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Unformatted text preview: Chem 210 / Wang / Chapter 15B
Equilibrium Calculations & LeChatelier's Principle Chapter 15B Chemical Equilibrium
15.1 The Concept of Equilibrium
15.2 The Equilibrium Constant; Kc and Kp
15.3 Interpreting and Working with Equilibrium
Constants
15.4 Heterogeneous Equilibria
15.5 Calculating Equilibrium Constants
15.6 Application of Equilibrium Constants
15.7 LeChatelier’s Principle Equilibrium Calculations Question (a)
K = 55.5 for: H2(g) + I2(g) ⇔ 2 HI(g) at 425°C
(a) If 0.050 mol of H2 and 0.050 mol of I2 are placed in a
1.00 L vessel and heated to 425°C and allowed to reach
equilibrium, what are the equilibrium values in moles
for all the species?
[HI]2
[H2] = nH2÷V= 0.050÷1.00= 0.050 M
Also [I2] = 0.050 M
H2 + I2
2 HI
Initial
0.050 0.050
0
Change  x
x
+2x
Equil. 0.050x 0.050x 2x Equilibrium Calculations Question (a) [HI] = 2x
= 2*0.0394
= 0.080M Final answer:
At equilibrium, the 1 L vessel contains:
0.011
0.011 mole of each H2 and I2
0.080
0.080 mole of HI Equilibrium Calculations Question (b)
Kc = [ HI ] 2
[ I2 ] [ H2 ] H2(g) + I2(g) ⇔ 2 HI(g)
initially 0.040M 0.025M
0
during x
x
+2x
at equil. 0.040x
0.025x
2x
x = 0.0228
or 0.0472
X = 0.0228 M
Final answer:
[H2] = 0.040 – x = 0.017 M
0.017
[I2] = 0.025 – x = 0.002 M
0.002
[HI] = 2x = 0.046 M
0.046 Fall 2009 UMSJTU JI 55.5= [0.050[0.050x]2 Equilibrium Calculations Question (b) [H2] = 0.040 M
H2 +
Initial
0.040
Change  x
Equil. 0.040x K= [I2] = 0.025 M
I2
2 HI
0.025
0
x
+2x
0.025x 2x [2x]2 = 55.5 [0.040x][0.025x] [ I2 ] [ H2 ] √55.5= 55.5 = [0.0638+2x]2 √55.5= solve for x:
7.45 * (0.050x) = 2x
(0.050x = 0.0394 M Solve for x (using
quadratic
equation) then
find M of each
gas. [ 0.0638+2x ]2
[ 0.00212x ]2
7.45 = x=  5.08 x 103 [0.050x]2 7.45 = [2x]
[0.050x] H2(g) + I2(g) ⇔ 2 HI(g)
initially 0.040M 0.025M
0
change x
x
+2x
at equil. 0.040x
0.025x 2x [ HI ] 2 [0.0638+2x]
[0.00212x] [ I2 ] [ H2 ] (4) * x2 (0.0010.065x+x2)
0.0721x2 = 0.001 – 0.065x + x2
0.928x2 – 0.065x +0.001 = 0
x= b±√b24ac
2a [0.00212x]2 [2x]2 Equilibrium Calculations Question (b) 55.5 = Equilibrium Calculations Question (c) Kc = [0.050x]2 Kc = (c) [H2] = [I2] = 2.12×103 M and [HI] = 1.58×102M. If
0.0480 mol of HI are added, calculate new equilibrium
concentrations.
H2(g)
+ I2(g) ⇔
2 HI(g)
Initial
0.00212M 0.00212M 0.0158+0.048
Change
x
x
+2x
Equil.
0.00212x
0.00212x
0.0638+2x [ HI ] 2 [2x]2 55.5= [2x]2 K = 55.5 for: H2(g) + I2(g) ⇔ 2 HI(g) at 425°C
(b) If 0.040 mol of H2 and 0.025 mol of I2 are placed in a
1.00 L vessel and heated to 425°C and allowed to reach
equilibrium, what are the equilibrium values for all the
species? I2
2 HI
0.050
0
x
+2x
0.050x 2x x = 0.0394 M
[H2] = [I2] = 0.050  x
= 0.050 0.0394
0.050= 0.011 M =55.5 Solve for x (take the squareroot on both sides)
(take
squarethen find moles of each gas!
moles K = 55.5 for: H2(g) + I2(g) ⇔ 2 HI(g) at 425°C
H2 +
Initial
0.050
Change  x
Equil. 0.050x [H2][I2] Equilibrium Calculations Question (a)
K = 55.5 for: H2(g) + I2(g) ⇔ 2 HI(g) at 425°C
(a) If 0.050 mol of H2 and 0.050 mol of I2 are placed in a
1.00 L vessel and heated to 425°C and allowed to reach
equilibrium, what are the equilibrium values in moles
for all the species?
H2(g) + I2(g) ⇔ 2 HI(g) x = 0.0228
or 0.0472 Equilibrium Calculations Question (c)
initially
during
at equil. Kc = H2(g)
+ I2(g) ⇔
2 HI(g)
0.00212M 0.00212M 0.0158+0.048
x
x
+2x
0.00212x
0.00212x
0.0638+2x [ HI ] 2
[ I2 ] [ H2 ] 55.5 = [ 0.0638+2x ]2
[ 0.00212x ]2 x=  5.08 x 103 [Negative value of x means the reaction mixture reached
equilibrium with more reactants and proportionally less
more
and
less
product than what was present in the “initial” condition.]
condition.] [H2] = 0.00212x = 7.20 x103M
[I2] = 0.00212x = 7.20 x103M
[HI] = 0.0638+2x = 5.36 x102 M page 1 Chem 210 / Wang / Chapter 15B
Equilibrium Calculations & LeChatelier's Principle Question: Consider the reaction A + B AB.
AB
The vessel on the right contains an equilibrium
mixture of A (red), B (blue), and AB molecules.
(red), (blue),
AB
If the stopcock is opened and the contents of the
two vessels are allowed to mix, predict how
predict
many AB molecules will be at the new
equilibrium?
at
at equilibrium K = [AB]/[A][B]
= [4/V]÷([2/V][2/V])
= V1
assume
assume V = 1 L
K=1
A +B
AB
AB
Initial
2/2
2/2 12/2
Change
+x
+x
x
Equil.
1+x
1+x
6x K = 1 = (6x)/(1+x)(1+x)
(61+2x+x2 = 6x
62 +3x 5 = 0
x
x = (3± 29)/2
(29)/2 Le Chatelier’s principle
Add more NH3 to this equilibrium
N2 + 3 H2
2 NH3 + 92 kJ
Too much NH3
Will react to use it up … speed up the reverse
reaction: “system shifts left”
As excess NH3 is being consumed:
Heat will be absorbed.
N2 will increase; H2 also increases.
Stoichiometry
Stoichiometry: 1 N2 to 3 H2 to 2 NH3 to 92 kJ
H2 increases at 3/2 the rate of decrease of NH3.
3/2
N2 increases at 1/2 the rate of decrease of NH3.
1/2 2 NH3 + 92 kJ (a) Heating the system:
• = raise temperature of system
• system shifts to left to use up extra heat.
• As a result: [N2] will …increase
[H2] will…increase 3X as fast as [N2]
[NH3] will…decrease 2X as fast as [N2] Fall 2009 UMSJTU JI x= b±√b24ac If a system at equilibrium is disturbed, that
equilibrium
system will respond by attempting to
minimize
minimize the effect of the disruption. 2a
2a
You can disturb an equilibrium by adding /
removing chemicals or heat / chill the system.
x = 1.2 6x = 4.8
AB = 4.8 × 2L
= 9.6 or round to 10
9.6 N2 + 3 H2 2 NH3 + 92 kJ Too much N2
Will react to use it up … the → reaction will
speed up more (system shifts right)
As N2 is being consumed:
Heat will be released
NH3 will increase,
at 2 times the rate of N2 decrease.
H2 will also decrease,
3 times faster than N2! What effect will there be if you added more
NH3? How about adding more N2 or heat? Le Chatelier’s principle
Temperature can also have an effect.
For exothermic reactions
reactants products + heat Raising the temperature shifts system to the left.
For endothermic reactions
heat + reactants 2 NH3 + 92 kJ (b) Add Helium gas – inert gas
Not a disturbance to the system. No change.
Presence of He does not change the pressure
of other gases… each gas behaves as if it is
alone in the container.
(c) Add a catalyst – reaction rates increase
Rate increases in both directions at the same
time. System stays in equilibrium. No change. products Raising the temperature shifts system to the right. Le Chatelier’s principle
N2 + 3 H2 2 NH3 + 92 kJ N2 + 3 H2 Le Chatelier’s principle
Add more N2 to this equilibrium Le Chatelier’s principle
questions
N2 + 3 H2 +1.193
+1.193
 4.193 Le Chatelier’s principle Le Chatelier’s principle
(d) Increasing pressure at the same temperature
Pressure affects an equilibrium involving
gaseous reactants or products.
3 H2(g) + N2(g)
2 NH3(g) + 92 kJ
92
4 mol
of gases 2 mol
of gases P↑ causes the system to shift to the side with the
least moles of gas (to reduce total pressure)
The
The above system shifts…to the right! page 2 Chem 210 / Wang / Chapter 15B
Equilibrium Calculations & LeChatelier's Principle Le Chatelier’s principle
N2 + 3 H2 2 NH3 +92 kJ (e) Increase volume –
Partial pressure of each gas decreases.
Each gas will try to “retain” original pressure;
System attempts to “restore” total pressure
Shifts left to make more gas molecules (2 4
molecules).
NOTE:
NOTE: Expansion of gas requires heat.
If
If volume increases suddenly, the system may shift
to
right to produce more heat. (If temperature of the
system is kept constant, the system will shift left.) Haber process:
N2 (g) + 3H2 (g) ↔ 2 NH3 (g) +92 kJ How does each
step maximize
production?
1) Compress
gases.
2) Preheat feed
gases.
3) Cool gases.
4) Recycle
unreacted
gases.
Ref. Page 651
Figure 15.12 This picture represents
two equilibrium
mixtures of solid CaCO3,
CaCO
solid CaO, and gaseous
CaO
CO2, as a result of the
CO
endothermic
endothermic
decomposition of
CaCO3: Predict and explain how temperature may affect this
equilibrium. Endothermic when T increases, forward
when
reaction speeds up per LeChatelier’s principle
More products (CO2 & CaO) formed at equilibrium
CaO)
Higher Temp
larger
larger K for endothermic
reactions!! Fall 2009 UMSJTU JI N2 + 3 H2 2 NH3 +92 kJ N2 is abundant in the
atmosphere, but
useless to plants and
people.
The
The above reaction
turns N2 to an useful
form: NH3.
Liquid
Liquid ammonia
(NH3) can then be
sprayed directly into
soil as fertilizer.
fertilizer.
How to maximize the yield of NH3?
Apply LeChatelier's principle!! N2 + 3 H2 2 NH3 +92 kJ Key stages in the Haber process for synthesizing ammonia.
Haber
for N2 + 3 H2 2 NH3 +92 kJ % yield of ammonia vs. temperature (0C) at five
temperature
at
pressures
different operating pressures. 2CrO42 + 2 H+ Cr2O72 + H2O
Cr Description: Yellow chromate ion is turned orange
orange
by addition of acid. The orange dichromate is
orange
returned to yellow by the addition of base. http://en.wikipedia.org/wiki/Fritz_Haber German chemist of Jewish ethnicity. Fritz Haber
(1868 –
1934) Nobel
Received 1918 Nobel Prize for the Haber process.
Credited as the "father of chemical warfare" for developing
/ deploying Cl2(g) and other poisonous gases during World
War I. Personally oversaw the first successful use of Cl2(g) at
the 2nd Battle of Ypres on April 22, 1915. First wife, Clara
Immerwahr (also a PhD chemist), opposed his work on
poison gas and committed suicide with his service weapon in
their garden.
He is also credited with proposing the BornHaber cycle as a
Bornmethod for evaluating the lattice energy of an ionic solid. Sample Problem 1 For the reaction: X(g) + Y2(g) + heat
XY(g) + Y(g)
the following molecular scenes depict different reaction
mixtures. (X = green, Y = purple) Yellow chromate ion + acid
orange
orange dichromate
2CrO42 + 2 H+
Cr2O72 + H2O
Cr Sample Problem 2 Effect of Temperature on the Equilibrium Position How does an increase in temperature affect Kc for the following
reactions?
(a) CaO(s) + H2O(l) (b) CaCO3(s) + 178kJ
(a) If Kc = 2 at the temperature of the reaction, which scene
represents the mixture at equilibrium?
(b) Will the reaction mixtures in the other two scenes proceed
toward reactant or toward products to reach equilibrium?
(c) For the mixture at equilibrium, how will a rise in
temperature affect [Y2]? (c) SO2(g) + 297kJ Ca(OH)2(aq + 82kJ CaO(s) + CO2(g) S(s) + O2(g) page 3 Chem 210 / Wang / Chapter 15B
Equilibrium Calculations & LeChatelier's Principle Question: The following pictures represent the
composition of the equilibrium mixture for the
AB at 300 K and at 400 K:
reaction A + B Is Co(H2 O)62+
O) CoCl endoCoCl42 endo or exothermic? Co(H2 O)62+ + heat ↔ CoCl42O) Is the reaction exothermic or endothermic?
Explain using Le Châtelier’s principle. 2NO2 (g) N2O4 (g) Room temp
Both present Heating up
Mostly CoCl42 2NO2 (g) N2O4 (g)
(g) Is the above reaction endo or exothermic? Low temp medium temp Cooling with ice
Mostly Co(H2O)6+2 Question: The following pictures represent equilibrium
mixtures at 325 K and 350 K for a reaction involving A atoms
atoms
atoms
(red), B atoms (blue), and AB molecules: Can you explain how heat is
involved in this reaction by
considering molecular
structures? 1. Write a balanced equation for the reaction.
2. Is the reaction exothermic or endothermic? Predict using
Le Châtelier’s principle.
3. If the volume of the container is increased, will the number
of A atoms increase, decrease, or remain the same?
atoms
Explain. Fall 2009 UMSJTU JI page 4 high temp ...
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