvents have concentrations that do not change during
reactions. • In place of terms for the concentrations of solids, pure liquids, and solvents (including
liquid water), a 1 is substituted in K expressions. By convention, if the concentration of a particle involved in a reaction is essentially
constant, that constant value is included in the value of the equilibrium constant.
Mathematically, the effect of writing constant concentrations as 1 moves the value for the
constant concentration into the K value and out of the K expression.
The concentrations that are assigned a value of 1 in a K expression are the same as those
that do not shift an equilibrium when applying Le Châtelier’s Principle.
Using the rules above, try the following example.
Q. Write the equilibrium constant expression for this reaction.
CaCl2
***** (s) Ca2+
+ 2 Cl―
(aq)
(aq)
K = [Ca2+] [Cl―]2 The reaction equation shows that the ions on the right are aqueous, meaning that the
solvent for the reaction is water. When dissolved in a solvent, the concentration of ions
can vary. Terms for concentrations that can vary are included in K expressions.
Because the reactant on the left is a solid, its concentration is assigned a value of 1.
Most K expressions will be fractions with a numerator and a denominator. For some
reactions, however, including the one above, the numerator or denominator of K will be
1. The 1 must be included if it is alone in the numerator (on top), but it is otherwise
omitted as a term when writing a K expression.
Substance states (solid, liquid, gas, or aqueous) must be written either in the balanced
equation or in the K expression. If the balanced equation is written and includes states, the
states can be omitted in the K expression.
Water: The “constant concentrations = 1” rule means that a [H2O] term will be given a
value of 1 when the water is a solid (ice), a pure liquid, or a solvent at high concentration.
However, if the water is present in a reaction in its gas phase, as vapor or steam, since all
gases are compressible, its concentration can vary, and the term for [H2O ] must be
(g)
included in the K expression.
In addition, when liquids are mixed with other liquids, they may dissolve in each other, as
occurs with alcohols and water. If the concentrations are relatively close in value, the ©2010 ChemReview.net v. 3d Page 842 Module 28 — Equilibrium liquid concentrations can vary, and neither liquid is considered as the solvent.
Concentrations that can vary are included in K expressions. The general rule is:
For H2O, [H2O(g)] is included in K expressions; but [H2O(s)] and [H2O(l) if pure or
a solvent] are replaced by a 1 .
Vapor Pressure: All solids and liquids have a vapor pressure: they will release gas particles
into a system at equilibrium. However, because vapor pressure depends only on
temperature, vapor pressure creates a constant pressure and gas concentration at each
temperature. Because it is constant, the vapor pressure of solids and liquids is omitted as a
factor in K expressions (but is a factor in the value of K). Practice B: Write the equilibrium constant expression for these reactions. 1. 2 C2H6 (g) + 7 O2 2. BaO + CO2
(g) (s) 4 CO2
+ 6 H2O
(l)
(g) (g) BaCO3 (s) 3. CH3COOH
+ OH―
(aq)
(aq)
4. Ca3(PO4)2 (s) 3 Ca2+ CH3COO―
+ H2O
(l)
(aq)
(aq) + 2 PO43―
(aq) Equilibrium Constant Calculations
To solve equilibrium calculations in a systematic fashion, we add a step or two to our
standard equation solving methods. Our rule will be,
For calculations involving K and concentrations, write the WRECK steps.
The WRECK steps are
1. W (WANTED): Write the WANTED unit and/or symbol.
2. R (Reaction): Write a balanced equation for the Reaction.
3. E (Extent): After the reaction, add the Extent of the reaction, such as: “(goes partially)”
or “(goes ~100%).”
4. [email protected]: List or calculate the concentrations at equilibrium for each of the particles
in the reaction equation.
5. K: If the reaction goes to equilibrium, write the K equation.
6. SOLVE the K equation for the WANTED symbol.
7. If a concentration is calculated using an equilibrium equation, its unit, moles/liter, must
be added to the answer.
Because K values do not include units, unit cancellation will not catch mistakes.
K calculations must therefore be solved with special care. ©2010 ChemReview.net v. 3d Page 843 Module 28 — Equilibrium Try the above 7step method on the following problem. If you get stuck, read a bit of the
answer until you are unstuck, and then complete your work.
Q. For the reversible reaction H2
+ I2
(g)
(g) 2 HI , if the concentrations at
(g)
equilibrium are [H2] = 0.020 M and [I2] = 0.32 M, and K = 25, find [HI]
.
at eq.
*****
Answer
For calculations involving K and concentrations, write the WRECK steps.
1. W (WANTED): ? = [HI]
in mol/L
eq.
2. R (Balanced Reaction Equation): H2
+ I2
(g)
(g) 2 HI (g) 3. E (Extent): If a reaction goes to equilibrium; it goes partially to completion.
4. C: [HI] eq. = ? (WANTED) [H2]
= 0.020 M
eq.
[I2] eq. = 0.32 M [HI]2
= 25
at equilibrium.
[H2][I2]
6. SOLVE the K equation for the WANTED symbol. 5. K: K= Until this point, our rules have been:
• always include units when you solve, and • solve in symbols before substituting numbers, because symbols move more
quickly than numbers with their units. For the special case of K calculations,
• units are omitted, since K values do not have units, and • you may plug numbers into the original equation, and move numbers to solve,
since numbers without units will likely move as easily as symbols. Because they are done without units, K calculations may solve quickly, but because unit
cancellation will not catch mistakes, your must carefully check your substitution and
algebra.
Substitute the data numbers directly into the K equation above, solve, then check below.
*****
[HI]2 = 25 at equilibrium. (0.020) (0.32)
[HI]2 = (25) (0.020) (0.32) = 0.160
[HI] eq. (Carrying an extra sf until the final step.) = square root of [HI]2 = ([HI]2)1/2 = (0.160)1/2 = 0.40 mol/L 7. When solving K for a concentration, M (or mol/L) must be added to the answer. ©2010 ChemReview.net v. 3d Page 844 Module 28 — Equilibrium Significant figures in K calculations: Coefficients are exact, and in terms such as [HI]2
above, since the 2 is based on coefficients, the 2 is exact. All numbers based on coefficients
have infinite sf and do not restrict the sf in an answer. Practice C
1. Given the reaction: 4 NH3 + 5 O2 4 NO + 6 H2O (all gases) At a certain temperature the equilibrium concentrations are:
[NH3] = 0.050 M, [O2] = 0.0020 M, [NO] = 0.50 M, [H2O] = 0.20 M.
What is the value of K at this temperature?
2. Given the reaction: 2A + B
4 C (all gases) , concentrations at equilibrium are
[B] = 0.125 M and [C] = 0.50 M. If K = 200. , find [A].
3. For the reaction at equilibrium: CO2 (g) + H2
(g) CO (g) + H2O (l) in a 2.0 liter container is found 0.40 mol CO2, 0.60 mol H2O, and 0.90 mol H2. If the
value of the equilibrium constant is 4.8, how many moles of CO are in the mixture? ANSWERS
Practice A
2
1. K = [CO2] [H2O]
[CH4] [O2]2 8
10
2. K = [CO2] [H2O]
[C4H10]2 [O2]13 Practice B
[CO2]4 1. K = 2. K = [C2H6]2 [O2]7 1 3. K = [CO2] [CH3COO―]
[CH3COOH] [OH―] 4. K = [Ca2+]3 [PO43―]2 Practice C
1. For calculations involving K and concentrations, write the WRECK steps.
WANT:
?=K
Rxn:
4 NH3 + 5 O2
4 NO + 6 H2O (all gases)
Extent: Goes to equilibrium, use K equation. [email protected] (See list in problem)
4
6
(Since this H2O is a gas, it is included in the K equation.)
K = [NO] [H2O]
4 [O ]5
[NH3]
2 K:
SOLVE: ?=K= The equation as written solves for the WANTED symbol. Plugging in numbers:
[0.50]4 [0.20]6
[0.050]4 [0.0020]5 ©2010 ChemReview.net v. 3d ―1 4
―1 6
―4
6
―6
= [5.0 x 10 ] [2.0 x 10 ] = (625 x 10 )(2.0) x 10
[5.0 x 10―2]4 [2.0 x 10―3]5 (625 x 10―8)(2.0)5 x 10―15 Page 845 Module 28 — Equilibrium
6
―10
= (2.0) x 10
(2.0)5 x 10―23 = 2.0 x 10+13 (K values are written without units.) The arithmetic may be done in any way that results in a correct answer.
If you need practice at exponential notation calculations, review Lessons 1C and 28B.
2. For calculations involving K and concentrations, write the WRECK steps.
WANT:
? = [A]
Rxn:
2A + B
4 C (all gases)
Extent:
Goes to equilibrium, use K.
[email protected] (See list in problem)
4
K:
K = [C]
[A]2 [B]
SOLVE: First solve K for the term that includes the WANTED symbol.
[A]2 = [C]4 =
K [B] (0.50)4
= 0.0625 = 0.00250
(200.)(0.125)
25 ? = [A] = ([A]2)1/2 = (0.0025)1/2 = 0.050 M Then solve for the WANTED symbol. (When solving K for concentration, add M) 3. For calculations involving K and concentrations, write the WRECK steps.
WANT: ? = mol CO (Goes to equilibrium, use K.)
CO + H2O
CO2 + H2
(l)
(g)
(g)
(g)
[email protected] [CO] = ?
[CO2] = ?
Data is given in moles, but the K equation requires concentration (mol/L). All of the gases are
in a 2.0 liter container.
If you needed that hint, adjust your work and continue.
*****
To find mol/L, divide mol by L. R+E: [CO2] = 0.40 mol in 2.0 liters = 0.20 mol/L
[H2] = 0.90 mol/2.0 L = 0.45 M
K:
SOLVE: K= [CO]
= 4.8
[CO2][H2] ( [Liquids] is left out of K expressions.) [CO] = K • [CO2] • [H2] = (4.8)(0.20)(0.45) = 0.432 M CO (carrying extra sf) Done? Always check your WANTED unit (especially after a long calculation).
? = mol CO = 2.0 L • 0.432 mol CO = 0.86 mol CO
L
A good habit at the end of each calculation is to (a) box your final answers, but (b) each time you make the
box , look back at the WANTED unit or symbol at the start of your answer to make sure that you found the
unit WANTED.
***** ©2010 ChemReview.net v. 3d Page 846 Module 28 — Equilibrium Lesson 28D: Equilibrium Constant Values
K Values And The Favored Side
Equilibrium constant values are always positive numbers, and the values are most often
written in scientific notation. At equilibrium
• If the substances on the right side of an equation have higher concentrations than
the substances on the left, the products are said to be favored, and the value of K
will be a number greater than one. • If the substances on the left have higher concentrations than those on the right, the
reactants are favored, and the value of K will be a number between zero and one (in
scientific notation, a positive number times a negative power of 10). • The more a reaction favors the products (goes to the right), the higher will be the
value of its equilibrium constant.
Examples: For these reactions at 25oC,
A. 2 N2 (g) + O2 2 NO (g) C. HSO4―(aq) K = 1.0 x 10―30 Ag(NH3)2+(aq) B. Ag+(aq) + 2 NH3(aq) (g) K = 1.7 x 107 H+(aq) + HSO42― (aq) K = 0.013 Reaction A has a K value that is positive, but is much less than one. At
equilibrium, the substances on the left side (the reactants) will be favored.
Reaction B has a K value that is much greater than one. At equilibrium, the
substances on the right side (the products) will be favored.
For K values much larger than one, the reactions go close to completion. For K
values much smaller than one, the reaction goes only slightly.
Reaction C has a K value that is smaller than one, favoring the left side, but
compared to most K values, K is not far from one. At equilibrium in Reaction
C, you would expect to find a more balanced mixture of reactants and products
than in Reaction A or B. K Values For Reversed Reactions
Equilibrium is the result of a reversible reaction. Reversible reactions can be written in
either direction.
For example, the conversion of nitrogen dioxide to dinitrogen tetroxide is reversible.
It can be written as either
2 NO2 (g) N2O4 (g) Kf = [N2O4]
[NO2]2 ©2010 ChemReview.net v. 3d or as N2O4
(g) 2 NO 2 (g) Kr = [NO2]2
[ N 2 O4 ] Page 847 Module 28 — Equilibrium The two equilibrium constants are different, but related:
Kf = [N2O4]
[NO2]2 = 1 = or 1
Kr [NO2]2 Kf = 1/Kr [N2O4]
The relationship Kf = 1/Kr will be true for all reactions. Stated in words:
If a reaction equation is reversed, the new K value is the reciprocal of the original K. K Values When Coefficients Are Multiplied
When writing a K value, either the balanced equation or the K expression must also be
written to indicate the coefficients used to balance the equation. Coefficients are ratios, and
only one set of ratios will balance an equation, but different coefficients can be used to
balance the equation so long as the coefficients are in the same ratio.
The value of K will depend on the coefficients used to balance the equation, but if the K
value is known for any one set of coefficients, the value for K for different coefficients can
easily be determined.
Example: Consider 2 NO2 (g) N2O4 (g) K1 = [N2O4]
[NO2]2 and 4 NO2 (g) 2 N2 O 4 (g) K2 = [N2O4]2 = ( K1 )2 [NO2]4
The coefficients of the second equation are double the first. The ratios are the same,
and both equations are balanced. The K values will be different, but related. If the
coefficients are doubled, the new K value for the K expression based on those doubled
coefficients is the original K value squared.
This relationship will be true for all K values:
If a value of K is known for a reaction with one set of coefficients, those coefficients can
be multiplied by any positive number, and the new value of K will be the original K to
the power of the multiplier.
In equation form, this rule can be written,
If K = # for the reversible reaction:
then K = (#)n for the equation: aA + bB
naA + nbB cC + dD
ncC + ndD The factor n can be any positive number: an integer, decimal, or fraction.
This rule illustrates why, when you write a K value, the reaction coefficients on which it is
based must also be shown by writing either the K expression or the balanced equation. ©2010 ChemReview.net v. 3d Page 848 Module 28 — Equilibrium Practice
For the following reactions at 25oC:
1. Cu(s) + 2 Ag+(aq) Pb2+(aq) + 2 Cl―(aq) 2. PbCl2(s)
3. AgI(s) Cu2+(aq) + 2 Ag(s) K = 1 x 1015
K = 2 x 10―5 Ag+(aq) + I―(aq) K = 1.5 x 10―16 a. Write the K expression for each reaction.
b. Which equilibrium most favors the substances on the right side of the equation?
c. Which reaction will form the least amount of product?
d. What will be the K expression and K value for the equation
Cu2+(aq) + 2 Ag(s) Cu(s) + 2 Ag+(aq) e. What will be the K expression and K value for the equation
1/2 Pb2+(aq) + Cl―(aq) 1/2 PbCl2(s)
f. What will be the K expression and K value for the equation
1/2 Cu2+(aq) + Ag(s) 1/2 Cu(s) + Ag+(aq) ANSWERS
2+
1a. K = [Cu ]
[Ag+]2 2. K = [Pb2+] [Cl―]2 3. K = [Ag+] [I―] 1b. Reaction 1, with the largest K value, most favors the right side (products).
1c. Reaction 3, with the K value much smaller than the others, will most favor the left side (reactants), and will
form the smallest concentrations of products.
1d. The Part d equation is reaction 2 written backwards. Both the K expression and K value will be the
reciprocals of reaction 1.
+2
K value = 1 =
K expression = [Ag ]
1
= 1 x 10―15
Kf
1 x 1015
[Cu2+]
1e. The Part e reaction is reaction 2 with all coefficients multiplied by 1/2. In the Part e K expression, the new
coefficients become the powers of the concentrations. The K value for Part e will be the reaction 2 K
value to the 1/2 power: the square root of the K in reaction 2.
K expression = [Pb2+]1/2 [Cl―] K value = (2 x 10―5)1/2 = (20 x 10―6)1/2 = 4.5 x 10―3 1f. Part f is reaction 1 written backwards and multiplied by 1/2. When writing the reaction backwards, invert
the K value. When multiplying coefficients by 1/2, take the original K value to the 1/2 power. ©2010 ChemReview.net v. 3d Page 849 Module 28 — Equilibrium +
K expression = [Ag ]
[Cu2+]1/2 K value = square root of 1 = ((1 x 1015)―1)1/2 = (10 x 10―16)1/2
Kf
= 3.2 x 10―8 ***** Lesson 28E: Kp Calculations
Prerequisites: Gas lessons 17A, 17D, 18D, and the prior lessons in Module 28.
***** Kc Versus Kp
For the special case of reactions that include gases but do not include substances dissolved
in a liquid solvent, an equilibrium constant can be calculated based on either the molar
concentrations or the partial pressures of the gases.
To distinguish between these two kinds of equilibrium constants,
• Kc is written instead of K for equilibrium constants based on concentration, and • Kp is written for equilibrium constants based on pressures. At a given temperature, for the general reaction
wA(gas) + xB(gas) yC(gas) + zD(gas) at equilibrium these ratios will be constant:
y
z
Kc = [C]y[D]z and Kp = (PC) •(PD)
where P represents partial pressure.
w•(P )x
[A]w[B]x
(P )
A B In most respects, the rules for Kc and Kp are the same.
• Coefficients of the balanced equation become powers in the Kp expression. • If a Kp value is listed, the coefficients and direction of the reaction must be shown. • If an equation is reversed, the Kp value will be the reciprocal of the original. • If coefficients are multiplied by a positive number, Kp will have a value of the
original Kp value to the power of the number. • Kp values are not assigned units. • Terms for pressures of solids and pure liquids are written as 1 in Kp expressions. The differences between Kc and Kp include
• Kp is calculated based on atmospheres as the pressure unit. In calculations that use
Kp, all pressures must be converted to atmospheres. • When calculating a partial pressure using Kp values, atmospheres must be added
as the unit of the answer. ©2010 ChemReview.net v. 3d Page 850 Module 28 — Equilibrium If K is written without a subscript, it is assumed to be a Kc. Because Kc can be used for
reactions that include both gases and substances dissolved in a solvent, and Kp cannot,
Kc is encountered more frequently.
Kp and Kc calculations are solved using the same steps. Use the rule: “For calculations
using K and concentrations or pressures, write the WRECK steps.”
Try the following problem. If you get stuck, read the answer until unstuck and try again.
Q. For the reaction PCl5(g) PCl3(g) + Cl2(g) at 250oC, Kp = 1.78 . If at equilibrium the partial pressures are 0.820 atm. for Cl2
and standard pressure for PCl5, calculate the partial pressure of PCl3.
*****
Answer
For calculations using K and concentrations or pressures, write the WRECK steps.
1. W (WANTED): ? = PPCl in atm.
3 2. R (Balanced Reaction Equation): PCl3(g) + Cl2(g) PCl5(g) 3. E (Extent): This reaction goes to equilibrium (only partially to completion).
4. [email protected] (In Kp calculations, use partial pressures in place of concentrations.)
PPCl 3 =? PCl = 0.820 atm. 2 PPCl 5 = standard pressure ≡ 1 atm. (exact) 5. K (Write the K equation): Kp = PPCl3 • PCl2
PPCl = 1.78 5 You may either plug the numbers into the equation above and solve or solve in symbols
then numbers.
PPCl
P
3 = Kp • PCl5 = 1.78 • 1 atm. = 2.17 atm.
PCl
0.820 atm.
2 When using Kp to find a partial pressure, all pressure units in the DATA must be
converted to atmospheres. If a pressure is WANTED, the unit atm. must be added to
the answer. Practice A
1. Write the Kp expression for each of these reactions.
a. SiO2(s) + 4 HF(g) SiF4(g) + 2 H2O (g) b. CuSO4•5H2O(s) CuSO4(s) + 5 H2O (g) ©2010 ChemReview.net v. 3d Page 851 Module 28 — Equilibrium 2. For the reaction CH4(g) + 2 H2S (g)
CS2(g) + 4 H2(g) at 700oC,
Kp = 8.9 x 10―4 . If the partial pressures at equilibrium are 0.32 atm. for H2 , standard
pressure for CH4, and 50.5 kPa for CS2, calculate the partial pressure of H2S. Converting Between Kc and Kp
When all of the variable terms in a K expression are gases, either Kc or Kp equations can be
applied to solve calculations, but the numeric values of Kc and Kp may or may not be the
same. The rules are:
1. If the two sides of a balanced equation have the same number of total moles of gas
(based on adding the coefficients of the gas terms), the Kc and Kp values are the same.
2. If the two sides of the equation have a different total number of moles of gas, Kc and
Kp will have different values.
3. If either the Kc or Kp value is known at a given temperature, the other value can be
calculated using
Kp = Kc (RT)Δn where • R is the Gas Constant using atmospheres (R = 0.0821 L·atm/mol·K) • T is absolute temperature in kelvins, and • Δn = (the sum of the coefficients of the gases in the products) MINUS (the
sum of the coefficients of the gases in the reactants)
The number for Δn may be positive or negative, and it may be a fraction. This conversion equation, including the Δn definition, must be memorized. As always
when using K equations, the coefficients and reaction direction must be indicated in the
calculation. In this Kp equation, the units for R and T must be as specified above, but units
do not cancel in the calculation.
To learn to use the Kp to Kc conversion equation, write it and the Δn definition until it is
in memory, then try the following problems.
Q1. Solve the conversion equation for the case in which the sum of the coefficients for
the gases on both sides of the reaction equation are equal.
*****
Kp = Kc (RT)Δn If the moles of gas on both sides are the same, Δn = 0, Kp = Kc (RT)0 = Kc (1) = Kc This matches rule 1 above: If the gas coefficient totals are equal on both sides, Kp = Kc
Q2. For the Haber Process reaction:
if Kc = 9.5 at 300.oC, find Kp .
***** ©2010 ChemReview.net v. 3d N2 (g) + 3 H2 (g) 2 NH 3 (g) Page 852 Module 28 — Equilibrium WANTED: Kp DATA: Kc = 9.5 The fundamental equation that relates Kp and Kc is Kp = Kc (RT)Δn
R = 0.0821 L·atm/mol·K . (Kp calculations solve in liters and atm.) T = 300oC + 273 = 573 K (T must be in kelvins) Δn = (right side total gas coefficients MINUS left) = 2 ― 4 = ― 2 (exact)
If needed, adjust your work and finish from here.
*****
Kp = Kc (RT)Δn SOLVE: = 9.5 • { (0.0821 L·atm/mol·K)(573 K) }―2 = 9.5 • (47.04)―2 = 9.5 • 4.52 x 10―4 = 4.3 x 10―3 In K calculations, units often do not cancel properly, and K values are not assigned
units.
***** NonIdeal Behavior
The K rules we have been using are based on ideal gas and solution behavior, including
the assumption that particles in gases or solutions do not attract when they collide. In
practice, all particles attract to some extent. This causes effective pressures and
concentrations to be less than predicted based on ideal behavior.
Effective concentrations and pressures should be used in K calculations, and corrections
can be made to reflect the real activity of particles. In most cases, however, the
consequences of nonideal behavior are small in comparison to other sources of
experimental error in K calculations. Practice B
1. Convert these to scientific notation. Use a calculator as needed. Try to do each
calculation two different ways.
a. 2. If (3.33)― 1 = b. (3.5 x 103)―2 = Kp = Kc (RT)Δn c. (4.8 x 10―3)―2 = and Δn = +2, which of these could be used to solve for Kc ? a. Kc = Kp • R2 • T2 b. Kc = Kp • R―2 • T―2 c. Kc = Kp
R2 T2 d. Kc = Kp
(RT)2 3. Calculate Δn for these reactions.
a. N2 b. NH3 c. CH4(g) + 2 H2S (g) (g) + 3 H2 (g) (g)
1/2 N2 ©2010 ChemReview.net v. 3d 2 NH3
(g) (g) + 3/2 H2 (g) CS2(g) + 4 H2(g)
Page 853 Module 28 — Equilibrium d. 1/2 N2 e. 2 C2H6 f. H2 (g) + O2
(g)
+ 7 O2 + I2 (g) (g) NO2 (g) (g) 4 CO2
+ 6 H2O
(g)
(g) (g)
2 HI (g) 4. For which reactions in Problem 3 will Kp = Kc ?
5. If, for the reaction 2 SO2(g) + O2(g)
6. If the Haber Process reaction is written as NH3 at 27oC, Kc = 1.3, calculate Kp. 2 SO3(g) 1/2 N2 (g) (g) + 3/2 H2 (g) with a Kp of 670 at 600.oC,
a. What is the value of Kp for N2 b. What is the value of Kc for N2 (g)
(g) + 3 H2 2 NH3 (g) + 3 H2 (g) 2 NH3 (g) (g) at 600.oC?
at 600.oC? ANSWERS
Practice A
1a. Kp = PSiF (PH O)2
4•
2 5
1b. Kp = (PH2O) (PHF)4
2. For calculations involving K and concentrations or pressures, write the WRECK steps.
WANT ? = PH S in atm.
2
R + E: CH4(g) + 2 H2S (g)
C: PCS = 50.5 kPa •
2 CS2(g) + 4 H2(g) (goes to equilibrium)
= 0.500 atm. 1 atm.
101 kPa PH = 0.32 atm.
2
PCH = standard pressure = 1 atm. (exact)
4
PH S = ?
2
In calculations using a Kp value, all pressures must be converted to atmospheres.
K: Kp =
(PH S)2 =
2 PCS (PH )4
2
2•
= 8.9 x 10―4
PCH • (PH S)2
4
2
PCS (PH )4
2
2•
PCH • Kp
4 = (0.500) • (0.32)4
1.0 • 8.9 x 10―4 P
P
2 1/2
1/2
H2S = ( ( H2S) )
= (5.89) ©2010 ChemReview.net v. 3d = 5.89 = 2.4 atm. (solving K for P, add atm. as unit) Page 854 Module 28 — Equilibrium Practice B
1a. 3.00 x 10―1 b. 8.2 x 10―8 c. 4.3 x 104 Kp
Kc = Kp =
= Kp R―2T―2
(RT)2
R2T2 ;
Answers b, c, and d are equivalent to the given equation, answer a is not. 2. Kp = Kc (RT)2 = Kc R2T2 ; 3. a. 2 ― 4 = ―2 b. 2 ― 1 = +1 d. 1 ― 1.5 = ―0.5 or ―1/2 c. 5 ― 3 = +2
e. 10 ― 9 = +1 f. 2 ― 2 = 0 4. Only 3f. Kp = Kc only if Δn = 0.
5. WANT: Kp
Kc = 1.3 DATA: The fundamental equation that relates Kp and Kc is Kp = Kc (RT)Δn .
R = 0.0821 L·atm/mol·K
T = 27oC + 273 = 300. K
Δn = (right side total gas coefficients MINUS left) = 2 ― 3 = ―1
Kp = Kc (RT)Δn SOLVE: = Kc (RT)―1 = (1.3) • { (0.0821 L·atm/mol·K)(300. K) }―1 = (1.3) • (24.63)―1 = (1.3) • (4.06 x 10―2) = 0.053
6a. The part a reaction is the Question 4 reaction reversed and doubled. The new Kp value will be the
reciprocal, squared, of the original Kp.
Kp new = ((670)―1)2 = (670)―2 = 2.2 x 10―6
6b. Kp was found in part a. Kc is WANTED. The equation relating Kp and Kc is Kp = Kc (RT)Δn .
WANTED: Kc DATA: Kp = 2.2 x 10―6
R = 0.0821 L·atm/mol·K
T = 600.oC + 273 = 873 K
Δn = (right side total gas coefficients MINUS left) = 2 ― 4 = ―2 SOLVE: Kp = Kc (RT)Δn = Kc (RT)―2 = Kp Kc = Kp (RT)2 = (2.2 x 10―6) • { (0.0821)(873) }2 = (2.2 x 10―6) • (71.67)2
= (2.2 x 10―6) • (5.14 x 103 ) = 0.011
***** ©2010 ChemReview.net v. 3d Page 855 Module 28 — Equilibrium Lesson 28F: K and Rice Moles Tables
So far in our K calculations, the concentrations or pressures at equilibrium have been known.
However, if pressures or concentrations are known for a mixture of reactants initially (with
no products yet formed), as well as for any one reactant or product after the reaction has
reached equilibrium, pressures and concentrations at equilibrium for all substances, and a
value for K, can be calculated.
To solve equilibrium calculations, we need a “chemistry accounting system.” For complex
reactions, our rule has been to make a rice moles table: rice for the labels of the rows:
Reaction, Initial, Change, End/Equilibrium,
and moles for the numbers that go into the table. We have previously used rice tables to
calculate mixture amounts at the end of a reaction (Lesson 10H). For reactions that go to
completion, the limiting reactant has a count of zero in the End row.
For reactions that go to equilibrium, all reactants and products must be present at
equilibrium, so none of the reactant or product counts go to zero. However, a rice table is
the preferred method to track changes in reactant and product amounts. Our rule will be,
Rice Tables: In calculations for reactions that go to equilibrium (or for any reaction that
is complex), to find moles, concentrations, or gas pressures, use a rice table.
Let’s illustrate this method with a variation on our previous “lab drawer” problem.
Q. The morning chemistry lab assistant is filling lab drawers. The initial inventory
contains 95 burners, 220 racks, and 2,500 test tubes. Into each top drawer is placed
one Bunsen burner, two test tube racks, and 20 test tubes. When the afternoon
assistant arrives, 60 racks remain in the inventory. How many drawers were filled?
How many burners and test tubes remain in the inventory at the end of the process? To solve, complete the following table, then check your answer below.
Reaction/Process ___ Burner ___ Racks __ Test Tubes ____ Drawer 1 Burner 2 Racks 20 Test Tubes 1 Drawer 95 220 2,500 0 Initial Count
Change (use + and ―)
At Equilibrium (End)
*****
The initial data:
Reaction/Process
Initial Count
Change (use + and ―)
At Equilibrium (End) 60 Adjust your work if needed, and then fillin all of the boxes in the table. ©2010 ChemReview.net v. 3d Page 856 Module 28 — Equilibrium *****
Calculate the one change for which there is data. From that number, use the ratios of the
process to complete the Change row. Include a ― sign for components used up and a + sign
for those formed. Then calculate the amount of each component present at the end.
*****
Reaction/Process 1 Burner 2 Racks 1 Drawer 20 Test Tubes 95 220 2,500 0 Change (use + and ―) ― 80 ― 160 ― 1,600 + 80 At Equilibrium (End) 15 60 Initial Count 900 80 *****
This same method can be used to find values for moles for chemical substances at
equilibrium. Cover the answer below and construct a rice moles table to solve this problem.
Q. A reaction that occurs at high temperatures (and can cause air pollution from car
engines) is: N2 + O2
2 NO (all gases). If 1.00 moles of N2, 2.00 moles of
O2, and no NO are initially mixed, and at equilibrium, 1.80 moles of O2 remains,
how many moles of N2 and NO are present at equilibrium?
*****
WANT: moles of N2 and NO at equilibrium.
Strategy: To solve for values at equilibrium, use a rice moles table. Initial 1 N2 1 O2 2 NO 1.00 mol Reaction 2.00 mol 0 mol Change (use + and ―)
At End/Equilibrium 1.80 mol • Calculate the change that you can. • Use the coefficients to complete the Change row. Coefficients show the ratios in
which the moles of reactants are used up and moles of products form. • Calculate the Equilibrium row.
*****
The one change that can be calculated is below. Finish from here.
Reaction
Initial
Change (use + and ―)
At Equilibrium 1 N2
1.00 mol 1 O2
2.00 mol 2 NO
0 mol ─ 0.20 mol
1.80 mol ***** ©2010 ChemReview.net v. 3d Page 857 Module 28 — Equilibrium From the O2 moles change and the ratios of reaction (the coefficients), the other changes
can be calculated. Be sure to include the + and ─ signs. Complete the Equilibrium row.
Reaction
Initial
Change (use + and ―) 1 O2 2 NO 1.00 mol 2.00 mol 0 mol ─ 0.20 mol ─ 0.20 mol 1 N2 At End/Equilibrium + 0.40 mol 1.80 mol *****
Using the row labels, the WANTED moles at equilibrium can be found.
Reaction
Initial
Change (use + and ―)
At End/Equilibrium 1 N2 1 O2 2 NO 1.00 mol 2.00 mol 0 mol ─ 0.20 mol ─ 0.20 mol + 0.40 mol 0.80 mol N2 1.80 mol O2 + 0.40 mol NO Practice A
H2O(g) + CO(g) , the initial gas mixture is
H2(g) + CO2(g)
composed of 2.00 moles of H2 and 1.00 moles of CO2. At equilibrium, 0.30 moles of
CO gas is found. Calculate the moles of the other substances present at equilibrium. 1. For the reaction Rice Tables Using Concentration and Pressure
Which units can be use in a rice moles table?
1. Moles can always be used. Rice calculations are based on coefficients, and coefficients
can always be read as moles.
2. Concentrations (molarity) and gas pressures can also be used in rice tables if all of the
moles and pressures in a problem are measured in the same volume.
Why? Coefficients are mole ratios. However, if all of the moles are contained in the
same volume, dividing each of the moles by that same volume will not change the mole
ratios. The mole and the mol/L ratios will be the same, and the rice table can be used to
calculate either the moles or the mol/L used up and formed.
Pressure ratios will also be the same as mole ratios if volume is held constant.
The logic? Since PV= nRT , P = n (constant 1/V)(gas constant R)(constant T). This
simplifies to P = (constant) n , so P and n are directly proportional (Lesson 18A) if
V and T are constant. T must be constant for the system to be at equilibrium. Since
coefficients are mole ratios, they will also hold for any ratios that are directly
proportional to moles (n), which P is in this case. ©2010 ChemReview.net v. 3d Page 858 Module 28 — Equilibrium 3. If gases or solutions are mixed together as part of a problem, measurements of the initial
moles per liter and gas pressures cannot be substituted directly into a rice table, because
the volume that the moles are in varies during the process.
For example, if 10 mL of 0.50 M Reactant A is mixed with 20 mL of 0.50 M Reactant B
to conduct a reaction, both A and B are diluted as the solutions are mixed. Using the
concentrations before mixing as concentrations that apply to a reaction that occurs
after mixing would cause an error.
4. If all initial amounts are converted to moles, the rice table can be used. Dilution does not
change the moles of substance present. Final mol/L can then be calculated if the total
solution volume at equilibrium is known.
Bottom line?
• Rice tables can always be solved in moles. • Rice tables can use values for molarity or gas pressure IF the volume in which the
substances are measured is the same in all parts of the problem. Practice B: Check your answers after each part. 1. For the reaction 2 SO2 + O2
2 SO3
, in a sealed glass vessel with
(g)
(g)
(g)
temperature held constant, the initial gas mixture contains [SO2] = 1.00 M,
[O2] = 0.60 M, and no SO3. At equilibrium, [SO3] = 0.40 M. a. Calculate the concentrations of all reactants and products at equilibrium.
b. Use your Part A answer to calculate a value for the equilibrium constant under
these conditions. Using Rice Tables With K Calculations
The values calculated in a rice table can be used in K calculations, as was done in Part B of
the Practice B problem above. However, care must be taken to write the units in the rice
table. Why?
• Rice moles tables can always be solved in moles, and moles are often supplied in
problem data. K equations, however, require consistent units of either mol/L for
solutions or gases or atmospheres for gases. K equations cannot be solved in moles. • If data in a K calculation is supplied in moles, as is often the case, at some point the
data must be converted to moles per liter at equilibrium before it is used in a Kc
equation. This means that in both the rice table and the DATA table that is used with K calculations, it
is important distinguish measurements in moles (mol) from moles/liter (M).
When data is supplied in moles, it is often easiest to solve the rice table in moles, then to
convert the moles at equilibrium found in the table to moles per liter, then to substitute
those values into K calculations.
©2010 ChemReview.net v. 3d Page 859 Module 28 — Equilibrium Using those hints, try the following problem, then check your answer below.
Q. For the reaction H2 + I2
2 HI (all gases), initial amounts are 0.100 moles of
H2, 0.090 moles of I2, and no HI. At equilibrium, 0.020 moles of H2 are present.
a. Calculate the moles of all of the substances present at equilibrium. b. If the reaction takes place in a 2.0 liter vessel, calculate the value of K. *****
a. WANT: moles of H2 , I2 , and HI at equilibrium.
Strategy: To solve for values at equilibrium, use a rice table.
*****
1 I2 1 H2 Reaction
Initial 2 HI 0.100 mol At Equilibrium ― 0.080 mol ― 0.080 mol + 0.16 mol 0.020 mol Change ( use + and ― ) 0.090 mol 0 mol 0.010 mol 0.16 mol In rice tables, the units must be stated and must all be the same.
The bottom row answers Part A.
If you have not already done so, complete Part B.
*****
b. Part B involves K and concentrations. A K equation relates those terms.
The rule is: for calculations using K equations, write the WRECK steps.
1. WANTED unit or symbol. If the type of K is not specified, assume K means Kc .
2. Reaction:
3. Extent: H2 + I2 2 HI (all gases) The reaction goes to equilibrium. Use a K equation to solve. 4. [email protected] The important rule is:
To find Concentrations at equilibrium (step C in the WRECK steps), calculate
the values in the bottom row of a rice table. Convert to moles/liter if needed.
In the rice moles table for this problem, we know moles at equilibrium, but a Kc
equation requires mol/L. If needed, adjust your work.
*****
All of these moles are in 2.0 L. In the DATA table, convert moles to the
unit that measures the symbol: mol/L at equilibrium.
[HI]eq. = 0.16 mol at eq./2.0 L = 0.080 M at eq.
[H2]eq. = 0.020 mol/2.0 L = 0.010 M
[I2]eq. = 0.010 mol/2.0 L = 0.0050 M ©2010 ChemReview.net v. 3d Page 860 Module 28 — Equilibrium 5. K (Write the K equation):
6. SOLVE: Kc = (0.080)2 K= (0.010)(0.0050) [HI]2
[H2][I2] all measured at equilibrium = 64 x 10―4
5.0 x 10―5 = 130 At the point when values are substituted into a K equation, the units are omitted, but
until that step, the units must be included with all DATA. Careful labeling of DATA
with units and symbols is essential in order to keep track of which data to use at each
step. Practice C
2 NOCl (all gases), 0.40 mol NO and 0.60 mol Cl2
1. For the reaction 2 NO + Cl2
are originally mixed in a 4.0 L sealed glass vessel. At equilibrium, 0.20 moles of NO
gas remains.
a. Calculate the moles of Cl2 and NOCl present at equilibrium.
b. Calculate the value for K under the above conditions. ANSWERS
Practice A
1. WANTED: moles of H2, CO2, and H2O at equilibrium. Strategy: To find values at equilibrium when some of the data is not at equilibrium, use a rice table. Initial data:
Reaction
Initial 1 H2 1 CO2 1 H2O 1 CO 2.00 mol 1.00 mol 0 0 Change + 0.30 mol At Equilibrium 0.30 mol Calculate the change row based on coefficients, then find the moles at equilibrium WANTED.
*****
1 CO2 1 H2O 1 CO 0 0 Reaction 1 H2 Initial 2.00 mol 1.00 mol ― 0.30 mol ― 0.30 mol + 0.30 mol + 0.30 mol 1.70 mol 0.70 mol 0.30 mol 0.30 mol Change
At Equilibrium ©2010 ChemReview.net v. 3d Page 861 Module 28 — Equilibrium Practice B
1a. WANTED: [ ]eq for all 3 substances, in mol/L .
DATA: measurements at equilibrium are WANTED, so a rice moles table is needed.
Since this reaction involves gases in a container with a fixed volume, moles, mol/L or gas pressures can be
used in the rice moles table. Since the data is in mol/L, and you want mol/L, use mol/L as the units in the
rice table.
2 SO 2 Reaction
Initial 1 O2 2 SO3 1.00 M
― 0.40 M At Equilibrium ― 0.20 M + 0.40 M 0.60 M Change (use + and ―) 0.60 M 0M 0.40 M 0.40 M 1b. For K calculations, write the WRECK steps.
WANTED = Kc
Reaction and Extent: 2 SO2 + O2
(g)
(g) 2 SO3
(g) (goes to equilibrium mixture) [email protected] Use the values in the bottom row of the rice table.
[SO3]2 Kc = These must be values at equilibrium, the values in the bottom row of the
rice table. [SO2]2 [O2]
SOLVE: Kc = =
(0.16)
= 1.1
(0.40)2
2 (0.40)
(0.60)
(0.36)(0.40) Practice C
1a. WANTED: moles of reactants and products at equilibrium.
DATA: To find measurements at equilibrium when some supplied data is not, use a rice table.
Since the WANTED unit is moles, solve the rice table in moles.
Reaction 2 NO 1 Cl2 Initial 0.40 mol 0.60 mol ― 0.20 mol ― 0.10 mol + 0.20 mol 0.20 mol 0.50 mol 0.20 mol Change (use + and ―)
At Equilibrium 2 NOCl
0 1b. For calculations using K, write the WRECK steps.
WANTED: Kc = ? Reaction and Extent: 2 NO + Cl2 2 NOCl (all gases) (goes to an equilibrium mixture) [email protected] To find concentrations at equilibrium, use the bottom row of the rice moles table.
In the above rice table are moles. A Kc equation requires mol/L at equilibrium. All of these
moles in the table are in 4.0 L. In the data table, convert moles to the unit of each symbol:
mol/L, then solve the K equation. ©2010 ChemReview.net v. 3d Page 862 Module 28 — Equilibrium [ NOCl ]eq. = 0.20 mol at eq./4.0 L = 0.0500 M at eq. (carry extra sf until end)
[ NO ]eq. = 0.20 mol/4.0 L = 0.0500 M
[ Cl2 ]eq. = 0.50 mol/4.0 L = 0.125 M
[NOCl]eq.2 Kc = [NO]eq2 [Cl2]eq.
Solve: K= (0.0500)2
(0.0500)2(0.125) = 1
0.125 = 8.0 ***** Lesson 28G: K Calculations From Initial Concentrations
In K equations, concentrations or gas pressures must be measured at equilibrium.
In some cases, knowing only concentrations or pressures in the initial mixture before the
reaction begins, the concentrations or pressures at equilibrium can be calculated with the
help of algebra. Let’s learn this method using an example.
Q. The reaction H2 + I2
2 HI (all gases) is carried out in a closed system with
constant volume, at a temperature where Kc = 144. In the initial mixture before the
reaction begins, [ H2 ] = [ I2 ] = 0.500 M. What will be the concentration of each
substance at equilibrium? K calculations are solved using the WRECK steps. Complete the following steps for the
problem above.
Steps 13: WRE. Write the WANTED unit, balanced Reaction equation, and Extent to
which the reaction goes to completion.
Step 4. [email protected] In finding the concentrations or pressures in a reaction mixture at
equilibrium, there are several variations in K calculations. Let’s compare.
a. When concentrations at equilibrium are supplied in a problem, the answers to this
step are supplied, and a rice table is not needed (Lesson 28C).
b. When concentrations or gas pressures at equilibrium are not known, we use a rice
table to find them.
i. If the initial moles, mol/L or gas pressures are known for reactants, and any one
value is known at equilibrium, the bottom row of the rice table (values at
equilibrium) can be solved using arithmetic (Lesson 28F). ii. If moles, mol/L or gas pressures are known for all of the initial reactants, the
bottom row of the rice table can often be solved using algebra. A key step is
assigning signs to x values in the Change row of the rice table. ©2010 ChemReview.net v. 3d Page 863 Module 28 — Equilibrium For the problem above, make a rice table. In the Change row, represent the change that
will take place to reach equilibrium using terms such as +x, ―x, +2x, ―3x, etc.
Fill in the Equilibrium row using terms that include x.
Complete Steps 14 above, and then check your answer below.
*****
Answer
1.
WANT: [H2], [I2], and [HI] at equilibrium. 2. Reaction: H2 + I2
2 HI (all gases)
Since the reaction has a K value, it goes to equilibrium. 3. Extent: 4. [email protected]
The concentrations are supplied for the initial reactants only, with no products
yet formed. As the reaction begins, some of the [H2] and [I2] reactants are
used up, and some products form. In the Change row, the signs for all of the
reactant terms must therefore be negative, and the signs for all of the product
terms must be positive.
If needed, use that hint to fill in the Change row, then complete the table. *****
According to the coefficients, if [H2] changes by ―x, [I2] must also change by ―x
as it reacts with H2, and [HI] must increase by +2x .
Reaction
Initial
Change 1 H2 1 I2 2 HI 0.500 M 0.500 M 0M ―xM ―xM + 2x M At Equilibrium
Complete the bottom row using terms with numbers and x’s.
*****
Reaction
Initial
Change
At Equilibrium 1 H2 1 I2 2 HI 0.500 M 0.500 M 0M ―xM ―xM + 2x M (0.500 ― x) M (0.500 ― x) M 2x M As a check, use this rule: The numbers in front of x’s in rows 3 and 4 must be the same as the
coefficients in row 1. Coefficients determine the Change ratios.
Step 5. K: Write the K equation and solve for x. Substitute the Equilibrium row terms
into the K equation. ***** ©2010 ChemReview.net v. 3d Page 864 Module 28 — Equilibrium Hint: you will need to take a square root.
*****
[HI]2
= 144
using mol/L measured at equilibrium.
K:
K=
[H2][I2]
(2x)2 K= = 144 Take the square root of both sides,
then solve for x. (0.500 ― x)2
*****
(2x) = (144)1/2 = 12.0 Finish solving for x. (0.500 ― x)
*****
2x =
2x = 12 (0.500 ― x)
6 ― 12 x 14x = 6 ;
Step 6. x = 6/14 = 0.429 Solve for the WANTED unit. Substitute x into the equilibrium row to find the
WANTED values for each equilibrium concentration. *****
H2
At Equilibrium
Step 7. 0.500 ― 0.429 = 0.071 M I2 HI 0.071 M + 2(0.429) = 0.858 M Check. Substitute those equilibrium concentrations into the K expression.
Calculate a value of K. Compare to the K value given in the problem. *****
7. Check: K = [HI]2 = (0.858)2 = 0.736 = 146 versus K = 144 in the data.
0.00504 [H2][I2]
(0.071)2
When your check K agrees with the K supplied in the problem, allowing for the doubtful
digit, your answer is likely correct.
***** Summary: Solving K Calculations – The WRECK Steps
13: WRE. Write the WANTED unit, balanced Reaction equation, and Extent to which the
reaction goes to completion.
4. [email protected]
a. If concentrations at equilibrium are given in a problem, use them. A rice table is
not needed. ©2010 ChemReview.net v. 3d Page 865 Module 28 — Equilibrium b. When concentrations or gas pressures at equilibrium are not known, make a rice
table. The bottom row values are the concentrations at equilibrium, or can be
used to calculate those concentrations, that are needed in the K equation.
i. If initial moles, mol/L or gas pressures are known for reactants, and any one
value is known at equilibrium, the bottom row of the rice table can be solved
using arithmetic. ii. If moles, mol/L or gas pressures are known for all of the initial reactants, the
Equilibrium row of the rice table can often be solved using algebra.
A key step is assigning signs and x values to the Change row of the rice table.
The Change row will have terms such as ―x, +2x, etc.
5. K: Write the K equation. Substitute the Equilibrium row terms into the K equation,
then solve for x if needed. 6. Solve for the WANTED unit. If needed, substitute x into the Equilibrium row to find
the values for each equilibrium concentration. 7. Check. Substitute the equilibrium concentrations into the K expression. Calculate a
value for K. Compare to the K value supplied in the problem. Practice: If you get stuck, read a part of the answer below, then try again.
(2x)2 1. Try solving without a calculator: = 64 (0.200 ― x)2
(2x)2 2. Use a calculator as needed: = 81 (0.600 ― x)2
3. Based on the following rice table,
Reaction 1 CO Initial 0.200 M 1 H2O
0.200 M 1 C O2 1 H2 0M 0M Change
At Equilibrium
a. Complete the Change row using terms with x’s.
b. Write the concentrations at equilibrium using terms with numbers and x’s.
c. Write the equilibrium constant expression for the reaction.
d. If K = 0.49, calculate the value of x.
e. Calculate values for the concentrations at equilibrium.
f. Check your answers by using them to calculate a value for K. ©2010 ChemReview.net v. 3d Page 866 Module 28 — Equilibrium 4. In a closed system with constant volume, for the reaction H2 + Cl2
2 HCl (all
gases) at 3050 K, Kp = 225. In the initial mixture before the reaction begins, PH2 =
PCl = 0.750 atm. What will be the partial pressure of each substance at equilibrium?
2 ANSWERS
1. (2x)2 = 64 (0.200 ― x)2 2x Taking the square root of both sides: = 8.0 0.200 ― x 2x = 8.0(0.200 ― x)
2x = 1.6 ― 8x
10x = 1.6, x = 0.16 (2x)2 2. (0.600 ― x)2 = 81 Taking the square root of both sides: 2x = 9.0 0.600 ― x 2x = 9.0(0.600 ― x)
2x = 5.4 ― 9x
11x = 5.4, x = 0.49 3. a. The numbers in front of x must match the coefficients of the balance equation.
Since the table shows reactants but no products in the initial mixture, to reach equilibrium, some of the
reactants must be used up. The reactants must therefore have negative signs in the Change row.
Some products must be formed, so the signs in the Change row for the products must be positive.
b. See table below.
Reaction 1 CO 1 H2O Initial 0.200 M 0.200 M Change ―xM At Equilibrium
K= K= [CO2] [H2] +xM +xM +xM [CO2] [H2]
[CO][H2O] d. 0M +xM 0.200 ― x M 1 H2 0M ―xM 0.200 ― x M c. 1 CO2 [CO][H2O] = (x)2 = 0.49 (0.200 ― x)2 Taking the square root of both sides: x
0.200 ― x = 0.70 x = 0.70(0.200 ― x) ©2010 ChemReview.net v. 3d Page 867 Module 28 — Equilibrium x = 0.14 ― 0.70x
1.7x = 0.14, x = 0.0824 e. 0.200 ― 0.0824 =
0.118 M At Equilibrium
f. 1 H2O 1 CO K= [CO2] [H2] (0.0824)2 = = (0.118)2 [CO][H2O] 1 CO2 1 H2 0.118 M 0.0824 M 0.0824 M 0.006790 = 0.49
0.01392 Matches K. Check. 4. Calculations using Kp values and gas partial pressures in atmospheres solve in the same manner as K
calculations using concentration. To solve K calculations, write the WRECK steps, solve, and check.
WANT: PH and PCl and PHCl at equilibrium.
2
2 R and E: H2 + Cl2 2 HCl (goes to equilibrium) [email protected] If pressures at equilibrium are not known, a rice table is needed.
Rice tables can be solved in partial pressures if volume is held constant.
1 H2 1 Cl2 2 HCl Initial 0.750 atm 0.750 atm 0 atm Change ― x atm ― x atm + 2x atm (0.750 ― x) atm (0.750 ― x) atm 2x atm Reaction At Equilibrium In the change row, the coefficient of x must match the coefficient for that term in the balanced
equation, and the reactant and product terms must have opposite signs.
If only the initial concentrations or pressures are known, the reactant concentrations or pressures must
decrease in going to equilibrium, and products must be formed. This means that the signs in the
Change row must be negative for the reactants and positive for the products.
K: Use the K equation to solve for x.
K= K= (PHCl)2
PH • PCl
2
2
(2x)2
(0.75 ― x)2 = 225 using partial pressures at equilibrium. = 225 Take the square root of both sides, then solve for x.
*****
(2x)
=
(0.75 ― x) (225)1/2 = 15.0 Finish solving for x. *****
2x = 15 (0.75 ― x)
2x = 11.25 ― 15x ©2010 ChemReview.net v. 3d Page 868 Module 28 — Equilibrium 17x = 11.25 ;
Step 6. x = 11.25/17 = 0.662 Solve for the WANTED unit. Substitute x into the equilibrium row to find the WANTED values
for each equilibrium concentration. *****
H2
At Equilibrium
Step 7. 0.750 ― 0.662 = 0.088 atm Check: K = Cl2 HCl 0.088 atm + 2(0.662) = 1.32 atm 2
(PHCl)2 = (1.32) = 1.742 = 225 versus K = 225 in the data.
PH • PCl
(0.088)2 0.007744
2
2 If the check K agrees with the supplied K, allowing for the doubtful digit, your answer is likely correct.
***** Lesson 28H: Q: The Reaction Quotient
Timing: Do this lesson when you are asked to calculate values for Q.
** * * *
The reaction quotient (Q) is the number that results when concentration or pressure values
for a mixture of reactants and products that may not be at equilibrium are substituted into
the K expression.
For a reversible reaction, if K is known, Q will identify the direction in which a reaction
mixture will need to shift (toward products or reactants) to reach equilibrium.
In a closed system, if a mixture of reactants and products is not at equilibrium, the reaction
will continue in the direction that takes the mixture toward equilibrium. The Q value will
shift continuously to get closer to K. When Q = K, net change stops.
The speed at which a reaction proceeds is determined by reaction kinetics, but the direction
that the reaction will shift is determined by Q and K.
To determine
• whether a mixture for a given reversible reaction is at equilibrium, or • the direction a reaction mixture must shift to reach equilibrium, the concentrations or pressures in the reaction mixture are substituted into the K expression.
This calculates a value for Q.
Then,
• If Q < K, the mixture must shift right, toward products, to reach equilibrium. • If Q > K, the mixture must shift left, toward the reactants, to reach equilibrium. • If Q = K, the mixture is at equilibrium. Those three cases must either be memorized or be derived when you need them. ©2010 ChemReview.net v. 3d Page 869 Module 28 — Equilibrium The logic of these shifts makes sense. For the general reaction
aA + bB cC + dD Q = [C]c[D]d = product of the [products]
[A]a[B]b
product of the [reactants] = K if measured at equilibrium, • If Q is a larger number than K, for the Q ratio to reach K, as it must at equilibrium,
product concentrations must go down, and reactant concentrations must go up.
That means the reaction mixture must “shift toward reactants” to reach K. • If Q < K, the [products] must go up, and the [reactants] down. The mixture must
shift toward the products. The equation for the expression for K and Q is the same, but the numeric values for K and Q
will often differ. Q can be calculated from measurements in any reaction mixture. A K
value must be based on measurements at equilibrium.
As with K values, Q is written as a number without units.
To summarize, this rule must be memorized.
The Q Rule: As a reaction proceeds, Q values shift to get closer to K.
To determine which direction a mixture will shift in order to reach equilibrium,
• substitute the mixture concentrations or pressures into the K expression to
calculate Q, then • compare the numeric values of Q and K. If Q < K, the mixture will shift to the right, toward the products.
If Q > K, the mixture will shift to the left, toward the reactants.
If Q = K, the mixture is at equilibrium and net changes do not occur.
If the Q < K rule for a shift in one direction is memorized, the cases for Q = K and Q > K
should be easy to write as needed.
Using the Q rule, try the following problem.
Q. For the reaction in a closed system:
N2 (g) + 3 H2 (g) 2 NH3 (g) At 400.oC, Kc = 3.8 x 104. A mixture of those gases not yet at equilibrium contains [N2] = 0.12 M,
[H2] = 0.030 M, and [NH3] = 0.60 M.
a. Calculate the reaction quotient.
b. As the mixture continues to react, will the equilibrium shift left or right?
c. As the mixture continues to react, will [NH3] increase or decrease?
***** ©2010 ChemReview.net v. 3d Page 870 Module 28 — Equilibrium Answer
a. WANT:
DATA:
Q= Q = number found by substituting the current concentrations or pressures
into the K expression.
mixture concentrations: [N2] = 0.12 M, [H2] = 0.030 M, [NH3] = 0.60 M.
[NH3]2 (0.60)2 = [N2][H2]3 = (0.36) = 1.1 x 105 (0.12)(2.7 x 10―5) (0.12)(0.030)3 b. To find the direction a mixture will shift to reach equilibrium, compare Q to K.
Since Q = 1.1 x 105 and Kc = 3.8 x 104 = 0.38 x 105, and Q is larger than Kc . This means
that the reaction direction must shift to the left to reach equilibrium.
c. The [NH3] in the products must decrease as the reaction continues and shifts left. The Q
value must shift toward a final value of Kc . Practice
1. Circle the largest numeric value of the three in the series.
a. 2.0 x 106 or 32.5 x 104 or 0.026 x 107 b. 2.0 x 10―6 or 32.5 x 10―4 or 0.026 x 10―7 2. For a given reversible reaction, what is the difference between
a. the K expression and the Q expression?
b. A K value and a Q value?
3. For a mixture, if Q = 0.0010, does the mixture have more reactants or more products?
4. For a given mixture in a reversible reaction, Q = 2.0 x 104 and Kc = 4.0 x 105 . To reach
equilibrium from this mixture, will the reaction shift toward the left or the right?
5. For the reaction PCl5 PCl3 + Cl2
, Kc = 0.56 , a gas mixture in a closed
(g)
(g)
(g)
system is found to contain [PCl5] = 0.50 M, [PCl3] = 0.20 M, and [Cl2] = 0.40 M. As the
mixture continues to react, will the reaction shift to the left or right? ©2010 ChemReview.net v. 3d Page 871 Module 28 — Equilibrium ANSWERS
1. To compare values, change to a consistent power of 10. Below we convert to the largest power in the
series. When the exponential terms are the same, the significand will determine which value is higher.
a. 0.20 x 107 or 0.0325 x 107 or 0.026 x 107 b. 0.020 x 10―4 or 32.5 x 10―4 or 0.000 026 x 10―4 2a. The K and Q expressions are the same.
2b. The K value for a given reaction at a given temperature is always the same, but it only applies to a reaction
mixture at equilibrium. The Q value can vary. Q can be measured for any mixture of reactants and
products, at any point in the reaction.
3. Q = (product of [products)/(product of [reactants]). If Q is less than one, the bottom term of the ratio must
be larger. This mixture must have higher concentrations of reactants than products.
4. Q is less than Kc. Based on the K and Q expression, to reach K, the [products] must increase, so the
reaction must shift to the right. Anytime Q < K, the reaction must shift right to reach equilibrium.
5. WANT: Which direction the reaction will shift.
To find which way a mixture must shift to reach equilibrium, compare Q to K.
The Kc value is known. To calculate Q, substitute the current concentrations into the Kc expression.
Q= [PCl3] [Cl2]
[PCl5] = (0.20) (0.40)
(0.50) = 0.16 , which is less than Kc = 0.56 in the data. Since Q is less than K, the [products] on top must increase to reach K. To get to equilibrium from the
current mixture, the reaction must shift to the right.
***** Lesson 28I: Calculations Using K and Q
Timing: Do this lesson when you are asked to calculate concentrations or gas pressures at
equilibrium starting from a mixture of reactants and products that is not at equilibrium.
*****
So far, we have solved calculations involving equilibrium concentrations or gas pressures
in four situations.
• If concentrations or gas pressures at equilibrium are known, K calculations are done
by substituting those values into the K expression. • If concentrations or gas pressures are known for an initial mixture that is all
reactants (with no products), as well as for any one reactant or product after the
reaction has reached equilibrium, values at equilibrium can be found using a rice
table and arithmetic. ©2010 ChemReview.net v. 3d Page 872 Module 28 — Equilibrium • If concentrations or gas pressures are known for an initial mixture that is all
reactants (with no products), values at equilibrium may be able to be found using a
rice table and algebra. • In a mixture of reactants and products, the direction that a reaction will shift to reach
equilibrium can be determined by comparing K and Q. In our fifth type of calculation, we will calculate concentrations or gas pressures in a
reaction at equilibrium, starting from a nonequilibrium mixture of the reactants and
products. In these cases, concentrations and gas pressures at equilibrium can often be
calculated using a rice table, Q and K values, and algebra. Equilibrium Calculations Based on Mixtures of Reactants and Products
In all of our K calculations to this point, our initial mixtures have been reactants only, with
no products. This has meant that the sign of the terms in the change row in the rice table
has always been negative for the reactants and positive for the products. This is logical: if
there are only reactants, the reaction has not yet started. If the reaction goes at all, some
reactants must be used up and some products formed.
However, if a closed system has a mixture of reactants and products, and the reaction is not
yet at equilibrium, whether the mixture will need to shift to the right toward the products
or left toward reactants to reach equilibrium is not obvious. To calculate which way the
equilibrium will shift, we must compare K and Q.
• If a reaction mixture must shift to the right to reach equilibrium, the signs in the
Change row of the rice table will be negative for the reactants, and positive for the
products. As a reaction shifts to the right, reactants are used up and products form. • If a reaction mixture must shift left to reach equilibrium, the signs for the terms in
the Change row will be positive for the reactants, and negative for the products,
because in shifting left, products are used up and reactants form. The above means that for K calculations involving mixtures of reactants and products, the
signs in the Change row of a rice table must be assigned carefully, based on K versus Q.
For K calculations based on an initial mixture of reactants and products, let’s learn to apply
these rules using an example.
Q. For the reaction H2 + I2
2 HI (all gases) , at a temperature where Kc = 16,
in a closed system with constant volume, if in the mixture [ H2 ] = [ I2 ] = 0.100 M
and [ HI ] = 0.500 M, what will be the concentrations of each substance at
equilibrium? K calculations are solved using the WRECK steps. Complete the following steps for the
problem above.
Steps 13: WRE.
Write the WANTED unit, balanced Reaction equation, and Extent to
which the reaction goes to completion. ©2010 ChemReview.net v. 3d Page 873 Module 28 — Equilibrium Step 4. [email protected] This step is where K calculations differ. If the concentrations at equilibrium are not known, make a rice table.
If the data in the problem is for a mixture of reactants and products, Q is
compared to K to determine in which direction the mixture must shift to reach
equilibrium. The direction of the shift determines the signs in the Change
row.
• If the reaction mixture must shift to the left to reach equilibrium, the
Change row for reactants will have + #x terms, and the products ― #x. • If the mixture must shift to the right, the Change row for products will
have + #x terms, and the reactants ― #x. In the Change row of a rice table, all of the reactant terms must have the
same sign, and all of the product terms must have the opposite sign.
Complete those steps, and then check your answer below.
*****
1.
WANT:
2,3. R+E:
4. [email protected] [H2], [I2], and [HI] at equilibrium.
H2 + I2
2 HI (all gases)
(goes to equilibrium. Use K).
Since the concentrations at equilibrium are not given, a rice table
is needed.
For a mixture of reactants and products, to decide the signs in the
Change row, compare Q to K to determine the direction the
mixture must shift.
Q= [HI]2 = [H2][I2] (0.500)2 = 25.0
(0.100)2 Since Q = 25.00 > K = 16 , the reaction must shift to the left,
toward the reactants, to get to equilibrium.
Now write the rice table. In the Change row, represent the change that will take
place to reach equilibrium using terms like +x, ―x, +2x, ―3x, etc.
Complete the Equilibrium row using terms that include x.
Finish Step 4, and then check your answer below.
*****
As the reaction shifts toward reactants to reach equilibrium, some of the [H2] and [I2]
reactants are formed. In the change row, the signs for all reactant terms must therefore be
positive, and the signs for all product terms must be negative.
*****
According to the coefficients, if [H2] changes by +x, [I2] must change by +x, and the change
in [HI] must be ―2x . For x terms, the coefficients in the top row and the numbers in front of
the x’s in the Change row must be the same. ©2010 ChemReview.net v. 3d Page 874 Module 28 — Equilibrium 1 H2 1 I2 2 HI 0.100 M 0.100 M 0.500 M +xM +xM ― 2x M 1 H2 1 I2 2 HI 0.100 M 0.100 M 0.500 M +xM +xM ― 2x M ( 0.100 + x ) M ( 0.100 + x ) M ( 0.500 ― 2x ) M Reaction
Initial
Change
At Equilibrium
Complete the bottom row.
*****
Reaction
Initial
Change
At Equilibrium
Step 5. K: Write the K equation, then solve for x. Substitute the equilibrium row terms
into the K equation. *****
K= [HI]2 = (0.500 ― 2x)2 = 16 (0.100 + x)2 [H2][I2] Take the square root of both sides,
then solve for x. *****
(0.500 ― 2x) = (16)1/2 = 4.0 Finish solving for x. (0.100 + x)
*****
0.500 ― 2x = 4.0 (0.100 + x) 0.500 ― 2x =
6x = 0.100 ;
Step 6. 0.400 + 4x (Don’t round to sf until the end of the calculation) x = 0.100/6 = 0.0167 Solve for the WANTED units and symbols. Substitute the x values into
equilibrium row terms. *****
H2
At Equilibrium
Step 7. 0.100 + 0.017 = 0.117 M I2 HI 0.117 M 0.500 ― 2(0.017) = 0.466 M Check. Substitute those equilibrium concentrations into the K expression.
Calculate a value of K. Compare to the K value given in the problem. ***** ©2010 ChemReview.net v. 3d Page 875 Module 28 — Equilibrium 7. Check: K = [HI]2 = [H2][I2] (0.466)2 = 15.9 versus K = 16 in the data.
(0.117)2 The check K agrees with the given K, allowing for the doubtful digit.
***** Summary: Solving K Calculations
To calculate concentrations or pressures at equilibrium from those that are not at
equilibrium, write the WRECK steps, solve, and check.
1,2,3. WRE. Write the WANTED symbol or unit, balanced Reaction equation, and
Extent of the reaction.
4. [email protected] Find all concentrations or partial pressures at equilibrium.
If values at equilibrium are not known, make a rice table.
If all known values are initial reactants, Change row reactant signs are all negative.
For a mixture of reactants and products not at equilibrium, the mixture will shift to
reach equilibrium. Find the reaction direction using Q. The direction of the shift
decides Change row signs (+ #x or ─ #x).
5. K: Solve K for x.
6. Find the WANTED unit using the x values.
7. Check. Calculate K using equilibrium values. Compare to K in the data. Practice: For hints, read a part of the answer. 1. When solving for an equilibrium concentration using a rice moles table, if K = 25 and
Q = 15,
a. will the reaction shift to the right or left to reach equilibrium?
b. Will the sign of the reactant terms in the change row of the table be positive or
negative?
2. A mixture for a reversible reaction has the following rice table.
Reaction 1A 1B 2C Initial 0.50 M 0.50 M 2.0 M Change +xM +xM ─ 2x M At Equilibrium
a. Write an expression for the equilibrium constant Kc .
b. Calculate a Q value for the initial mixture.
c. Will the mixture shift toward reactants or products to reach equilibrium?
d. In the table, complete the concentrations at equilibrium in terms of x .
e. If the value for the equilibrium constant is 4.0, calculate the value of x.
©2010 ChemReview.net v. 3d Page 876 Module 28 — Equilibrium f. Calculate the concentrations of A, B, and C at equilibrium. g. Check your answer by using the part f answers to calculate a K value.
3. In the reaction CO + H2O CO2 + H2 all gases in a closed system, if Kp = 0.36 and the initial concentrations are 0.50 M for each reactant and 1.00 M for each product,
a. What will be the value of Kc ?
b. What will be concentration of each substance at equilibrium? ANSWERS
1a. Since Q is lower than K, to reach equilibrium, Q must increase. The reaction mixture must shift to the right,
toward higher product concentrations, to reach equilibrium.
1b. When the reaction must shift to the right to reach equilibrium, the product terms in the Change row of the
rice table will have positive signs, because their concentration will increase during the shift.
2a. The rice table coefficients show that the reaction is A+B 2C , so K = [C]2
[A][B] 2b. To calculate Q, substitute the initial concentrations into the K expression.
Q= [C]2
[A][B] = [2.0]2
= 4.0
[0.50][ 0.50]
0.25 = 16 2c. Since the signs for the reactants in the change row are positive, the reactant concentrations will be
increasing as the mixture shifts toward equilibrium. The reaction is shifting to the left toward reactants.
2d. At Equilibrium 0.50 M + x 0.50 M + x 2.0 M ─ 2x 2e. To calculate x, substitute the equilibrium concentrations into the K equation.
K = 4.0 = [C]2
[A][B] = 2 [2.0 M ─ 2x]
[0.50 M + x]2 = 4.0 To find x, begin by taking the square root of both sides.
[2.0 M ─ 2x]
[0.50 M + x] = 2.0 2.0 M ─ 2x = 2.0(0.50 M + x)
2.0 M ─ 2x = 1.0 M + 2x
4x = 1.0 M ; x = 0.25 M
2f. Substitute for x in the bottom row of the rice table.
[A] = 0.50 M + 0.25 M = 0.75 M = [B] ; [C] = 2.0 M ─ 2(0.25 M) = 1.5 M
2g. K= [C]2
[A][B] = [1.5]2
=
[0.75][0.75] ©2010 ChemReview.net v. 3d 2.25
0.562 = 4.0 This agrees with the K supplied in part e. Page 877 Module 28 — Equilibrium 3a. Because the total moles of gas are the same on both sides, Kp = Kc = 0.36
3b. To calculate equilibrium measures from nonequilibrium measures, do the 7 steps: WRECK, solve, check.
1. WANT: The four concentrations at equilibrium.
2,3. R+E: CO + H2O CO2 + H2 (all gases) (goes to equilibrium, use K to solve). 4. [email protected] If concentrations at equilibrium are not known, a rice table is needed.
For a mixture of reactants and products, to determine the signs in the Change row,
determine the direction the equilibrium must shift by comparing Q and K.
Q= [CO2] [H2]
[CO][H2O] = (1.00)(1.00) = 1
(0.50)(0.50)
0.25 = 4.0 Since 4.0 = Q > 0.36 = K, the mixture is shifting toward reactants to reach equilibrium.
In the Change row, if the reaction is shifting toward reactants, reactants must get a +
sign, and products a ―. All reactant terms must have the same sign, and all product
terms must have the opposite sign. Write Equilibrium row in terms of x.
1 H2O Reaction 1 CO Initial 0.50 M Change +xM At Equilibrium 1 CO2
1.00 M 0.50 M
+xM 0.50 + x M 1 H2
1.00 M ―xM 0.50 + x M 1.00 ― x M ―xM
1.00 ― x M 4. K: Solve the K equation for x using the equilibrium row terms.
Kc = [CO2] [H2] (1.00 ― x)2 = 0.36 = (0.50 + x)2 [CO][H2O] Taking the square root of both sides:
Solving for x: (1. 00 ― x)
(0.50 + x) = (0.36)1/2 = 0.60 (1.00 ― x) = 0.60 (0.50 + x)
1.00 ― x = 0.30 + 0.60x
0.70 = 1.60 x
x = 0.438 M (carry an extra sf until the last step) 6. Solve for the WANTED units.
WANTED: [reactants] = [CO] = [H2O] = 0.50 M + x = 0.50 M + 0.438 = 0.94 M
[products] = [CO2] = [H2] = 1.00 M ― x = 1.00 M ― 0.438 = 0.56 M
(Hundredths has doubt in both. Adding and subtracting, the place decides the doubtful digit.)
7. Check: K = [CO2] [H2]
[CO][H2O] 2
= (0.56) = 0.35 calculated ≈ 0.36 in original data. Check!
(0.94)2 ***** ©2010 ChemReview.net v. 3d Page 878 Module 28 — Equilibrium Lesson 28J: Solving Quadratic Equations
Timing: Do this lesson if you are asked to solve quadratic equations as part of K or other
calculations. Even if you feel confident about the math of quadratic equations, this lesson
may contain information on the use of quadratics in science problems that will be helpful to
review.
***** Quadratic Equations
In the calculations in the prior lesson, the numbers and formulas were chosen to allow
solving for K values by simply taking a square root. In other K calculations, you may need
to solve a quadratic equation for x.
The general format for a quadratic equation is ax2 + bx + c = 0 . A quadratic equation
has x2 an term, but no x powers higher than 2. In a quadratic equation, the terms a, b, and c
will be numbers and x is the unknown.
Once a quadratic equation is in the general format, the numbers a, b, and c are known, and
x can be solved using quadratic formula:
x = ─b ± (b2 ─4ac)1/2
2a
The quadratic formula will result in two values for x , both of which solve the quadratic
equation.
Both the general format and the quadratic formula must be memorized. The quadratic
formula may be best memorized by repeated recitation: “x equals minus b plus or minus
the square root of b squared minus 4ac, all over 2a.”
Tools available online (search quadratic formula calculator) as well as some calculators will
solve a quadratic equation for x after you have values for a, b, and c. However, you may
not be allowed to use those tools on quizzes and tests. This lesson will review how to solve
quadratic formulas with minimal calculator use.
The following steps will solve the type of quadratic equations most often encountered in
firstyear chemistry. Let’s begin with a simple example.
Q. 4 x 2 + (x ─ 2 ) = 3 Solve for x using the steps below. Steps To Solve Quadratic Equations
1. Move all terms to one line and eliminate parentheses.
2. Group the terms into the general quadratic format: ax2 + bx + c = 0
Write the values for a, b, and c.
Complete those steps, then check below.
***** ©2010 ChemReview.net v. 3d Page 879 Module 28 — Equilibrium 1. All terms are on one line. Removing parentheses:
4x2 + (x ─ 2) = 3
4x2 + x ─ 2 = 3
2. Move the terms into the general format: ax2 + bx + c = 0 by rearranging the terms to get a zero on the right side.
4x2 + x ― 5 = 0
Write the values: a = 4, b = 1, c = ― 5 .
Now try step 3, then check your answer below.
3. Substitute the values for a, b, and c into the quadratic formula and solve:
x = ─b ± (b2 ─4ac)1/2
2a
*****
3. Substituting a, b, and c into the quadratic formula:
x = ─b ± (b2 4ac)1/2 = ─(1) ± {(1)2 ─ 4(4)( ― 5)}1/2
2a
2(4)
= ─1 ± (81)1/2 = ─1 ± 9 8 8 = 8 and 8 = ─1 ± (1 + 80)1/2 =
8 ― 10 = +1 and ― 1.25 8 When solving a quadratic, the result will be two values for x.
Now, if using a calculator is permitted in your class when solving quadratics, use the
values of a, b, and c found above to solve. See if you get the same two values for x.
4. Check the two values by substituting them into the original equation.
*****
4(1)2 + (1 ─ 2) = 4 ─ 1 = 3 and 4(― 1.25)2 + (― 1.25─ 2) = 6.25 ― 3.25 = 3 Both values are solutions to the original equation. Practice A. Solve for x, then substitute your answers into the original equation to check
them. The second problem is more challenging.
1. x2 ― 3x ― 28 = 0 2. 2= (1 + 2x)2
(2.5 ― x) (1.5 ― x) Solving Quadratic Equations in Science Problems
The above four steps will solve quadratic equations. The result is nearly always two values
for x. In science problems, however, one of those x values will usually result in an “unreal”
quantity. An example would be an x that results a negative value for mass or volume:
quantities that must be positive in real measurements. The other x will be a number that,
when plugged into the values in the problem, does result in measurements that are real.
The x value that makes sense is the one that is used to complete the problem. ©2010 ChemReview.net v. 3d Page 880 Module 28 — Equilibrium When solving quadratic equations for use in science problems, we will add Step
5. Substitute the two x values into the data one at a time, and see which one makes sense.
Use the value that makes sense to solve the science calculation.
Use the five steps above to solve for x in the following problem. If you get stuck, read the
below answer until unstuck, adjust your work, and finish.
Q. For the reaction in a closed system: A + B
C , K = 10. At equilibrium,
concentrations are [C] = (1.40 + x) M, [B] =(0.40 ― x) M, and [A] = (1.00 ― x) M.
a. Solve for x. b. Solve for [A], [B], and [C]. *****
Answer
a. For K calculations, write the WRECK steps, solve, and check.
1. WANT: x 2,3. R&E: A+B 4. [C] = (1.40 + x) M, [B] =(0.40 ― x) M, and [A] = (1.00 ― x) M [email protected] C (Since a K value is given, reaction goes to equilibrium) Since the [email protected] values are known, a rice table is not needed.
5. K: Write the K expression, then solve for x.
K= [C] ; so:
[A][B] 10. = (1.40 + x)
(1.00 ― x)(0.40 ― x) If a K equation cannot be solved by taking a square root, see if the equation has x2
terms that indicate a quadratic equation. In firstyear chemistry, assigned
problems will nearly always be those that can be solved by either taking a square
root or solving a quadratic.
In this case, the denominator, when multiplied out gives an x2 term, an indication
that you can use the quadratic formula to solve.
The arithmetic shown below can be done in different ways, but the five general steps to
solve a quadratic given above should be followed. Apply those steps to find x.
*****
1. Move all terms to one line.
10 (0.40 ― x) (1.00 ― x) = (1.40 + x)
Eliminate parentheses. 10 ( 0.40 ― 1.40x + x2 ) = 1.40 + x
4 ― 14x + 10x2 = 1.40 + x 2. Group terms into the general quadratic format. On the left side, write a single x2
term, a single x term, and a single number term. On the right side must be zero.
*****
General: ax2 + bx + c = 0 In this problem: 10x2 ― 15x + 2.6 = 0 3. Substitute the values for a, b, and c into the quadratic formula and solve for x.
To begin, fill in these blanks: a = ______ , b = _______ , c = ___________
***** ©2010 ChemReview.net v. 3d Page 881 Module 28 — Equilibrium a = 10 , b = ― 15 , c = 2.6 .
At this point, if during tests you are allowed to use a calculator that will solve
quadratic equations, use the manual for the calculator to determine which steps to
use, then solve the quadratic above.
If calculator quadratic solving is not allowed, solve by substituting a, b, and c into
the quadratic formula.
** * * *
x = ─b ± (b2 ─4ac)1/2
2a = ─(― 15) ± {(― 15)2 ─ 4(10)( 2.6)}1/2
2(10) = = + 15 ± (225 ─ 104)1/2 = +15 ± (121)1/2 = 15 ± 11 = 26 and 4 = 1.3 and 0.2
20
20
20
20
20
If you have computer access, search online for quadratic formula calculator. Use it to
check the answers calculated above. You may want to compare several different
online tools.
4. Check: both values work to solve the K equation, but
5. Only one x value will work to solve the science problem. Substitute the two
resulting answers for x into the problem data and see which one makes sense.
** * * *
Since [ B ] = (0.40 ― x) , if x = 1.3, [B] would equal ― 0.90 Molar. The concentration
of a substance cannot be negative. The x value that gives a positive number for every
concentration is x = 0.20 . Use that x value to calculate the concentrations and
finish the problem.
This answers Part a of the problem. Complete part b.
*****
b. [A] = (1.40 + x) = (1.40 + 0.20) = 1.60 M (when adding, track the place with doubt) [B] = (0.40 ― x) = (0.40 ― 0.20) = 0.20 M
[C] = (1.00 ― x) = (1.00 ― 0.20) = 0.80 M
In problems that calculate a value for K, you can check your work by substituting the
calculated answers back into the K expression, then comparing the calculated K to the K
given in the problem. Let’s add this to our to our quadratic solving process as Step
6. Check: If K was given in the problem, substitute your answers into the K equation and
calculate K. The result should be the K given in the problem.
Do step 6, then check your answer below.
*****
Check: K= [A] =
[B][C] 1.60
=
(0.20)(0.80) 1.60 = 10.
0.16 This matches the original K value.
The value x = 0.20 M works. ***** ©2010 ChemReview.net v. 3d Page 882 Module 28 — Equilibrium SUMMARY: Solving Quadratic Equations In Science Calculations
1. Move all terms to one line and eliminate parentheses.
2. Group the terms into the general quadratic format: ax2 + bx + c = 0
3. Substitute the values for a, b, and c into the quadratic formula. Find two values for x .
4. Check: Substitute your two answers into the original equation.
5. Substitute the two x values into the data one at a time, and see which one makes sense.
Use that value to complete solving the science calculation. Practice B
1. For this reaction in a closed glass cylinder H2 + I2
2 HI (all gases)
at a temperature where Kc = 2.0, initial concentrations are [H2] = 2.5 M, [I2] = 1.5 M, and
[HI] = 1.0 M. What will be the concentration of each substance at equilibrium?
PCl5 (all gases) in a container with a fixed volume,
2. For the reaction PCl3 + Cl2
at an equilibrium temperature where Kp = 20., initial partial pressures are 0.20 atm. for
PCl3, 0.040 atm. for Cl2, and 0.58 atm for PCl5. What will be the partial pressure of
each gas at equilibrium? ANSWERS
Practice A
1. a. x2 ― 3x ― 28 = 0
This equation is already in the general format of a quadratic equation: a = 1, b = ― 3, c = ― 28 .
Substitute into the quadratic formula.
x = ─b ± (b2 ─4ac)1/2
2a = ─(― 3) ± {(― 3) 2 ─ 4(1)( ― 28)}1/2
2(1) + 3 ± (121)1/2 =
2 = + 3 ± 11
2 = 14
2 and ―8
2 =
= + 3 ± (9 + 112)1/2
2
+7 and ― 4 Check: (7)2 ― 3(7) ― 28 = 49 ― 21 ― 28 = 0
(― 4)2 ― 3(― 4) ― 28 = 16 + 12 ― 28 = 0
2. 2= (1 + 2x)2
(2.5 ― x) (1.5 ― x) Group terms on one line and eliminate parentheses. 2 (2.5 ― x) (1.5 ― x) = (1 + 2x)2
2 (3.75 ― 4x + x2) = (1 + 4x + 4x2)
7.5 ― 8x + 2x2 = 1 + 4x + 4x2
Move terms into the general format: ax2 + bx + c = 0 2x2 + 12x ― 6.5 = 0 ; a = 2, b = +12, c = ― 6.5 . ©2010 ChemReview.net v. 3d Page 883 Module 28 — Equilibrium Substitute a, b, and c into the quadratic formula and solve for the two x values.
x = ─b ± (b2 ─4ac)1/2 = ─(12) ± {(12)2 ─ 4(2)( ― 6.5)}1/2
2a
2(2)
=
Check: = ─12 ± (144 + 52)1/2
4 = ─12 ± (196)1/2 = ─12 ± 14 = 2 and ― 26 = +0.5 and ― 6.5 = x
4
4
4
4 =
(1 + 2(0.5))2
(1 + 2x)2
(2.5 ― x) (1.5 ― x)
(2.5 ― 0.5) (1.5 ― 0.5) = 4
=2
(2) (1) (1 + 2x)2
=
(1 + 2(― 6.5))2
= (―12)2
(2.5 ― x) (1.5 ― x)
(2.5 ― (― 6.5) (1.5 ― (― 6.5))
(8) (9) = 144
72 =2 Practice B
1. For K calculations, do the WRECK steps, solve, and check.
1. WANTED: [H2], [I2], and [HI] at equilibrium. 2 HI (all gases) (goes to equilibrium. Use K to solve.)
H2 + I2
4. [email protected] If concentrations at equilibrium are not known, a rice table is needed. 2,3. R+E: For a mixture of reactants and products, to determine the signs in the change row,
determine the direction the equilibrium will shift by comparing Q and K.
(1.0)2
[HI]2
=
= 1.0
= 0.27
Q=
[H2][I2]
(2.5) (1.5)
3.75
Since Q = 0.27 < K = 2.0 , the reaction must shift to the right to get to equilibrium.
This means that the changes in the [products] must be positive ( + # x terms).
Reaction 1 H2 1 I2 2 HI Initial 2.5 M 1.5 M 1.0 M Change ―xM ―xM + 2x M At Equilibrium
Each change box has an x. The number in front of x is the coefficient at the top of that column. Signs
must all be positive on the side the reaction is shifting to, and all negative on the other side.
Complete the equilibrium row using terms that include x.
*****
Reaction 1 H2 1 I2 2 HI Initial 2.5 M 1.5 M 1.0 M Change ―xM ―xM + 2x M (2.5 ― x) M (1.5 ― x) M (1.0 + 2x) M At Equilibrium 5. Solve K for x. Substitute the equilibrium row terms into the K equation. ©2010 ChemReview.net v. 3d Page 884 Module 28 — Equilibrium (1.0 + 2x)2
[HI]2
=
=2
[ H2 ][ I2 ]
(2.5 ― x) (1.5 ― x) K= This cannot be solved by a square root. Since the top and bottom when multiplied out have x2
terms, try the quadratic steps to solve.
*****
This quadratic equation is solved in Problem 1c above.
The answers are x = +0.5 and x = ― 6.5
6. Solve. Substitute the value of x to find the WANTED eq. row values.
[ H2 ]eq. = 2.5 ― x . Either 2.5 ― 0.5 = 2.0 M or 2.5 ―6.5 = ―4.0 M Since real concentrations cannot be negative, use
[ I2 ]eq = 1.5 ― x x = 0.5 . = 1.5 ― 0.5 = 1.0 M [ HI ]eq = 1.0 + 2x = 1.0 + 2(0.5) = 2.0 M
7. Check: substitute those [ ]eq. into the K expression. Calculate a K to compare to the data.
K= (2.0)2 = 4.0 = 2.0 versus 2.0 in the problem. Check.
[HI]2 =
2.0
[H2][I2]
(2.0) (1.0) 2. For K calculations, do the WRECK steps, solve, and check.
1. WANT: 2,3. ? = PPCl , PPCl , and PCl , in atm at equilibrium
3
5
2 R+ E: 4. PCl5(g)
(goes to equilibrium, use K).
PCl3(g) + Cl2(g)
[email protected] If pressures at equilibrium are not known, a rice table is needed.
For a mixture of reactants and products, to determine the signs in the Change row,
determine the direction the equilibrium will shift by comparing Q and K.
PPCl
=
0.58 atm.
= 72
Q=
5
PPCl • PCl
(0.20 atm.)(0.040 atm.)
3 2 Since Q = 72 > K = 20. , the reaction must shift to the left to reach equilibrium.
The means the reactants in the Change row will have positive signs.
Write rice table with Change row terms such as +x, ―x, +2x, ―3x, etc.
Complete the Equilibrium row using those x terms.
*****
Reaction
Initial
Change
At Equilibrium 1 PCl3 1 Cl2 1 PCl5 0.20 atm 0.040 atm 0.58 atm. +x +x ―x (0.20 + x) atm. (0.040 + x) atm. (0.58 ― x) atm. Each Change box has an x. The number in front of x is the coefficient in that column. Signs must all
be positive on the side the reaction is shifting to (left), and all negative on the other side. ©2010 ChemReview.net v. 3d Page 885 Module 28 — Equilibrium 5. Solve K for x. Substitute the equilibrium row terms into the K equation.
Kp = PPCl
5
PPCl • PCl
3 =
2 (0.58 ― x) = 20 (0.20 + x)(0.040 + x) The problems you are assigned will be solved either by taking a square root or solving the
quadratic. This equation cannot be solved by taking the square root, but it does have x2 terms if
the denominator parentheses are removed. Arrange the terms to fit the general quadratic format.
*****
Group all terms on one line and eliminate parentheses.
20 (0.20 + x) (0.040 + x) = (0.58 ― x)
20 (0.0080 + 0.24x + x2) = 0.58 ― x
0.16 + 4.8x + 20x2 = 0.58 ― x
Group terms into the general format: ax2 + bx + c = 0
20x2 + 5.8x ― 0.42 = 0
Substitute a, b, and c into the quadratic formula and solve for the two x values.
x = ─b ± (b2 ─4ac)1/2 = ─(5.8) ± {(5.8) 2 ─ 4(20)( ― 0.42)}1/2 = ─5.8 ± (33.64 +33.6)1/2 =
2a
2(20)
40
= ─5.8 ± (67.24)1/2 = ─5.8 ± 8.2
40
40 = 2.4 and ― 14 = +0.060 and ― 0.35
40
40 Which x value makes sense?
PPCl = 0.20 + x = 0.20 + 0.060 = 0.26 atm. or 0.20 ― 0.35 = ― 0.15 atm.
3
Since pressure cannot be negative, the valid value of x = 0.060.
6. SOLVE for Eq. values. Substitute the value for x to find the WANTED eq. row values.
PCl = 0.040 + x = 0.040 + 0.060 = 0.100 atm.
2
PPCl = 0.58 ― x = 0.58 ― 0.060 = 0.52 atm.
5 7. Check: substitute those atm. into the K expression. Calculate a K to compare to the data
PPCl
Kp =
5
=
0.52
= 20. and the Kp value in the problem is 20. Check!
PPCl • PCl
(0.26)(0.100)
3 2 ***** ©2010 ChemReview.net v. 3d Page 886 Module 28 — Equilibrium Summary – Equilibrium
If you have not already done so, you may want to organize this summary into charts,
numbered lists, and flashcards.
1. Reactions can be divided into three types: those that go nearly 100% to completion,
those that don’t go, and reactions that are in practice reversible and go partially to
completion. Reversible reactions continue until both the forward and reverse reactions
are going at the same rate, and no further change seems to take place. The reaction is
then said to be at equilibrium.
For equilibrium to exist:
• All reactants and products must be present in at least small quantities, and • The reaction must be in a closed system: no particles or energy can be entering or
leaving the system. 2. Le Châtelier’s Principle: If a system at equilibrium is subjected to a change, processes
occur which tend to counteract that change.
Le Châtelier’s Principle predicts shifts in concentration, temperature, and pressure.
To apply Le Châtelier’s Principle, write the reactants and products of the reversible
reaction with “twoway arrows” inbetween. Then,
a. Increasing a [substance] which appears on one side of a equilibrium equation shifts
an equilibrium to the other side. The other substance concentrations on the same
side as the [increased] are decreased, and the substance concentrations on the other
side are increased.
b. Decreasing a [substance] which appears on one side of a equilibrium equation shifts
the equilibrium toward that side. The other [substances] on the same side are
increased, and the [substances] on the other side are decreased.
c. Adding energy shifts the equilibrium away from the side with the energy term, and
removing energy shifts the equilibrium toward the side with the energy term.
d. Energy can be added to a system by increasing its temperature. Energy can be
removed by cooling. When energy is produced by a shift in equilibrium, the
temperature of the system goes up. When energy is used up, the system’s
temperature goes down.
e. Increasing the pressure on a gas, such as by reducing the volume of the container, will
shift the equilibrium toward the side with fewer total moles of gas, based on adding
the gas coefficients on each side of the balanced equation.
f. Decreasing the pressure on a gas, such as by increasing the volume of the container,
will shift the equilibrium toward the side with more moles of gas. g. If gas moles are equal on both sides, pressure changes will not shift an equilibrium.
h. Adding or removing a solid, pure liquid, or solvent does not shift an equilibrium.
because shifting an equilibrium does not change the concentration of a solid, pure
liquid, or solvent (though it may change the amounts present). ©2010 ChemReview.net v. 3d Page 887 Module 28 — Equilibrium 3. The Equilibrium Constant
a. For the general reaction aA + bB
at equilibrium, the ratio cC + dD [C]c[D]d = K
[A]a[B]b at a given temperature, will be constant. b. The ratio with powers and symbols for concentrations is called the equilibrium
constant (K) expression. The K value is the number that is the ratio. If the
temperature changes, the K value will change, but the K expression does not.
c. Generally, only concentrations that can change are included in K expressions. Terms
for solids, pure liquids, and solvents (including liquid water) are written as 1 in K
expressions. This moves the constant value of those terms into the value of K.
4. K Values
a. An equilibrium constant value is a positive number without units. At equilibrium,
• If the substances on the right side of an equation have higher concentrations
than those on the left, the value of K will greater than one. • If the substances on the right side of an equation have lower concentrations than
those on the left, the value of K will be a number between zero and one (in
scientific notation, a positive number with a negative power of 10). • The more a reaction goes to the right, the higher will be the value of K. b. If a value of K is known for a reaction written in one direction, the value of K for
the reverse reaction will be the reciprocal of the original K. Kf = 1/Kr
c. If a value of K is known for a reaction with one set of coefficients, those
coefficients can be multiplied by any positive number, and the new value of K
will equal the original K to the power of the multiplier. K = (#)n
d. In K calculations, the units often do not cancel properly.
5. Kp Equations
At a given temperature, for a reaction in which all substances are gases
wA(gas) + xB(gas) yC(gas) + zD(gas) at equilibrium these ratios will be constant:
Kc = [C]y[D]z
[A]w[B]x and y
z
Kp = (PC) •(PD)
where P represents partial pressure.
w•(P )x
(PA)
B Kc and Kp calculations are done in the same manner, except
• The values of Kc and Kp will not be the same if the two sides of the balanced
equation have a different number of total moles of gases. • In calculations using Kp , gas pressures must be measured in atmospheres. • When calculating a gas pressure using Kp values, atmospheres must be added
as the unit of the answer. ©2010 ChemReview.net v. 3d Page 888 Module 28 — Equilibrium 6. Converting Between Kc and Kp
When all terms in a K expression are gases, if either a Kc or a Kp value is known at a
given temperature, the other value can be calculated using
Kp = Kc (RT)Δn where • R is the Gas Constant, using R = 0.0821 L·atm/mol·K • T is absolute temperature in kelvins, and • Δn = (the sum of the coefficients of the gases in the products) MINUS (the
sum of the coefficients of the gases in the reactants). The number for Δn may be positive or negative and may be a fraction.
If the total for the gaseous coefficients on the two sides of the reaction equation are
equal, Δn = zero, and Kp = Kc .
7. Rice Moles Tables. In reaction calculations, we can track the counts of particles
before, during, and at the end of a reaction with a rice moles table. Rice tables have 4
rows: balanced Reaction equation, Initial, Change, and End/Equilibrium. Rice
tables can always be solved in moles, and can also be solved in concentration units if
the volume is the same during all measurements.
8. The reaction quotient (Q) is the number that results when concentration or pressure
values for a mixture that may not be at equilibrium are substituted into the K expression.
9. To determine the direction a mixture will shift to reach equilibrium, compare Q to K.
If Q > K, the mixture will shift toward reactants. At equilibrium, Q = K.
10. For calculations involving K values and concentrations or gas pressures,
write the WRECK steps, solve, and check.
1,2,3. WRE. Write the Wanted unit or symbol, balanced Reaction equation, and
reaction Extent. 4. [email protected] Find all concentrations or partial pressures at equilibrium.
If values at equilibrium are not known, solve with a rice table.
If all known values are for initial reactants, Change row reactant signs are negative.
If known values are for a mixture of reactants and products not at equilibrium, find
the reaction direction using Q. Direction decides Change row signs (─#x or +#x).
If the reaction must shift right to reach equilibrium, Change row signs for products
will be positive. If reaction must shift left, reactant Change row signs are positive.
In the Change row, all of the reactant terms must have the same sign, and all of
the product terms must have the opposite sign.
5. K: Solve K for x using algebra (e.g. take square root or solve the quadratic formula).
6. Solve for the WANTED unit using the x values.
7. Check. Calculate K using the solved values. Compare to K in the data. #####
©2010 ChemReview.net v. 3d Page 889 ...
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This note was uploaded on 11/16/2011 for the course CHM 1025 taught by Professor J during the Summer '09 term at Santa Fe College.
 Summer '09
 J
 Equilibrium

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