**Unformatted text preview: **Module 35 — Solubility Equilibrium Calculations In Chemistry
Modules 19 and above have been re-numbered.
The former Module 35 on Electrochemistry is now Module 37
The former Module 36 on Batteries is now Module 38
If you are looking for those topics, check those modules
At www.ChemReview.Net *****
Module 35: Solubility Equilibrium
And
Module 36: Thermodynamics
Module 35 – Solubility Equilibrium ........................................................................ 1091
Lesson 35A:
Lesson 35B:
Lesson 35C:
Lesson 35D:
Lesson 35E: Slightly Soluble Ionic Compounds ............................................................... 1091
Ksp Calculations............................................................................................... 1094
Solubility and Common Ions ......................................................................... 1102
pH and Solubility ............................................................................................ 1109
Quantitative Precipitation Prediction........................................................... 1113 Module 36 – Thermodynamics .................................................................................. 1125
Lesson 36A:
Lesson 36B:
Lesson 36C:
Lesson 36D:
Lesson 36E:
Lesson 36F:
Lesson 36G: Review: Energy and Heats of Reaction ....................................................... 1125
Entropy and Spontaneity .............................................................................. 1130
Free Energy ...................................................................................................... 1138
Standard State Values ..................................................................................... 1144
Adding ΔG° Equations ................................................................................... 1149
Free Energy at Non-Standard Conditions ................................................... 1152
Free Energy and K ........................................................................................... 1157
For additional modules, visit www.ChemReview.Net © 2011 ChemReview.Net v1r Page i Module 35 — Solubility Equilibrium Module 35 — Solubility Equilibrium
Prerequisites: Before this module, complete Lessons 7B and 7C on ionic compounds,
Lesson 12B on the REC steps, and Lessons 28C, D, and F on equilibrium calculations.
***** Lesson 35A: Slightly Soluble Ionic Compounds
Pretest: If you think you know this topic, try the last two problems on both problem sets.
If you can do those problems, skip the lesson.
**** Slightly Soluble Salts
When any ionic compound (salt) is mixed with water, it dissolves to at least a slight extent.
In Module 13, our solubility scheme divided ionic compounds into soluble and insoluble,
based on whether 0.10 moles of a compound dissolved per liter of aqueous solution at 25ºC.
For many problems, this general division is all we need to make useful predictions about
chemical behavior.
However, the ionic compounds labeled insoluble in solubility schemes are more accurately
described as slightly soluble. The extent of their solubility, even if it is quite small, can be
important in many chemical processes.
When a salt with low solubility is mixed with water, some formula units leave the surface
and form separated ions that move about freely in the solution. As these ions accumulate,
the reverse reaction occurs as well: dissolved ions return to the solid, combining to form
neutral formula units on the surface of the ionic crystal.
Over time, the rate at which the ions return to the solid becomes equal to the rate at which
they dissolve. Though both reactions continue, there is no net change. As long as some
solid is present and the temperature remains constant, the mixture will remain in this
equilibrium condition.
At equilibrium, the solution above the solid is said to be saturated: the dissolved ion
concentrations are theoretically as high as they can be in an aqueous solution at that
temperature (exceptions occur that we will discuss later).
An example of this equilibrium is the slight solubility of lead chloride. In solubility
schemes, the combination of Pb2+ and Cl─ ions is predicted to be insoluble, but the solid
lead chloride does dissolve slightly in water. This reaction can be represented as
PbCl2(solid) Pb2+(aqueous) + 2 Cl─(aq) (goes slightly) With mixing, the reaction reaches equilibrium. Thereafter, as long as the temperature
remains constant, no further net change in the solid-solution mixture takes place.
For reactions that go to equilibrium, an equilibrium constant expression can be written.
Write the K expression for the reaction above, then check your answer below.
***** © 2011 ChemReview.Net v1r Page 1091 Module 35 — Solubility Equilibrium K= [Pb2+] [Cl─]2 = [Pb2+] [Cl─]2 = Ksp
1 As with all equilibrium constants,
• the general form is “product of the concentrations of the products over product of
the concentrations of the reactants.” • The lowest-whole-number coefficients of the balanced equation become exponents in
the K expression. • The concentration of a solid is assigned a value of 1 in K expressions because solids
have a constant concentration. The concentrations of the ions are included in the K
expression because ion concentrations in a solution can vary. Terms with values
that can vary are included in K expressions, but terms that are constant are omitted. • All Ksp expressions have a 1 in the denominator. The 1 is omitted as understood
when writing the Ksp expression. For the special case of a slightly soluble ionic solid dissolving in water, the following
rules are used.
Ksp Rules
If a problem lists a Ksp, or if the reaction is a salt dissolving slightly in water,
1. The balanced equation is always written with one particle of the ionic solid on the
left and its ions on the right.
2. The K expression for the reaction is termed the solubility product, symbol Ksp.
3. A Ksp expression has ion concentrations to powers multiplied in the numerator, but
no term (a 1) in the denominator.
Applying those rules, solve this problem.
Q. The Ksp value for Co(OH)2 is 2.5 x 10─16. Write the balanced equation for the
reaction that this is a K value for, and then the Ksp expression.
*****
Answer
This reaction is 1 Co(OH)2(s) 1 Co2+ + 2 OH─ (goes slightly) • In Ksp problems, for the separated ions in both the reaction equation and the Ksp
expression, the aqueous (aq) state of the ions can be omitted as understood. • The low K value means that the reaction goes only slightly: equilibrium
strongly favors the reactants. • The Ksp expression is: © 2011 ChemReview.Net v1r Ksp = [Co2+][OH─]2 Page 1092 Module 35 — Solubility Equilibrium Practice A: Do every other problem now and the rest during your next study session.
Check answers at the end of the lesson. 1. Write the Ksp expression for these reactions.
Hg2+ + S2─ a. HgS(s) 2 Al3+ + 3 CO32─ b. Al2(CO3)3(s) 2. Write the reaction and Ksp expression for these salts dissolving slightly in water.
a. AgI b. Sr(OH)2 c. Calcium phosphate d. Copper (II) sulfide 3. Label these compounds as soluble or slightly soluble.
a. AgNO3 b. Fe(OH)3 c. Potassium phosphate d. Silver chloride Ksp Math
Solving calculations that include Ksp values will require taking powers and roots of
exponential notation. This math was covered in Lesson 28B. The following problems
apply those rules to Ksp -type calculations. If you need a rule review, see Lesson 28B. Practice B: Solve in your problem notebook. Convert final answers to scientific
notation. Try every other problem, and more if you need more practice.
Do not use a calculator on these first four. 1. (104)─3 = 3. (27 x 10─6)1/3 = 2. (2.0 x 10─5)3 =
4. (16 x 10─12)1/4 = On 5-8,
• first write an estimated answer and convert the estimate to scientific notation; • then use a calculator as needed get a final answer in scientific notation. 5. The square root of 2.25 x 10―10 = 6. The cube root of 6.4 x 10―8 = 7. (5.1 x 103)2 = 8. (1.6 x 10─15)1/4 = Solve the following using a calculator as needed.
9. (exact 3)(9.6 x 10―11)1/2 = 10. (exact 9)(8.1 x 10―11)1/4 = 11. (exact 4 )(7.5 x 10―4)3 =
Solve problems 12-16 for x .
12. 4x3 = 5.0 x 10―16 13. (108)x5 = 3.46 x 10―17 14. (2x)2 (0.025) = 4.9 x 10―14 15. (x) (0.20)2 = 4.8 x 10―19 16. (3x)3(2x)2 = 2.62 x 10―36 © 2011 ChemReview.Net v1r Page 1093 Module 35 — Solubility Equilibrium ANSWERS
Practice A
1a. Ksp = [Hg2+][S2─] 1b. Ksp = [Al3+]2 [CO32─]3 Ag+(aq) + I─(aq) 2a. AgI(s) Sr2+(aq) + 2 OH─(aq) b. Sr(OH)2(s) Ksp = [Sr2+] [OH─]2 3 Ca2+(aq) + 2 PO43─(aq) c. Ca3(PO4)2(s)
d. Ksp = [Ag+] [I─] Cu2+(aq) + S2─(aq) CuS(s) Ksp = [Ca2+]3[PO43─]2 Ksp = [Cu2+][S2─] 3. Soluble are a. AgNO3 and c. Potassium phosphate; Slightly soluble are b. Fe(OH)3 and d. Silver
chloride. See the solubility scheme in Lesson 13A.
Practice B
1. 10―12 3. 3.0 x 10―2 2. 8.0 x 10―15 6. = (16 x 10―16)1/4 = 4.0 x 10―3
10. 2.7 x 10―2 4. 2.0 x 10―3 5. 1.50 x 10―5 8. 2.0 x 10―4 7. 2.6 x 107 9. 2.9 x 10―5 11. 4 (7.5 x 10―4)3 = 4 (422 x 10―12) = 1690 x 10―12 = 1.7 x 10―9 12. 4x3 = 5.0 x 10―16 ; x3 = 1.25 x 10―16 = 125 x 10―18 ; x = 5.0 x 10―6 13. (108)x5 = 3.46 x 10―17 ; x5 = 3.46/108 x 10―17 = 32.2 x 10―20 ; x = 2.00 x 10―4
14. (2x)2 (0.025) = 4.9 x 10―14 ; 4x2 = 198 x 10―14 ; x2 = 49 x 10―14 ; x = 7.0 x 10―7
15. (x) (0.20)2 = 4.8 x 10―19 ; x(0.040) = 4.8 x 10―19 ; x = 120 x 10―19 = 1.2 x 10―17
16. (27x3)(4x2) = (108)x5 = 2.62 x 10―36; x5 = 0.0243 x 10―36 = 243 x 10―40 ; x = 3.0 x 10―8
***** Lesson 35B: Ksp Calculations
Pretest: If you think you know this topic, try the last two problems at the end of the
lesson. If you can do those problems, skip the lesson.
**** Ksp Values
For ionic solids that dissolve only slightly in water, the concentration of the ions in the
solution is low at equilibrium. Ksp values will therefore be less than one: positive
numbers with negative exponents when written in scientific notation.
For example, in the reaction of slightly soluble silver chloride dissolving in water,
AgCl(s) Ag+(aq) + Cl─(aq) Ksp = [Ag+] [Cl─] = 1.6 x 10―10 © 2011 ChemReview.Net v1r (goes slightly) at 25ºC Page 1094 Module 35 — Solubility Equilibrium Ksp values for a sample of slightly soluble ionic compounds are listed in the table below.
As with all equilibrium constants,
•
•
• Ksp values have units, but the units are not included
with Ksp values or used in Ksp calculations.
If a concentration is calculated based on a Ksp, the
unit mol/L must be added to the concentration.
K values are difficult to measure precisely, and
values may vary among textbooks. To match
textbook answers, use the Ksp values in that text. Solid Ksp at 25ºC AgBr 5.0 x 10―13 AgI 1.5 x 10―16 Fe(OH)2 1.8 x 10―15 Ba3(PO4)2 5.0 x 10―23 Because a Ksp value varies with temperature, it must have a temperature attached, but by
convention in Ksp problems, temperature is assumed to be 25ºC unless otherwise noted. Ksp WRECK Steps
To solve Ksp calculations, use the fundamental rule for reactions that go to equilibrium
rather than completion: Write the WRECK steps.
For the following example, write the WRE part of the WRECK steps, and then check your
answer below.
Q. When solid silver sulfate is mixed with water, what is the [Ag+] in the solution at
equilibrium? (Ksp Ag SO = 1.2 x 10―5)
2 *****
WANTED: 4 [Ag+] Rxn. & Extent: 1 Ag2 SO4(s) 2 Ag+ + 1 SO42─ (goes slightly) Next, to find the [email protected], we can either use a rice table or write the
bottom row of the rice table by inspection. Let’s try the rice table first.
In Ka calculations, for a weak acid ionizing slightly in water, x was defined as the small
amount of reactant that reacts. For an ionic solid ionizing slightly in water, we define x in
the same way. Since x is reactant used up, the reactant’s x has a negative sign in the
Change row. We also assume that initially the reaction has not started, so there are no
products.
Based on the reaction and rules above, fill in the empty boxes of the following table, then
check your answer below.
Reaction
Initial
Change (+,―) mol/L solid 0M ─xM At Equilibrium
***** © 2011 ChemReview.Net v1r Page 1095 Module 35 — Solubility Equilibrium Reaction 1 Ag2SO4 (s) 2 Ag+ 1 SO42─ Initial mol/L solid 0M 0M Change (+,―) ─xM + 2x M +xM At Equilibrium mol/L solid ─ x + 2x M +xM This table follows the standard rice rules from Lesson 28F for reactions that go to
equilibrium.
• Because coefficients show the ratios in which reactants are used up and products
form, the coefficients in Reaction row 1 must match the numbers in front of the x
terms in rows 3 and 4. • In the Change row, each reactant term must have the same sign, and each product
term must have the opposite sign. • The units in the table must be consistent: moles or mol/L. Mol/L can be used as a
unit if all particles are in the same solution, as they are when an ionic solid dissolves
slightly to form an aqueous solution. Once the bottom row of the rice table is written in terms of x, complete the K step of the
WRECK steps:
• Write the K equation: the expression (with [ ] symbols) and K value (if known). • Write the exact K equation: substitute the x terms from the bottom rice row into the
K expression. • Solve the exact K equation to find the unknown value: K or x . • If needed, use x to solve for the WANTED symbol. Try those steps based on the rice table above, then check your answer below.
*****
The K expression and value:
The exact K equation: Ksp = [Ag+]2 [SO 2─] = 1.2 x 10―5
4
Ksp = ( 2x )2 ( x ) = 1.2 x 10―5 Solve for x: ( 4x2 ) ( x ) = 1.2 x 10―5
4x3 = 1.2 x 10―5 x3
x
Solve for the WANTED symbol: = 0.30 x 10―5 = 3.0 x 10―6
= 1.44 x 10―2 [Ag+] = 2x = 2.9 x 10―2 M Significant figures: Since coefficients are exact, for terms such as ( 2x )2 above in which the
numbers are based on coefficients, those numbers are exact. Numbers based on coefficients
do not limit the significant figures in an answer.
***** © 2011 ChemReview.Net v1r Page 1096 Module 35 — Solubility Equilibrium The Solubility: Small x
For the reaction of a slightly soluble ionic solid dissolving in water, x is defined as the
solubility: the small number of moles per liter that leave the solid.
For a slightly soluble ionic solid dissolving in water: the solid’s solubility = small x
Since solubility is the moles of solid that dissolve per liter, the unit of solubility is mol/L,
and the unit mol/L (or its abbreviation M) must be added when solubilities are calculated
using K equations.
In rice tables, the key definition is x. The rule is
In Ksp calculations, when writing rice tables or WRECK steps,
• Let ─ x represent the small mol/L of the reactant used up in the reaction.
• At equilibrium, let [solid]eq. = mol/L solid ─ x
• Define the product concentrations using positive x terms.
The term mol/L solid can mean either the moles of solid added per liter of solution, or it can
mean the density of the solid, measured in mol/L. The values will differ, and which value
applies depends on the question being asked. However, in Ksp calculations, the value does
not matter, because the value for mol/L solid is not included in Ksp equations.
Note that the solubility is not the same as the solubility product.
Solubility of an ionic solid = small x in mol/L units.
Solubility product = Ksp = [ions] multiplied, with no denominator and no units. Writing [Ions]eq. By Inspection
Compared to general K calculations, Ksp calculations are simplified because an ionic solid
formula is always on the reactant side of the balanced equation, and its coefficient is always
one. On the right side are always its separated ions.
Because reactions with a Ksp have this simple and consistent form, at the C step of the
WRECK steps, we can write the bottom rice row by inspection, rather than by writing a
complete rice table. This is the same method that we used in Ka and Kb problems.
To simplify Ksp calculations: write the first term in the C row of the WRECK steps, which is
the same as the first term in the bottom row of the rice table, as mol/L solid ─ x (see the
rice table above).
Using that rule, complete the REC steps for this ionic solid dissolving slightly in water.
Rxn. & Extent: 1 Fe(OH)3(s)
^ [email protected]: ▐ *****
© 2011 ChemReview.Net v1r Page 1097 Module 35 — Solubility Equilibrium Rxn. & Extent:
[email protected]: ^
mol/L solid ─ x 1 Fe 3 + 1 Fe(OH)3(s)
▐ ^
+x + 3 OH─
│ (goes slightly) ^
+ 3x Try one more.
Q. Write the REC steps for the slight ionization of silver phosphate.
*****
Tip: To determine the solid formula when given a name, it helps to write and balance the
separated ions on the products side first.
*****
Rxn. & Extent: 3 Ag+ 1 + 1 PO43─ (goes slightly) Complete the REC steps by inspection.
*****
Rxn. & Extent:
[email protected]: + 1 PO43─ 3 A g+ 1 Ag3PO4(s)
^
mol/L solid ─ x ^
3x (goes slightly) ^
x As was done above based on the rice table, a value of x can be solved by substituting the x
terms into the Ksp equation.
Try this example. Write the complete WRECK steps. Write the C-step terms by inspection
as done in the two REC step problems above. Find the value for x .
Q. What is the solubility of strontium chromate (Ksp = 3.6 x 10―5).
*****
WANTED: ? = Solubility = small x in mol/L
*****
Rxn. & Extent: Sr2+ + _________________ CrO42─ (goes slightly) *****
Rxn. & Extent:
[email protected]: ^
mol/L solid ─ x 1 Sr2+ + 1 CrO42─ 1 SrCrO4(s)
▐ ^
x │ (goes slightly) ^
x If needed, finish the calculation.
*****
At the K step:
• Write the K equation: the K expression and K value. • Write the exact K equation, substituting the x terms above into the K expression. • Solve the exact K equation to find the unknown value. • Solve for the symbol WANTED.
***** © 2011 ChemReview.Net v1r Page 1098 Module 35 — Solubility Equilibrium Ksp = [Sr2+] [CrO 2─] = (x)(x) = x2 = 3.6 x 10―5 = 36 x 10―6
4
WANTED = x = 6.0 x 10―3 M = solubility = mol/L of SrCrO4 that dissolves Summary
Nearly all K calculations are solved the same way.
1. Write the WRECK steps.
2. Find the Concentrations at equilibrium, usually in terms of x, either using a rice table
or by inspection.
3. Solve the K equation, then solve for the WANTED symbol.
Ksp is the special K for a slightly soluble salt dissolving in water.
1. In Ksp WRECK steps, write a balanced equation with one particle of the ionic solid
on the left, its aqueous separated ions on the right, and (goes slightly).
2. Ksp = solubility product. A Ksp expression is written with [ions] multiplied in the
numerator, but no term in the denominator.
3. For slightly soluble salts, x is the solubility: the small moles per liter that dissolve
from the solid. At equilibrium, [solid]eq. = mol/L solid ─ x
4. Solubility of an ionic solid = small x . Add units of moles/liter.
Solubility product = Ksp = [ions] multiplied, with no denominator and no units. Practice: Do every other problem today, then the rest over the next few days. 1. For each of these slightly soluble salts dissolving in water, write the Reaction, its Extent,
the Concentrations at equilibrium in terms of x, the Ksp expression in symbols, and the
exact Ksp equation in terms including x . Then combine and simplify the x terms in the
exact equation.
a. CaF2 b. BaSO4 c. Pb3(PO4)2 2. If the solubility of PbI2 is 1.5 x 10―3 at 25ºC, what is its Ksp?
3. Find [Ca2+] in a mixture of water and Ca3(PO4)2 (Ksp = 1.3 x 10―32) at equilibrium.
4. The Ksp of MgCO3 is 6.8 x 10―6.
a. Find the solubility of MgCO3 . b. Convert the solubility to nanomoles/mL . 5. Which has higher solubility: AgCl (Ksp = 1.6 x 10―10) or Ag2CO3 (Ksp = 8.1 x 10―12)? © 2011 ChemReview.Net v1r Page 1099 Module 35 — Solubility Equilibrium ANSWERS
Practice A
1a. Rxn. & Extent: 1 CaF2(s)
^
mol/L solid ─ x [email protected]: ▐ 1 Ca2+ + 2 F─
^
^
x
│ 2x (goes slightly) Ksp = [Ca2+] [F─]2 = (x)(2x)2 = 4x3
1b. Rxn. & Extent: 1 BaSO4(s)
^
mol/L solid ─ x [email protected]: ▐ 1 Ba2+ + 1 SO42─
^
^
x
│
x (goes slightly) Ksp = [Ba2+] [SO 2─] = x2
4
1c. Rxn. & Extent:
[email protected]: 1 Pb3(PO4)2(s)
^
mol/L solid ─ x ▐ 3 Pb2+ + 2 PO43─
^
^
3x
│
2x (goes slightly) Ksp = [Pb2+]3[PO 3─]2 = (3x)3(2x)2 = (27x3)(4x2) = (108)x5
4
2. In calculations for reactions that go to equilibrium, write the WRECK steps.
WANTED: Ksp
Rxn. & Extent:
[email protected]: 1 PbI2(s)
^
mol/L solid ─ x 1 Pb2+ +
^
x 2 I─
^
2x (goes slightly) Ksp = [Pb2+] [I─]2 = (x)(2x)2 = 4x3
DATA: 1.5 x 10―3 M = solubility = x
Ksp = 4x3 = 4 (1.5 x 10―3)3 = 4 (3.38 x 10―9) = 14 x 10―9 = 1.4 x 10―8 = Ksp
3. In calculations for reactions that go to equilibrium, write the WRECK steps.
WANTED: [Ca2+]
Rxn. & Extent:
[email protected]: 1 Ca3(PO4)2(s)
^
mol/L solid ─ x 3 Ca2+ +
^
3x 2 PO43─
^
2x (goes slightly) Ksp = [Ca2+]3[PO 3─]2 = (3x)3(2x)2 = (27x3)(4x2) = (108)x5
4
(108)x5 = 1.3 x 10―32 ;
x = 1.64 x 10―7 ; x5 = 0.0120 x 10―32 ; x5 = 12 x 10―35 [Ca2+] = 3x = 4.9 x 10―7 M = [Ca2+] © 2011 ChemReview.Net v1r Page 1100 Module 35 — Solubility Equilibrium 4a. In calculations for reactions that go to equilibrium, write the WRECK steps.
WANTED: solubility = x
Rxn. & Extent:
[email protected]: in mol/L 1 MgCO3(s)
^
mol/L solid ─ x 1 Mg2+ + 1 CO32─
^
^
x
│
x ▐ (goes slightly) Ksp = [Mg2+] [CO 2─] = (x)(x) = x2 = 6.8 x 10―6
3
x = 2.6 x 10―3 M = solubility
The units of solubility, moles that dissolve per liter, must be added to the answer.
4b. ? nanomoles = 2.6 x 10―3 mol •
mL
L 10―3 L
1 mL • 1 nanomole =
10―9 mol 2.6 x 103 nmol
mL 5. AgCl has the higher solubility product ( Ksp ), but that may not mean it has the higher solubility (x).
Solve for the solubility of each compound.
a. WANTED: solubility AgCl = x
Rxn. & Extent:
[email protected]: 1 AgCl (s)
^
mol/L solid ─ x 1 Ag+ + 1 Cl─
^
^
x
│
x ▐ (goes slightly) Ksp = [Ag+] [Cl─] = (x)(x) = x2 = 1.6 x 10―10
x = 1.3 x 10―5 M = solubility AgCl
b. WANTED: solubility Ag2CO3 = x
Rxn. & Extent:
[email protected]:
K equation:
Exact K in terms of x :
Solving for x: 1 Ag2CO3(s)
^
mol/L solid ─ x ▐ 2 Ag+ + 1 CO32─
^
^
2x │
x (goes slightly) Ksp = [Ag+]2 [CO 2─] = 8.1 x 10―12
3
Ksp = ( 2x )2 ( x ) = 8.1 x 10―12
( 4x2 ) ( x ) = 8.1 x 10―12
4x3 = 8.1 x 10―12
x3 = 2.02 x 10―12
x = 1.3 x 10―4 M = solubility Ag2CO3 The Ag2CO3, though it has a lower solubility product than AgCl, has the higher solubility.
***** © 2011 ChemReview.Net v1r Page 1101 Module 35 — Solubility Equilibrium Lesson 35C: Common Ions and Solubility
Prerequisites: If needed, review Lesson 28A on Le Châtelier’s Principle.
***** Adding Solid
When a mixture of a solid salt and its dissolved ions is at equilibrium, the solution is
saturated: the ion concentration is as high as theoretically possible. What happens to the
[ions] in the solution if more of the same ionic solid is added to this solution?
The amount of the solid on the bottom of the solution increases, but the [ions] in the
solution is already at its maximum when the solution is at equilibrium, so the [ions] does
not shift. Changing the amount of the solid does not change the concentration of the solid
(determined by its density), and does not shift the equilibrium. Adding A Common Ion
In the laboratory, for mixtures of water and slightly soluble solids, we often want to cause
ions that are dissolved in the solution to return to the solid. One way to cause a dissolved
ion of interest to return to its solid is to add a soluble salt that contains the other ion in the
ionic compound. Since that other ion that is the same in both the slightly soluble and the
added soluble salt, it is a common ion.
For example, in a mixture of solid silver bromide (AgBr) and water at equilibrium,
some silver ion is dissolved in the solution above the solid. The equation for the
solubility equilibrium is
1 AgBr(s) 1 Ag+ + 1 Br─ (goes slightly: Ksp = 5.0 x 10―13) If solid sodium bromide (NaBr) is added to the mixture of AgBr and water, it will
dissolve ~100% and separate ~100% into Na+ and Br─ ions.
1 NaBr(s) 1 Na+ + 1 Br─ (goes ~100%) For AgBr and NaBr, Br─ is a common ion. Adding a substantial amount of NaBr
substantially increases the [Br─] in the solution. Using Le Châtelier’s Principle
(Lesson 28A), write answers to the following questions.
1. As Br─ ions from NaBr are added to the solution, in which direction will the
AgBr solubility equilibrium shift: left or right?
2. Will the [Ag+] ions dissolved in the solution increase or decrease?
3. How will this shift change the amount of AgBr solid in the bottom of the
beaker?
4. How will this shift change the concentration of AgBr solid?
***** © 2011 ChemReview.Net v1r Page 1102 Module 35 — Solubility Equilibrium Answers
According to Le Châtelier’s Principle, if the concentration of a component that appears on
one side of an equilibrium equation is increased, the equilibrium shifts toward the opposite
side of the equation.
In this case, adding Br─ ions shifts the AgBr equilibrium to the left, decreasing the [Ag+] in
solution. For the [Ag+] to go down, silver ions must combine with Br─ ions and return to
the surface of the solid, so the amount of solid AgBr goes up. However, the concentration of
a solid is constant and is not changed by shifts in equilibrium.
In a solubility equilibrium, if a common ion is added:
the common ion shifts the non-common ion of the slightly soluble salt out of the solution
and into the solid. Practice A
1. Which of these compounds would you add to a mixture of solid AgI and water at
equilibrium to reduce the [Ag+] in the solution?
a. AgBr b. KI c. AgI d. AgNO3 e. NaI 2. Which of these compounds, when added to a solution of solid AgI at equilibrium with
its dissolved ions, will reduce the [I─] in the solution?
a. AgBr b. KI c. AgI d. AgNO3 e. NaI Common Ion Ksp Calculations
To take advantage of the common ion effect to shift a solubility equilibrium, the [common
ion] must be relatively high. A soluble salt that contains the common ion is required to
create this high [common ion].
When a common ion is added to an equilibrium solution of a slightly soluble solid and its
ions,
• Le Châtelier’s Principle predicts the direction that the equilibrium will shift. • Ksp values can calculate how much the equilibrium will shift. To solve Ksp calculations with common ion added, apply these rules:
• If you see a Ksp value for a substance, write the WRECK steps for its slight
ionization. • If a problem involves [ions] of two compounds, write the REC steps for both. Common ion Ksp problems have two compounds, one with a K that calls for WRECK
steps, the other a soluble ionic compound that calls for REC steps. Begin by writing the
WANTED unit, and then the REC steps for each compound. © 2011 ChemReview.Net v1r Page 1103 Module 35 — Solubility Equilibrium Fill in the blanks in the following table of the reactions that occur in this mixture with a
common ion, then check your answers below.
Q. To a mixture of Ba3(PO4)2 (Ksp = 5.0 x 10―23) and water at equilibrium, 0.25 mol
per liter of solid K3PO4 is added. Once the new equilibrium is established, what
will be the solubility of the Ba3(PO4)2?
WANTED: _____________
Rxn. & Extent: 3 Ba2+ + 2 PO43─ 1 Ba3(PO4)2(s)
^
mol/L solid ─ x [email protected]: ▐ 3 K+ 1 K3PO4(s) Rxn. & Extent: ^
0.25 M [email protected]:
Ksp = [ ______ ] __ 0M [ ______ ] __ ^
____ ▐ │ ^
_____ + (goes ____________) 1 PO43─ ^
______ │ (goes _________) ^
______ = 5.0 x 10―23 [Ba2+] = ______ M
[PO43─] = ____________________ M (exact) ≈ ___________ M (approximate)
*****
Answer
Since barium phosphate has a small Ksp value, it is slightly soluble. From our solubility
scheme (Lesson 13A), we know that potassium compounds are water soluble. We also
know that compounds with metal atoms, including potassium, in an aqueous solution
will separate ~100% into ions. If needed, adjust your work and complete the table.
*****
WANTED: solubility Ba3(PO4)2 = x in mol/L
Rxn. & Extent: 3 Ba2+ + 1 Ba3(PO4)2(s)
^
mol/L solid ─ x [email protected]: ▐ 3 K+ 1 K3PO4(s) Rxn. & Extent: ^
0.25 M 0 M [email protected]: ^
3x ▐ │
+ ^
0.75 M │ 2 PO43─ (goes slightly) ^
2x
1 P O4 3 ─ (goes ~100%) ^
0.25 M Ksp = [Ba2+]3[PO43─]2 = 5.0 x 10―23
[Ba2+] =
[PO43─] = 3x
0.25 M + 2x (exact) ≈ 0.25 M (approximate) For problems with two or more reactions and/or lots of numbers, writing a DATA table
below the equation, with each equation symbol and its DATA, is a good idea. © 2011 ChemReview.Net v1r Page 1104 Module 35 — Solubility Equilibrium The common ion is supplied to the solution by both ionic compounds, but the largest
contribution is due to the soluble salt.
For the [common ion], use the approximation rule: a small number added or subtracted
from a large number does not substantially change the large number (see Lesson 30B).
This rule applies to a small x and 2x and 3x.
The steps for solving K equations are
• Solve the K equation using the approximation.
• Apply the 5% test to see if the approximation can be used.
• Then solve for the WANTED symbol.
Find the Ba3(PO4)2 solubility in this solution that has common ion added.
*****
Ksp = [Ba2+]3 [PO43─]2 = 5.0 x 10―23
Ksp = (3x)3 (0.25 M + 2x) ≈ (3x)3 (0.25) = 5.0 x 10―23
^ Exact ^Approximate We could solve the exact equation for x using the quadratic formula, but the approximation
solves more quickly. Solve for x using the approximation first.
*****
(3x)3 (0.25) = 5.0 x 10―23
27x3 = 20. x 10―23
x3 = 0.741 x 10―23 = 7.41 x 10―24
x= 1.9 x 10―8 M = solubility Ba3(PO4)2 When solving Kc, Ka, Kb, or Ksp calculations for concentration or solubility, always add
mol/L (M) as the unit of the answer.
The value for solubility above is based on an approximation. K values have relatively high
uncertainty, and as with Ka calculations, a 5% error in calculations is usually within the
range of experimental error.
In a common ion calculation, error is introduced by ignoring the x term in the exact
[common ion] at equilibrium, and this error can be measured by
% Ksp Common Ion Error = 5% test ≈ x or 2x or 3x ● 100%
[common ion]mixed By using this equation, we can also use a quick test as the 5% test.
If the difference between the exponents of the [common ion] and the x term in the exact
[common ion], with both written in scientific notation, is 3 or greater, the 5% test is
passed, and the approximation may be used. © 2011 ChemReview.Net v1r Page 1105 Module 35 — Solubility Equilibrium In this problem,
• [common ion]exact = 0.25 + 2x • [common ion] = 0.25 M = 2.5 x 10―1 M and 2x = 3.8 x 10―9 M so the exponent difference is 8, the ionization is much smaller than 5%, and the
approximation is acceptable. Summary: Adding a Common Ion to A Solubility Equilibrium
Rule: In a mixture of a slightly soluble ionic solid and its dissolved ions, to reduce the
solution concentration of one ion, add a soluble salt that contains the other ion in
the solid. To solve a Ksp calculation with common-ion added,
• As in all K calculations, write the WRECK steps. • Write REC steps for the ionization of both the slightly soluble and highly soluble
salt. • Write a DATA TABLE under the Ksp equation. Include the exact and approximate
[common ion]. • Solve the Ksp equation first using the approximate [common ion] first, then do the
5% test. Practice B: Do half now. Save a few for your next two study sessions. 1. Change the equation to an approximation, then solve for x .
a. (x) (0.50 + 2x) = 2.0 x 10―12 b. (2x)2 (0.25 + 3x) = 6.4 x 10―7 2. Into a 0.025 M Na2CrO4 solution, solid Ag2CrO4 (Ksp = 9.0 x 10―12) is added. At
equilibrium after mixing, some solid remains. Find the [Ag+] in the solution.
3. Solid AgBr (Ksp = 5.0 x 10―13) is added to a 0.30 M AgNO3 solution. After equilibrium
is reached between the solid and its ions, what is the [Br─] in the solution?
4. What is the solubility of PbI2 (Ksp = 1.4 x 10―8) in a 0.20 M KI solution? ANSWERS
Practice A
1. In the solution above a slightly soluble ionic solid, to reduce the equilibrium concentration of one ion, add a
soluble salt that contains the other ion. Above solid AgI, to reduce [Ag+], increase the [I─] substantially.
The two compounds that ionize ~100% to form I─ are (b) KI and (e) NaI . © 2011 ChemReview.Net v1r Page 1106 Module 35 — Solubility Equilibrium 2. Above solid AgI, to reduce [I─], increase the [Ag+] substantially. The one compound that ionizes ~100%
to form Ag+ is (d) AgNO3 .
Practice B
1a. (x) (0.50 + 2x) = 2.0 x 10―12
To approximate, x or 2x or 3x can be dropped if x is added or
subtracted from a much larger number.
(x) (0.50) = 2.0 x 10―12 ; x = 2.0/0.50 x 10―12 = 4.0 x 10―12 = x 1b. (2x)2 (0.25 + 3x) = 6.4 x 10―7
( 4x2 )( 0.25 ) = 6.4 x 10―7 ; (x2 )( 4 x 0.25 ) = 6.4 x 10―7
x2 = 6.4 x 10―7 = 64 x 10―8 ; x = 8.0 x 10―4 2. All Na compounds ionize ~100% in water. The small Ksp means that Ag2CrO4 is only slightly soluble.
Solve K calculations using the WRECK steps. If two compounds are involved, write REC steps for both.
Wanted: [Ag+] [email protected]: 1 Ag2CrO4(s)
^
mol/L solid ─ x Rxn. & Extent: 1 Na2CrO4(s) Rxn. & Extent: ^
0.025 M 0 M [email protected]: ▐ (goes slightly) 1 CrO42─ (goes ~100%) 2 Na+ +
▐ Ksp = [Ag+]2 [CrO42─] = 9.0 x 10―12
[Ag+] = 2 Ag+ + 1 CrO42─
^
^
2x
│
x
^
^
0.050 M │ 0.025 M Make a DATA TABLE with the equation symbols. 2x [CrO42─] = 0.025 M + x (exact) ≈ 0.025 M (approximate)
Solve the K equation using the approximation first.
Ksp = (2x)2 (0.025 M) = 9.0 x 10―12
4x2 = 360 x 10―12
x2 = 90 x 10―12
x = 9.5 x 10―6
? = [Ag+] = 2x = 1.9 x 10―5 M
Since (x/0.25)(100%) = (9.5/0.25) x 10―6 x 102% = 3.8 x 10―3% is <5% , the approximation is acceptable.
3. All nitrate compounds ionize ~100% in water. The small Ksp means that AgBr is only slightly soluble.
Solve K calculations using the WRECK steps. If two compounds are involved, write REC steps for both.
Wanted: [Br─] © 2011 ChemReview.Net v1r Page 1107 Module 35 — Solubility Equilibrium Rxn. & Extent:
[email protected]:
Rxn. & Extent:
[email protected]: 1 AgBr(s)
^
mol/L solid ─ x
1 AgNO3 (s)
^
0.30 M 0 M ▐ 1 Ag+ +
^
x
│ ▐ 1 Ag+
^
0.30 M 1 Br─
^
x
+ │ 1 NO3─
^
0.30 M (goes slightly) (goes ~100%) Ksp = [ Ag+ ][ Br ─ ] = 5.0 x 10―13
[Ag+] = 0.30 M + x ≈ 0.30 M
[Br─] = x Solve the K equation using the approximate value first.
Ksp = 5.0 x 10―13 ≈ (x)(0.30)
x = 16.7 x 10―13 = 1.7 x 10―12 M = [Br ─]
The difference between [common ion] = 3.0 x 10―1 and x = ~10―12 is 11 which is greater than 3, so
the approximation is acceptable.
4. All potassium compounds ionize ~100% in water. The small Ksp means that PbI2 is only slightly soluble.
Solve K calculations using the WRECK steps. If two compounds are involved, write REC steps for both.
Wanted: solubility of PbI2 = small = x
Rxn. & Extent:
[email protected]: 1 PbI2(s)
^
mol/L solid ─ x Rxn. & Extent: 1 KI(s)
^
0.20 M 0 M [email protected]: Ksp = [Pb2+] [I─]2 = 1.4 x 10―8
x
[Pb2+] =
[I─ ] = 0.20 M + 2x (exact) ≈ ▐ 1 Pb2+
^
x
1 K+ +
^
0.20 M │ ▐ +
│ 2 I─
^
2x
1 I─
^
0.20 M (goes slightly) (goes ~100%) Make a DATA TABLE with the equation symbols.
0.20 M (approximate) When x is small compared to the larger number, x or 2x or 3x added or subtracted may be omitted to
obtain the approximation that simplifies calculations.
Solve the K equation using the approximation first.
Ksp = (x) (0.20)2 = 1.4 x 10―8
x(0.040) = 1.4 x 10―8
x = 35 x 10―8
? = x = solubility = 3.5 x 10―7 M
The difference between [common ion] = 2.0 x 10―1 and x = ~10―7 is 6 which is greater than 3, so the
approximation passes the 5% test.
***** © 2011 ChemReview.Net v1r Page 1108 Module 35 — Solubility Equilibrium Lesson 35D: pH and Solubility
Adding Acid to Solutions of Slightly Soluble Salts
Some ions found in slightly soluble salts do not react as acids and bases. These ions include
• Cl─, Br─, and I─.
Those three ions are conjugates of the strong acids HCl, HBr, and HI. Conjugates of strong
acids are such weak bases that they are essentially pH neutral.
Other ions found in slightly soluble salts can act as acids and bases. Most of these ions are
basic. The basic ions include
• OH─, F─, S2─, CO 2─, and PO 3─.
3 4 Adding acid to an equilibrium that includes a base will shift the equilibrium. The acid
will use up basic particles and reduce the [base]. This reduction in the [base] will shift
the equilibrium as predicted by Le Châtelier’s Principle.
Based on the above, write answers to this problem, then check your answers below.
Q. Solid CaCO3 is slightly soluble in water.
1 CaCO3 (s) 1 Ca2+ + 1 CO32─ (goes slightly) If a strong acid is added to the above system at equilibrium, what will happen to the
a. [CO32─] in the solution?
b. [Ca2+] in the solution?
c. The amount of solid CaCO3 in the mixture?
* * ** *
Answer
a. Carbonate ions are bases (see Lesson 14E), and bases react with strong acids.
Adding acid will therefore use up CO32─ in the solution and cause the [CO32─] to
decrease.
b. If the concentration of a component that appears on one side of an equilibrium
equation is decreased, the equilibrium shifts toward that side of the equation. Since
adding acid decreases the [CO32─], the equilibrium above shifts to the right,
causing the [Ca2+] to increase.
c. Shifting the equilibrium to the right, the only way that the [Ca2+] can increase is for
more solid CaCO3 to dissolve. © 2011 ChemReview.Net v1r Page 1109 Module 35 — Solubility Equilibrium Practice A
1. For mixtures of water and these slightly soluble salts, which will have its solubility
equilibrium shifted by the addition of a strong acid to the solution?
a. CaCl2 b. CaF2 c. AgI d. Ag2CO3 e. Ba3(PO4)2 2. To a mixture of water and solid Sr3(PO4)2 at equilibrium, strong acid is added. What
will be the change in the
a. [PO43─] in the solution? b. [Sr2+] in the solution? pH and Slightly Soluble Hydroxides
For slightly soluble salts that contain hydroxide ions, calculations can solve for unknown
quantities including salt solubility, cation concentrations, solution pH, and the Ksp of the
salt.
Using the rules for pH and Ksp, try this calculation. If you get stuck, read the answer until
unstuck, then complete the problem.
Q. In a mixture of solid Mg(OH)2 (Ksp = 1.5 x 10―11) and water at equilibrium,
a. [OH―] = ? b. pH = ? c. Will adding strong acid increase or decrease the [Mg2+] in the solution?
d. What will be the [Mg2+] if acid is added to adjust the pH to 10.0?
(Try this step without a calculator.)
* * ** *
Answer
a. In calculations for reactions that go to equilibrium, write the WRECK steps.
WANTED: [OH―]
Rxn. & Extent: ^
mol/L solid ─ x [email protected]: 1 Mg2+ + 2 OH─ 1 Mg(OH)2(s)
▐ ^
x │ (goes slightly) ^
2x Ksp = [Mg2+] [OH─]2 = 1.5 x 10―11
[Mg2+] = x [OH─] = 2x Ksp = (x) (2x)2 = 4x3 = 1.5 x 10―11
x3 = 0.375 x 10―11 = 3.75 x 10―12
x = 1.55 x 10―4
? = [OH─] = 2x = 3.1 x 10―4 M = [OH―] © 2011 ChemReview.Net v1r Page 1110 Module 35 — Solubility Equilibrium Know [OH─] from part a. b. WANTED = pH Write: pOH ≡ ─ log[OH─] and [OH─] ≡ 10─pOH and pH + pOH = 14.00 pOH ≡ ─ log[OH─] = ─ log (3.1 x 10―4) = 3.51; pH = 14.00 ─ 3.51 =
c. For the reaction: 1 Mg2+ + 2 OH─ 1 Mg(OH)2(s) 10.49 = pH
(goes slightly) adding acid uses up OH─, shifting the equilibrium to the right, increasing the [Mg2+].
d. WANTED: [Mg2+]
WRECK steps – see above.
DATA: pH = 10.0, so pOH = 4.0 and [OH─] ≡ 10─pOH = 10―4.0 = 1.0 x 10―4 M
Ksp = [Mg2+] [OH─]2 = 1.5 x 10―11
[Mg2+] = ? [OH─] = 1.0 x 10―4 M Ksp = [Mg2+] (1.0 x 10―4)2 = 1.5 x 10―11
[Mg2+] (1.0 x 10―8) = 1.5 x 10―11
[Mg2+] = 1.5 x 10―11 = 1.5 x 10―3 M
1.0 x 10―8 Practice B
1. In a mixture of solid AgOH (Ksp = 2.0 x 10―8) and water,
a. [Ag+] = ? b. [OH―] = ? c. pH = ? d. If a strong acid is added to the solution, will the [Ag+] increase or decrease?
e. What will be the [Ag+] if acid is added to adjust the pH to 9.5?
f. Compare answers to parts a and e. Was your prediction in answer d correct? ANSWERS
Practice A
1a. CaCl2 contains Cl─ ion which is not basic, so acid will not shift the solubility equilibrium. 1b. CaF2 forms F─ ion which is basic and will react with acids, so acid will shift the equilibrium. 1c. AgI in solution forms I─ ion which is not basic, so acid will not shift the solubility equilibrium.
Ag CO contains CO 2─, and 1e. Ba (PO ) contains PO 3─. Both of those ions are basic and will 1d. 2 3 3 3 42 4 react with acids, so acid will shift their solubility equilibrium. © 2011 ChemReview.Net v1r Page 1111 Module 35 — Solubility Equilibrium 2. In water, the initial solubility reaction is
3 Sr 2+ + 2 PO43─ 1 Sr3(PO4)2(s) (goes slightly) PO43─ is a basic ion and will be used up when combined with acids, so the [PO43─] will decrease.
The decrease of [PO43─] on the right side shifts the equilibrium to the right, increasing the [Sr 2+]
dissolved in the solution. Practice B
1a,b. In calculations for reactions that go to equilibrium, write the WRECK steps.
WANTED: [Ag+]
Rxn. & Extent:
[email protected]: 1 Ag+ + 1 OH─ 1 AgOH(s)
^
mol/L solid ─ x Ksp = [Ag+] [OH─] = 2.0 x 10―8
Equation DATA table: [Ag+] = x ▐ ^
x │ (goes slightly) ^
x = [OH─] Ksp = (x) (x) = x2 = 2.0 x 10―8
x = 1.4 x 10―4 M = [Ag+] = [OH─] = part b answer
1c. WANTED = pH Know [OH─] Write: pOH ≡ ─ log[OH─] and [OH─] ≡ 10─pOH and pH + pOH = 14.00
pOH ≡ ─ log[OH─] = ─ log (1.4 x 10―4) = 3.85 ; pH = 14.00 ─ 3.85 = 10.15 = pH
1d. Since AgOH has a very small Ksp, it is slightly soluble in water. The solubility reaction is:
AgOH(s) 1 Ag+ + 1 OH─ (goes slightly) Adding acid causes [OH─] to decrease. According to Le Châtelier’s Principle, if the concentration of a
component that appears on one side of an equilibrium equation is decreased, the equilibrium shifts
toward that side of the equation.
Adding acid therefore shifts the equilibrium above to the right and causes the [Ag+] to increase.
1e. WANTED: [Ag+]
WRECK steps – see above.
DATA: pH = 9.5, so pOH = 4.5 and [OH─] ≡ 10─pOH = 10―4.5 = 3.16 x 10―5 M = [OH─]
Ksp = [Ag+] [OH─] = 2.0 x 10―8
[Ag+] = ?
[OH─] = 3.16 x 10―5 M
Ksp = [Ag+] (3.16 x 10―5) = 2.0 x 10―8
[Ag+] = 2.0 x 10―8 = 0.633 x 10―3 = 6.3 x 10─4 M = [Ag+]
3.16 x 10―5 © 2011 ChemReview.Net v1r Page 1112 Module 35 — Solubility Equilibrium 1f. Without added acid, [Ag+] = 1.4 x 10―4 M . After adding acid to lower the pH from 10.15 to 9.5,
[Ag+] = 6.3 x 10―4 M . The [Ag+] increased with added acid, as predicted. ***** Lesson 35E: Quantitative Precipitation Prediction
In Module 13, we learned a simplified scheme to predict whether precipitation will take
place when two solutions are combined. Our predictions were based on an arbitrary but
convenient division: if a possible product has a solubility of less than 0.1 mol/L in water, it
is termed insoluble and will precipitate. These general rules allowed us to make accurate
precipitation predictions in cases when solutions with concentrations at or near 0.2 M are
combined.
In this lesson, our goal is to predict whether precipitation will occur when solutions that
have a wide range of concentrations are combined. Precipitation Prediction
By definition, when solutions of dissolved ions are combined, precipitation occurs if a
precipitate both forms and persists after stirring.
When two solutions of ions are combined, to determine whether a precipitation occurs, use
these steps.
1. Since solubility predictions involve a K (the Ksp), write the WRECK steps. Include the
Ksp equations for precipitate(s) that are predicted based on solubility schemes (see
Lesson 13C).
2. Find Q: substitute into the Ksp expression the ion concentrations in solution after the
solutions are combined, but before the ions react, then calculate a value for Q.
3. If Q > Ksp, the precipitate forms and persists after stirring.
If Q < Ksp, persistent precipitate will not form.
When Q = Ksp, the precipitate begins to persist after stirring.
These rules rely on the reaction quotient (Q) defined in Lesson 28H : Q is the number
obtained by substituting into the K expression the concentrations in a mixture that may not
be at equilibrium.
Apply those three steps to the following problem, then check your answer below.
Q. If 1.0 x 10―3 moles of Ag+ ions are combined with 1.0 x 10―5 moles of Cl― ions in 1.0
liters of solution, will AgCl (Ksp = 1.6 x 10―10) precipitate?
***** © 2011 ChemReview.Net v1r Page 1113 Module 35 — Solubility Equilibrium Answer
For K calculations, write the WRECK steps. If the reaction has a Ksp, that reaction is always
one formula unit of an ionic solid dissolving slightly in water to form separated ions.
To predict precipitation, compare Q for the mixture to Ksp for the precipitate.
WANTED: Q
Rxn. & Extent:
[email protected]: 1 Ag+ 1 AgCl (s)
^
mol/L solid ─ x + 1 Cl─ ^
x │ ^
x ▐ (goes slightly) Ksp = [Ag+] [Cl─] = 1.6 x 10―10
Q = ( [Ag+]after combining, before reacting )( [Cl─]after combining, before reacting )
[Ag+] = 1.0 x 10―3 moles/liter [Cl─] = 1.0 x 10―5 moles/liter Q = (1.0 x 10―3) (1.0 x 10―5) = 1.0 x 10―8
In calculations involving Q, as with K, the numbers have units but by convention the units
are omitted from Q values and during Q calculations.
Since Q = 1.0 x 10―8 is greater than Ksp = 1.6 x 10―10, AgCl precipitate persists.
***** Predicting the Point When Precipitation Begins
Comparing Q to Ksp can predict both weather a precipitate will form and the point at which
precipitate will form, when solutions is combined.
Apply the 3 steps for precipitation prediction to the following problem. If you get stuck,
read a little of the answer, adjust your work, and then complete the problem.
Q. Solid sodium sulfate is slowly stirred into a solution of 0.020 moles of Ag+ dissolved
in 1.0 liters of solution. At what [SO42─] will Ag2SO4 precipitate form?
(Ksp Ag2SO4 = 1.2 x 10―5)
*****
Answer
In this problem, it is helpful to write the sodium sulfate reaction. Sodium compounds
dissolve and ionize ~100% in water. Given a name rather than a formula for a salt, write
first the separated ions, then the solid formula (for review, see Lesson 7C).
1 Na2SO4(s) 2 Na+ + 1 SO42─ (goes ~100%) By adding solid sodium sulfate that then dissolves, we do not cause a substantial change in
the volume of the solution. © 2011 ChemReview.Net v1r Page 1114 Module 35 — Solubility Equilibrium For K calculations, begin by writing the WRECK steps. Then, to find the point at which
precipitation persists after stirring, apply the
Rule: Precipitate forms and persists when the substance being added reaches the
concentration at which Q = Ksp .
[SO42─] when Q = Ksp WANTED: Rxn. & Extent:
[email protected]: 2 Ag+ + 1 Ag2SO4(s)
^
mol/L solid ─ x ▐ ^
2x │ 1 SO42─ (goes slightly) ^
x Ksp = [Ag+]2 [SO 2─] = 1.2 x 10―5 = Q at start of precipitation
4
+] = 0.020 mol/L = 2.0 x 10―2 M ;
[Ag
[SO42─] = ? = 1.2 x 10―5 = 1.2 x 10―5 = 1.2 x 10―5 =
(2.0 x 10―2)2
4.0 x 10―4
[Ag+]2 3.0 x 10―2 M Summary: To predict whether combining ions will produce a precipitate,
1. Since Ksp is involved, write the WRECK steps for the possible precipitate.
2. Calculate Q, using [ions] after combining, but before reacting.
3. Compare Q to Ksp. If Q is equal or larger, precipitation is predicted. Practice A: Problem 2 is more challenging. 1. How many moles of solid AgNO3 must be mixed into a 1.0 liter solution containing
0.0200 moles of Na2CrO4 before precipitate forms? (Ksp Ag2CrO4 = 9.0 x 10―12)
2. How many grams of solid NaI must be mixed into a 1.0 liter solution containing 0.040
moles of Pb(NO3)2 before persistent solid forms? (Ksp PbI2 = 1.4 x 10―8) Solution Concentrations After Combining
In the problems above, the data lists moles of substances, plus the volume of the final
solution after the substances are combined but before they react. Given such data, the
concentrations needed to find Q values are easily calculated.
However, in predicting precipitation, most problems supply the concentrations and
volumes of the two solutions before they are combined. Combining solutions changes the
ion concentrations. To solve such calculations, the rules are:
1. For solution reactions, in Q and K and rice calculations, mol/L must be
concentrations after solutions are combined, but before they react. © 2011 ChemReview.Net v1r Page 1115 Module 35 — Solubility Equilibrium To find concentrations after combining, we will need rule
2. When solutions of different dissolved substances are combined, each substance is
diluted.
To find those diluted concentrations, the rule is:
3. When solutions are combined, volumes add.
For all but highly concentrated solutions, we can safely assume that 30.0 mL of one aqueous
solution added to 20.0 mL of another will result in a volume of 50.0 mL, even if water is a
product or reactant of a reaction that occurs when the solutions are combined.
Calculating the diluted concentrations can be done in several ways (see Lesson 12B). Two
methods are 1) solve by inspection and 2) use the dilution equation.
1. If the solutions that are combined have simple volume ratios, the molarity of the
substances or ions after combining can often be written by inspection.
The easiest case: for two solutions that contain different substances and ions, if equal
volumes of the two solutions are combined, the concentration of each substance or ion
in the combined solution is one-half its concentration before they were combined.
Try this example: If 50.0 mL of 0.60 M Ag+ is mixed with 50 mL of 0.20 M Cl―,
[Ag+]after combining, before reacting = _________________
[Cl─]after combining, before reacting = _________________
*****
If the volumes combined are equal, the concentrations are cut in half.
[Ag+]after combining, before reacting = 1/2 x 0.60 M Ag+ = 0.30 M Ag+
[Cl─]after combining, before reacting = 1/2 x 0.20 M Cl― = 0.10 M Cl― 2. If the dilution does not involve easy volume multiples, the quickest way to find the
diluted concentrations is to use the dilution equation:
(Volume concentrated)(Molarity concentrated) = (Volume diluted)(Molarity diluted)
written in symbols as VC x MC = VD x MD and memorized by recitation: “In dilution, volume times molarity equals volume times molarity.”
For precipitation predictions in which solutions of two substances are combined, we are
usually given the volume and molarity of each solution before they are combined, when
both solutions are more concentrated.
To find the ion concentrations after combining (MD) that are needed to calculate Q, first
calculate the volume of the one solution that exists after the solutions are combined.
That diluted volume (VD) is the volume in which all of the ions are dissolved after the © 2011 ChemReview.Net v1r Page 1116 Module 35 — Solubility Equilibrium solutions are combined. To find VD, simply add the volumes of the solutions that are
mixed together.
Then, to find the concentration of any particle after combining (MD), use the dilution
equation solved for MD:
? MD = VC x MC
VD Try those steps on this sample calculation: Q. If 150.0 mL of 0.0400 M Pb(NO3)2 is combined with 50.0 mL of 0.0500 M CaCl2
solution, in the solution after combining but before any reaction takes place,
a. [Pb(NO3)2] = ? c. [Cl─] = ? b. [CaCl2] = ? d. After the reaction takes place, will there be a precipitate of PbCl2 ?
(Ksp PbCl2 = 1.5 x 10―5)
*****
Answer
a. First calculate the volume of the single solution that exists after combining.
150.0 mL + 50.0 mL = 200.0 mL = VD for all particles – after combining. Then,
WANTED: ? = [Pb(NO3)2]diluted = MD
The dilution equation is VC x MC = VD x MD ? MD = VC x MC = 150.0 mL x 0.0400 M
VD = 0.0300 M = [Pb(NO3)2]diluted 200.0 mL *****
b. WANTED: ? = [CaCl2]diluted = MD
? MD = VC x MC = 50.0 mL x 0.0500 M
VD = 0.0125 M = [CaCl2]diluted 200.0 mL *****
c. WANTED: ? = [Cl─]diluted
The chloride ion is created by the separation of soluble, diluted CaCl2 into its ions.
Either solve by inspection or use the REC steps below.
Rxn. & Extent:
[email protected]: ^
0.0125 M 0 M 1 Ca2+ 1 CaCl2 (s) ? = [Cl─]diluted = ▐ + 2 Cl─ (goes ~100%) ^
^
0.0125 M │ 0.0250 M 0.0250 M *****
d. WANTED: Q compared to Ksp. If Q > Ksp, a precipitate will form.
In calculations that involve a K, write the WRECK steps. © 2011 ChemReview.Net v1r Page 1117 Module 35 — Solubility Equilibrium Rxn. & Extent: 1 Pb2+ + 1 PbCl2(s)
^
mol/L solid ─ x [email protected]: ▐ ^
x │ 2 Cl─ (goes slightly) ^
2x Ksp = [Pb2+] [Cl─]2 = Q expression
To calculate Q, use the concentrations in the reaction mixture after combining and
diluting, but before reacting.
[Pb2+] = 0.0300 M by inspection from [Pb(NO3)2] in answer a. [Cl─] = 0.0250 M from part c.
Q = (0.0300)(0.0250)2 = (3.00 x 10―2)(2.50 x 10―2)2 = (3.00 x 10―2)( 6.25 x 10―4) =
= 18.8 x 10―6 = 1.9 x 10―5 which is > Ksp = 1.6 x 10―5 The mixture will form persistent precipitate.
*****
In the problem above, we calculated the diluted concentrations, then wrote the WRECK
steps. In most problems, the steps will not be in a requested order, and it will be easier to
write the WRECK steps first. The Ksp expression will then identify the specific
concentrations after dilution that must be calculated to find the value for Q.
Summary: To predict whether combining ion solutions will produce a precipitate:
1. Since a K (Ksp) is involved, write the WRECK steps for the possible precipitation(s).
2. Calculate Q using the [diluted ions] after combining, but before reacting.
3. Compare Q to Ksp. If Q is larger, the precipitate is predicted to form. Practice B
1. If 100.0 mL of 0.010 M Ag+ is combined with 100.0 mL of 0.40 M SO42─, will a
precipitate form? (Ksp Ag2SO4 = 1.2 x 10―5)
2. If 150.0 mL of 0.10 M Pb2+ is combined with 100.0 mL of 0.0200 M Cl─, will there be a
precipitate of PbCl (Ksp = 1.5 x 10―5)?
2 Saturated Solutions Not At Equilibrium
If a solid is at equilibrium with its ions in an aqueous solution, the saturated solution can be
decanted (carefully poured off, leaving the solid behind) into a separate container. The
decanted solution is still saturated with its ions, but if there is no solid in the container, the
solution in the new container is not at solubility equilibrium. For equilibrium to exist, all
of the products and reactants must be present. © 2011 ChemReview.Net v1r Page 1118 Module 35 — Solubility Equilibrium Similarly, if an ionic solid is mixed with water and all of the solid dissolves, the system
cannot be at solubility equilibrium. A solubility equilibrium must have some solid present. Theoretical Versus Actual Results
In Ksp calculations, as with most calculations in general chemistry, our goal is to get close to
an accurate prediction of what experimental results will be. Error can be introduced from
many sources, including:
• The equations we use are often based on models that assume ideal behavior, when
actual behavior is not ideal. • Calculations may not take into account all of the factors that may have impact on
results. For example,
• We assume that salts predicted to be soluble by our solubility rules ionize ~100% in
solution, but some in some solutions the salt may effectively ionize less than 100%. • In common-ion calculations, if an added soluble salt is not pH neutral, the
equilibrium shifts will be affected by pH changes as well as [common-ion]. • Some solutions can become supersaturated. In those cases, the dissolved particle
concentrations are higher than they would be theoretically, based the predictions of
when precipitation should occur based on Ksp values or solidification should occur
based on melting points. In solutions with liquid components, nearly always, no
solid is present, so the solution is not at solubility equilibrium. Supersaturated
solutions will generally go to the conditions predicted at equilibrium if a “seed
crystal” of the solid can be added. The crystal serves as a template on which
particles in the solution can fit into geometry needed to form the solid. Without a
seed crystal, persuading a solution to crystallize when it should can be one of the
interesting challenges of laboratory chemistry. In upper-level chemistry courses, we consider these factors in more detail. In the
meantime, our general chemistry rules will result in generally accurate predictions of what
truth (experimental results) will be. Practice C
1. When 25.0 mL of 2.0 x 10―4 M K2SO4 is combined with 75.0 mL of 1.0 x 10―4 M BaCl2,
if BaSO4 (Ksp = 1.1 x 10―10) does not precipitate, is the solution supersaturated? As
always, show your work and explain your reasoning. © 2011 ChemReview.Net v1r Page 1119 Module 35 — Solubility Equilibrium ANSWERS
Practice A
1. WANTED: mol AgNO3 Strategies: There are three reactions to consider. First, the two soluble compounds dissolve and
separate into ions ~100%. Then, as one substance is added, persistent precipitate forms when Q =
Ksp .
Since a K is involved, write the WRECK steps for the formation of the precipitate.
Since the problem supplies solid formulas, and ion formulas are needed for K and Q, write the REC
steps for the formation of the ions.
If needed, use those steps to complete the problem, then check below.
*****
First, the two soluble solids ionize.
Rxn. & Extent: ^
0.0200 M 0 M [email protected]:
Rxn. & Extent: ^
?? M 0M (goes ~100%) ^
^
▐ 0.0400 M │ 0.0200 M
1 Ag+ 1 AgNO3(s) [email protected]: 1 CrO42─ 2 Na+ + 1 Na2CrO4(s) 1 NO3─ + ^
▐ ? M (Wanted) │ (goes ~100%) ^
?? M Then precipitate forms and ionizes slightly. For slight reactions, complete the WRECK steps.
Rxn. & Extent:
[email protected]: ^
mol/L solid ─ x 2 Ag+ + 1 Ag2CrO4(s)
▐ ^
2x 1 CrO42─ │ (goes slightly) ^
x Ksp = [Ag+]2 [CrO42─] = 9.0 x 10―12
The goal is moles of AgNO3 added when Q = Ksp .
Solving Q = Ksp for [Ag+] will get us close to moles AgNO3.
Make a DATA TABLE with the equation symbols.
[Ag+]as combined, before reacting = 2x
[CrO42─]formed in solution = 0.0200 mol/L from the first reaction above.
Q = Ksp = (2x)2 (0.0200 M) = 9.0 x 10―12
4x2 = 450 x 10―12
x2 = 112 x 10―12
x = 10.6 x 10―6
? = [Ag+] = 2x = 21.2 x 10―6 M = 2.1 x 10―5 M when persistent precipitation starts.
From the ionization equation for AgNO3 above:
[Ag+]as mixed = 2.1 x 10―5 M = [AgNO3]mixed = mol AgNO3 added per 1.0 L soln. = WANTED © 2011 ChemReview.Net v1r Page 1120 Module 35 — Solubility Equilibrium 2. In calculations for reactions that go to equilibrium, write the WRECK steps.
WANTED: g NaI DATA: 149.9 g NaI = 1 mol NaI (grams prompt) Strategies: To find grams, find moles first.
Precipitation begins when Q = Ksp .
Since the moles of NaI will be in a 1.0 L solution, the moles NaI will equal the [NaI]. Since
Na compounds ionize ~100% in water, [NaI]mixed = [I─]formed in solution.
If you know some of the steps, but maybe not all at the beginning of a problem, do some
steps that get you closer to the answer unit or symbol, then look for ways to finish, working
forward from what is known and backward from what is WANTED.
*****
Rxn. & Extent: 1 PbI2(s)
^
mol/L solid ─ x [email protected]: ▐ 1 Pb2+ +
^
x
│ 2 I─
^
2x (goes slightly) Ksp = [Pb2+] [I─]2 = 1.4 x 10―8 = Q at start of persistent precipitation
[Pb2+] = 0.040 M To find [I─] , find [I─]2 in equation above first [I─]2 = 1.4 x 10―8 = 1.4 x 10―8 = 0.350 x 10―6 = 35.0 x 10―8
4.0 x 10―2
[Pb2+]
[I─] = (35.0 x 10―8)1/2 = 5.92 x 10―4 M = [I─]
But what was wanted was grams NaI added. Find moles NaI added first.
*****
Moles of soluble NaI added per 1.0 L = [NaI]mixed = [I─]in solution = 5.92 x 10―4
From the moles NaI that need to be added, find grams.
*****
? g NaI = 5.92 x 10―4 mol NaI • 149.9 g NaI = 8.9 x 10―2 g NaI = 0.089 g NaI
1 mol NaI
Since the supplied Ksp had 2 sf, round the final answer to 2 sf. Practice B
1. Steps: To predict precipitation after combining, write the WRECK steps for the possible pcpt.; find Q using
diluted concentrations, compare Q to Ksp .
WANTED: Q Rxn. & Extent:
[email protected]: 1 Ag2SO4(s)
^
mol/L solid ─ x ▐ 2 Ag+ +
^
2x │ 1 SO42─
^
x (goes slightly) Ksp = [Ag+]2 [SO 2─] = 1.2 x 10―5 = Q
4
When solutions are combined, the dissolved substances are diluted. © 2011 ChemReview.Net v1r Page 1121 Module 35 — Solubility Equilibrium To calculate Q, we need the [ions] after combining, but before reacting.
Since the two solutions have equal volumes, combining them cut their [ions] in half.
[Ag+]after combining, before reacting = 1/2 x 0.010 M Ag+ = 0.0050 M Ag+
[SO42─]after combining, before reacting = 1/2 x 0.40 M SO42─ = 0.20 M SO42─
Q = [Ag+]2 [SO42─] = ( 5.0 x 10―3 )2 ( 2.0 x 10―1 ) = 50. x 10―7
= 5.0. x 10―6 which is less than Ksp = 1.2 x 10―5 , so pcpt. should not form.
2. WANTED: Precipitation prediction: compare Q to Ksp . Rxn. & Extent:
[email protected]: 1 PbCl2(s)
^
mol/L solid ─ x 1 Pb2+ + 2 Cl─
^
^
x
│ 2x ▐ (goes slightly) Ksp = 1.5 x 10―5 = [Pb2+] [Cl─]2 = Q to calculate using [diluted ions]
The dilution equation: VC x MC = VD x MD
[Pb2+]diluted = ? MD = VC x MC = 150.0 mL x 0.10 M = 0.060 M Pb2+
VD
250.0 mL total
[Cl─]diluted = ? MD = VC x MC = 100.0 mL x 0.020 M = 0.0080 M Cl─
VD
250.0 mL total
Q = [Pb2+] [Cl─]2 = (0.060)(0.0080)2 = (6.0 x 10―2)(8.0 x 10―3)2
= (6.0 x 10―2)(64 x 10―6) = 384 x 10―8 = 3.8 x 10―6 = Q
which is less than Ksp = 1.5 x 10―5 so the mixture will not form a precipitate. Practice C
1. To be supersaturated means that the solution should precipitate, but it does not. If this solution does not
precipitate, the question is whether it should. Begin by calculating whether the solutions should precipitate
when combined.
Steps: To predict precipitation after combining, write the WRECK steps for the possible pcpt.; find Q using
diluted concentrations, compare Q to Ksp .
WANTED: Precipitation prediction: compare diluted Q to Ksp Rxn. & Extent:
[email protected]: 1 BaSO4 (s)
^
mol/L solid ─ x ▐ 1 Ba2+ + 1 SO42─
^
^
x
│
2x (goes slightly) Ksp = 1.1 x 10―10 = [Ba2+] [SO 2─] = Q to calculate using [diluted ions]
4
The dilution equation is VC x MC = VD x MD
BaCl2 is in solution (soluble) and ionizes ~100%, so: 1 BaCl2 1 Ba2+ + 2 Cl─ 1 [Ba2+]diluted therefore equals 1 [BaCl2]diluted
© 2011 ChemReview.Net v1r Page 1122 Module 35 — Solubility Equilibrium [Ba2+]dil. = [BaCl2]dil. = ? MD = VC x MC = 75.0 mL x 1.0 x 10―4 M = 7.5 x 10―5 M Ba2+
VD
100.0 mL total
2 K+ + 1 SO 2─
Since all potassium compounds are water soluble & ionize ~100%: 1 K SO
2 4 4 [SO42─]diluted therefore equals [K2SO4]diluted
[SO42─]dil.= [K2SO4]dil. = ? MD = VC x MC = 25.0 mL x 2.0 x 10―4 M = 5.0 x 10―5 M SO42─
VD
100.0 mL total
Q = [Ba2+] [SO42─] = (7.50 x 10―5)(5.00 x 10―5) = 37.5 x 10―10 = 3.75 x 10―9 = Q
which is greater than Ksp = 1.1 x 10―10 . This means that the mixture should precipitate. If it does
not, the solution would be supersaturated.
*** ** Summary: Solubility Equilibrium
1. Nearly all K calculations are solved in the same way.
a. Write the WRECK steps.
b. Find the Concentrations at equilibrium for the terms in the K expression, usually
in terms of x, either using a rice table or by inspection.
c. Solve the K equation, then solve for the WANTED symbol.
2. Ksp is the special K for a slightly soluble salt dissolving in water.
a. In the Ksp WRECK steps, write a balanced equation with one particle of the ionic
solid on the left, its aqueous separated ions on the right, and (goes slightly).
b. Ksp = solubility product. A Ksp expression is written with [ions] multiplied in
the numerator, but no term in the denominator.
3. For a slightly soluble ionic solid dissolving in water: the solid’s solubility = small x
Solubility is the moles of solid that dissolve per liter. The unit mol/L (or M) must be
added to calculated solubilities.
4. Solubility of an ionic solid = small x . Add units of moles/liter.
Solubility product = Ksp without units. During K calculations, units are omitted.
5. When writing the rice table or the WRECK steps in Ksp calculations,
a. Let ─ x represent the small mol/L of the reactant that is used up in the ionization
reaction.
b. At Equilibrium, [solid]eq. = mol/L solid ─ x
c. Define the product concentrations using positive x terms. © 2011 ChemReview.Net v1r Page 1123 Module 35 — Solubility Equilibrium 6. When a common ion is added to a solubility equilibrium:
a. Rule: In a mixture of a slightly soluble ionic solid and its dissolved ions, to reduce
the solution concentration of one ion, add a soluble salt that contains the other ion
in the solid.
b. To solve a Ksp calculation with common-ion added,
• As in all K calculations, write the WRECK steps. • Write REC steps for the ionization of both the slightly soluble and soluble salt. • Write a DATA TABLE under the Ksp equation. Include the exact and
approximate [common ion]. • Solve the Ksp equation first using the approximate [common ion]. • If the difference between the exact and approximate [common ion] is greater
than 5%, solve the exact quadratic. 7. Adding acid to a solution at equilibrium that includes basic particles will reduce the
[base] and shift the equilibrium in accord with Le Châtelier’s Principle.
Basic ions include OH─, F─, S2─, CO32─, and PO43─.
8. To predict whether combining solutions will produce a precipitate:
a. Since Ksp is involved, write the WRECK steps for the possible precipitate.
b. Calculate Q using the [diluted ions] after combining, but before reacting.
c. Compare Q to Ksp. If Q is equal or larger, precipitate is predicted to form and
persist.
##### © 2011 ChemReview.Net v1r Page 1124 Module 36 — Thermodynamics Module 36 — Thermodynamics
Prerequisites: If needed, review Lessons 21A and 22A on phases, energy, and ΔH.
Pretest: If you think you know a lesson topic in this module, try the last two problems in
each problem set in the lesson. If you can do those problems, skip the lesson.
***** Lesson 36A: Review: Energy and Heats of Reaction
Timing: Calorimetry, heat and heat of reaction (∆H) calculations were covered in Modules
21 and 22. Begin this Module when you are assigned problems that involve entropy (S) or
free energy (G).
***** Thermodynamics, Energy, and Enthalpy
Thermodynamics includes the study of energy (E), enthalpy (H), entropy (S) and free
energy (G). Let us briefly review the rules for energy and enthalpy changes (from Modules
21 and 22).
1. To study thermodynamics, we divide the universe into two parts:
• The system is the particles of interest, which may be molecules or ions. • The surroundings is the environment outside of the system.
Universe = system + surroundings In chemical changes, the impact on the system and its surroundings is often
accounted for separately.
2. Energy is the capacity to do work. The Law of Conservation of Energy is: Energy
can neither be created nor destroyed (except in nuclear reactions). This means that in
any physical or chemical process,
ΔEuniverse = Efinal ─ Einitial = 0
However, energy can be transferred between substances and to and from the
surroundings. Energy can also change its form during chemical or physical processes. 3. Forms of energy include potential energy, defined as stored energy, and kinetic energy,
defined as energy of motion. Kinetic energy (KE) = ½ (mass) (velocity)2 4. Chemical substances can store energy in the attractions (bonds) between atoms,
molecules, and ions. During chemical reactions and phase changes, when bonds
break and form, energy can be stored or released. 5. One way to store energy in a substance is to change its phase. The solid phase of a
substance always has less stored (potential) energy than its liquid phase, which
always has less potential energy than its gas phase.
Potential energy of a substance: solid < liquid < gas © 2011 ChemReview.Net v1r Page 1125 Module 36 — Thermodynamics 6. When a substance is in one phase (all solid, liquid, or gas), adding or removing energy
will change the average kinetic energy of its particles (its temperature), but not its
potential energy. 7. During a phase change, when two phases are present, adding or removing energy
changes the potential energy, but not the average kinetic energy (temperature), of the
particles. 8. Energy (E), heat (q), and work (w) in the SI system are measured in joules (J). 9. In any physical or chemical change: ∆Esystem = q + w 10. In chemistry, measurements of heat, work, and changes in energy and enthalpy are
assigned signs from the perspective of the system.
•
• 11. If heat is added to the system, q is given a positive sign. If a reaction or process
releases heat from the system to the surroundings, q is given a negative sign.
If work is done on the system, such as in compressing a gas, w is positive. If a
system does work, such as an expanding gas moving a piston that is under
pressure, the system loses energy and w is negative. In the case of PV work by a gas, w = ─ Pexternal ∆Vsystem In converting from PV-work units to energy units: 1 liter • atm = 101 joules
12. If a thermodynamic symbol has no subscript, assume the subscript is system. 13. Enthalpy (H) is defined as H = E + PV and the change in enthalpy in a process = Hfinal ─ Hinitial =
if work is limited to PV work.
14. ΔH = ΔE + Δ(PV) For reactions in which the external pressure on a system is held constant and work is
limited to PV work, ∆H will measure the heat flow into or out of the system.
ΔH = q = “heat of reaction” . 15. In a reaction, if pressure and temperature are held constant and the change in volume
is small, and/or if the work term is much smaller than q (true for most reactions), then
∆H will approximate the change in the potential energy of the system, which will
approximate the heat flow into or out of the system. ΔH ≈ ΔEpotential ≈ q Practice A: Refresh your memory on the rules and sign conventions above, then answer
these questions from memory.
1. From the perspective of the particles in a system,
a. if heat is removed from the particles, is the sign of q positive or negative ?
b. If gas particles are compressed by a piston that is under external pressure, is the
sign of w positive or negative ? © 2011 ChemReview.Net v1r Page 1126 Module 36 — Thermodynamics 2. If a substance is melted, but no work is done,
a. is the sign of q positive or negative ?
b. Is the sign of ∆E positive or negative ? c. Is the reaction exothermic or endothermic?
16. In equations that include energy terms, the coefficients are in moles. 17. Because the energy content of a substance varies with its phase, in energy equations
the phase of each particle must be shown: (s), (1), (g), or (aq). 18. In exothermic reactions, energy is released into the environment, and the energy term
•
• 19. EITHER is shown with a positive sign on the products side;
OR (preferred) with a negative ΔH value written after the equation. In endothermic reactions, energy must be added, and the energy term
• EITHER is shown with a positive sign on the reactants side of the equation, • OR with a positive ∆H written after the reaction. 20. Reactions involving energy or heat can be reversed. To represent the reversed
reaction, write equation is backwards), and if a ∆H is included, change its sign. 21. All reaction coefficients and energy terms can be multiplied or divided by a number.
If a ∆H notation is attached, do the same to the value of ∆H. 22. Hess’s law: Equations with energy terms can be added to produce a new equation.
•
• 23. Like particles on the same side in different equations can add, and on opposite
sides can cancel.
∆H values add or subtract according to their signs. When ∆H for an equation is not known, it can be found by adding together equations
for which ∆H is known.
• Write the reaction equation WANTED, then a dotted line below it. • Write the first coefficient and substance formula below the dotted line. • Find an equation with a known ∆H that includes that first formula. Adjust its
direction and coefficients to put the dropped particle and its coefficient on the side
where it is WANTED. Modify and include ∆H. • Add other equations and their known ∆H values arranged to cancel particles not
wanted, and add to result in the equation WANTED. 24 In a formation equation, the reactants are all elements in their standard state at 25°C and
1 atm pressure, and the product is one mole of a compound. 25. The heat of formation (ΔH°f) of a compound is the heat required or released when one
mole of the compound is formed from its elements in their standard state. 26. All elements in their standard state are assigned a heat of formation of zero kJ/mole. 27. The standard enthalpy value for a compound (ΔH°) is its ΔH°f. © 2011 ChemReview.Net v1r Page 1127 Module 36 — Thermodynamics Practice B: If you need help on these, review Module 22. 1 Use these “known” reactions to fill in the blanks below.
H2 (g) + 1/2 O2(g) H2O(l) ΔH = ― 285.8 kJ (1) H2(g) + 1/2 O2(g) H2O(g) ΔH = ― 241.8 kJ (2) 1/2 N2 (g) + O2 (g) NO2 (g) ΔH = + 33.8 kJ (3) a. 2 N2(g) + 4 O2 (g)
b. 3 H2O(g) 4 NO2(g) ΔH = _______ 3 H2(g) + 3/2 O2(g) ΔH = _______ 2. Which of the three known reactions above are endothermic?
3. For n-octane, the heat of combustion is
C8H18(l) + 25/2 O2(g) 8 CO2(g) + 9 H2O(g) Use the above equation plus the table values
below to find the heat of formation for n-octane. ∆H = ─ 5450. kJ
Formula ΔH°f in kJ/mole H2O(g)
CO2(g)
28. ─ 241.8
─ 393.5 ΔH Summation Equation
By definition: ΔH reaction = ΔHproducts ─ ΔHreactants
If heats of formation are known for all of the substances in a reaction, the ΔH of the
reaction can be found by substituting ΔH°f values into:
ΔH = [sum of (coefficient x ΔH°) of products] ─ [sum of (coefficient x ΔH°) of reactants]
which can also be written as
ΔHreaction = ∑ cproduct ΔHproduct ─ ∑ creactant ΔHreactant Practice C: If you need additional review, see Lesson 22D. 1. The equation for the burning of n-butane is:
C4H10(g) + 13/2 O2(g) 4 CO2(g) + 5 H2O(g) Use the summation equation and the values at the
right to calculate ∆H. Formula
C4H10(g) ΔH˚f in
kJ/mole
─ 30.0
─ 393.5 H2O(g) © 2011 ChemReview.Net v1r CO2(g) ─ 241.8 Page 1128 Module 36 — Thermodynamics ANSWERS
Practice A
1 a. If heat is removed from the particles, the sign of q is negative. The particles lose energy.
b. If a gas is compressed by a piston that is under pressure, the sign of w is positive . The environment
has done work on the system, and in the process the system gains internal energy. In the equation
w = ─ Pexternal ∆Vsystem , ∆V is negative, and P must be positive, so w is positive. 2. a. q is positive . In melting, the system gains heat.
b. ∆E = q + w . If q is positive and w is 0, ∆E must be positive. If the system gains heat with no work
involved, the system gains energy, and ∆E for the system must be positive. c. If the system gains heat, the reaction is endothermic. Practice B
1a. 2 N2 (g) + 4 O2 (g) 4 NO2 (g) ΔH = + 135.2 kJ Table reaction # 3 is quadrupled.
SF: Multiplying by an exact number does not change the place with doubt.
1b. 3 H2O(g) 3 H2 (g) + 3/2 O2(g) ΔH = + 725.4 kJ Table reaction #2 is tripled and written backwards, so the sign is reversed. Be careful to distinguish the
gas/liquid/solid states.
2. Only (3) 3. The heat of formation of n-octane is WANTED:
C8H18(l) 8 C(s) + 9 H2(g) ∆H = ? *****
In burning (combustion) reactions, the O2 coefficients should work as a check if the other added
equations are correct. Here, the 25/2 O2 total on both sides cancels.
8 C(s) + 9 H2(g)
C8H18(l)
∆H = ?
-----------------------------8 CO2(g) 8 C(s) + 8 O2(g)
8 CO2(g) + 9 H2O(g) ∆H = ─ 3,148.0 kJ C8H18(l) + 25/2 O2(g) ∆H = + 5450. kJ (wanted equation)
(table rxn. x8 )
(supplied rxn reversed) 9 H2(g) + 9/2 O2(g)
∆H = ─ 2,176.2 kJ
9 H2O(g)
______________________________________________________ (table reaction x9) 8 C(s) + 9 H2(g) (wanted equation) C8H18(l) ∆H = + 126 kJ Practice C
1. C4H10(g) + 13/2 O2(g) 4 CO2(g) + 5 H2O(g) ΔH = (sum of ΔH° values of products) ─ (sum of ΔH° values of reactants)
= [ 4(─ 393.5 kJ) + 5(─ 241.8) ] ─ [ (─ 30.0 + 13/2(0) ] = ─ 1574.0 ─ 1209.0 + 30.0 = ─ 2,753.0 kJ
***** © 2011 ChemReview.Net v1r Page 1129 Module 36 — Thermodynamics Lesson 36B: Entropy and Spontaneity
Introduction To Thermodynamics
Thermodynamics is the study of why things in the physical universe happen as they do:
why some processes go to completion, some go partially, and others do not go at all.
Thermodynamics suggests ways to adjust conditions so that chemical reactions go in a
direction that adds value to materials. Theoretical Versus Empirical Relationships
Some of the laws of thermodynamics can be derived from kinetic-molecular theory:
Newton’s laws applied to molecules as they move and collide. Other thermodynamic
equations remain in essence empirical: found by trial and error to predict the behavior of
objects around us.
Though the why of thermodynamics may not be clear in all cases, many of the relationships
in thermodynamics can be expressed by relatively simple equations. If you commit these
fundamentals to memory, you will be able to solve problems. At this point, that’s the goal.
Additional courses in science will increase your understanding. Entropy
All chemical reactions are, in theory, reversible: In a closed system, they all proceed to
equilibrium, though the equilibrium mixture can strongly favor the reactants or the
products.
In determining whether or not a reaction will go, or the direction in which a reaction
mixture will shift, we must consider three factors in the reaction: its change in enthalpy
(ΔH), its temperature (T), and its change in entropy (ΔS).
To begin to understand entropy (symbol S), it helps to have experience with the
mathematics of probability and statistics. Let us begin with words and analogies that may
be more familiar.
Comparing two systems, the system with higher entropy is the one in which the energy is
more dispersed. This is generally the one in which the particles are in arrangements that
are more probable in terms of statistics. This condition may be described as
• more mixed, random, or chaotic; • less well-ordered; • having more freedom of motion; • having more options in terms of positions for the particles to occupy. For example: For a given amount of a substance, with other variables being equal,
• the gas phase has higher entropy than the solid. Why? In the solid, the particles are
locked in place in a highly ordered crystal. In the gas, the particles have more
freedom to move about and mix. • Particles of a gas have a higher entropy in a larger container than a smaller one. In
the larger, they have more freedom to move about and more positions to occupy. © 2011 ChemReview.Net v1r Page 1130 Module 36 — Thermodynamics • Gas particles have more entropy at lower pressure than at higher pressure. At lower
pressure there is more empty space between the particles, so the particles have more
available space to occupy. • In a chemical reaction, the side of the balanced equation that has more gas molecules
(as determined by adding the gas particle coefficients) generally has higher entropy.
More particles offer more options for mixing, and the arrangements of the particles
can be more random. • Larger molecules tend to have more energy than similar smaller molecules. In
larger molecules, there are more ways to distribute the energy of the atoms. Practice A: Learn the rules above, then try these problems. 1. Circle the system in each pair below that has lower enthalpy during the change
between the two phases.
a. Steam or liquid water b. CO2 gas or solid CO2 2. Circle the system or side in each pair below that is predicted to have higher entropy.
a. Ice or liquid water b. CO2 gas or solid CO2 c. At the same temperature: 1 mole of gas at 1 atm or d. At the same temperature: 1 mole of gas in 1.0 liters or e. A deck of cards in order f. 4 heads in a row g. N2O4(g) i. 1 mole of gas in 2.0 L CH4 or C4O10 or
2 NO2(g) or 1 mole of gas at 2 atm a deck of cards shuffled
2 heads and 2 tails in any order
h. CaO(s) + CO2(g) CaCO3(s) Spontaneity
If a chemical process occurs without intervention from its surroundings, the process is said
to be spontaneous. The direction that is spontaneous is the direction that the reaction will
go. In many reactions, the direction that is spontaneous will depend on the temperature
and pressure conditions under which the reaction is run.
In other usages, spontaneous may connote something that happens fast. In chemistry, a
spontaneous process will go, but it may go quickly or slowly. The laws of thermodynamics
determine the direction of a reaction, but the rules of kinetics determine its speed. Spontaneous Direction: The Second Law
All systems tend to go to minimum potential energy. For example, in a vacuum, without
opposing forces, all objects fall toward the center of the earth. As an object falls, it loses
gravitational potential energy.
However, this tendency does not mean that systems will always go to lower potential
energy. © 2011 ChemReview.Net v1r Page 1131 Module 36 — Thermodynamics For example, the gas phase of a substance has higher potential energy than its liquid
phase (Lesson 21A), yet a drop of a liquid left in the open evaporates, going from the
lower potential energy liquid phase to the higher potential energy gas phase.
Why would a system go to higher potential energy? The potential energy change of the
system is a factor, but in many cases it is not the controlling factor deciding whether a
process occurs.
The rule which predicts the direction in which all changes occur is termed
The Second Law of Thermodynamics
A process will be spontaneous if it increases the entropy of the universe.
Other ways to state the second law may help with understanding:
• In a spontaneous process, energy is always dispersed. • During change, the energy of the universe always goes toward the more probable
condition. • Every change increases the entropy of the universe. • Processes that do not increase the entropy of the universe don’t happen. In equation form, the second law can be represented as
ΔSuniverse = ΔSsystem + ΔSsurroundings = Positive for spontaneous change (Eq. 1) One implication of this equation is that during a spontaneous change, a system may or may
not go to higher entropy. If the change takes the system to lower entropy, the change may
be spontaneous if the change takes the surroundings to higher entropy.
Each physical and chemical change can be assigned a positive or negative value for the
resulting change in entropy (ΔS) of the system, its surroundings, and the universe.
• If ΔSuniverse is positive, the change, process, or reaction is spontaneous. • If ΔSuniverse is negative, the change, process, or reaction is spontaneous in the
opposite direction. • If ΔSuniverse is zero, net change does not occur, and the system is at equilibrium. In thermodynamics, we must be attentive to changes in both a system and its surroundings.
However, by convention in chemistry, if a subscript is not attached to an H, S, ΔH, or ΔS
symbol, it is understood that the subscript is system. ΔSsurroundings
If a process or reaction is carried out with no net change in temperature and pressure, the
change in the entropy of the surroundings is determined by the change in enthalpy of the
system and the temperature during the change. This relationship is:
ΔSsurroundings = ─ ΔH
T © 2011 ChemReview.Net v1r when P and T are constant,
and T is absolute temperature. (Eq. 2) Page 1132 Module 36 — Thermodynamics Let’s consider some implications of this equation.
• If a reaction or process is exothermic, the sign of its ΔH is negative. Since absolute
temperature (T) is always positive, for any exothermic reaction the overall sign on
the right side of the equation must be positive, and the change in the entropy of the
surroundings on the left must be positive.
This can be thought of as, if a system loses energy, the surroundings assume a more
probable state, such as becoming more mixed up and random. • The larger is the positive or negative change in enthalpy (ΔH) of a system during a
process, the larger is the positive or negative change in entropy of its surroundings.
However, because the T term is in the denominator, as the temperature at which a
process occurs is increased, the effect of ΔH on ΔSsurroundings decreases. To summarize:
Exothermic processes increase the entropy of the surroundings. Endothermic
processes decrease the entropy of the surroundings.
Increasing temperature decreases the impact of ΔH on ΔSsurroundings.
Equation 2 also supplies units that measure a change in entropy. Try this example.
Q. When referring to one substance, the SI units of ΔH are joules per mole. For a process
involving one substance, in equation 2, what must be the units of ΔS?
*****
Since T is measured in kelvins, and the units on both sides of an equation must be the
same:
ΔSsurr. = ─ ΔH = ─ ΔH • 1
T
T = joules •
1
= J/mol·K
mole kelvins or J·mol─1·K─1 In general:
The SI units of ΔS when referring to one substance are joules per (mol · K) ,
written as J/mol·K or J·mol─1·K─1.
In the case of a chemical reaction involving more than one substance, the SI unit of ΔS
is written as J/K or J·K─1, with reaction coefficients understood to be in moles.
Use equation 2 for the following calculation.
Q. For the melting of one mole of ice: H2O(s) H2O(l) ΔH = + 6.03 kJ/mol Find ΔSsurroundings for melting carried out at 0ºC and standard pressure.
***** © 2011 ChemReview.Net v1r Page 1133 Module 36 — Thermodynamics Answer
WANTED: ΔSsurr.
DATA: ΔH = + 6.03 kJ/mol
0ºC = 273 K T The equation that relates these symbols, when P and T are constant, is
ΔSsurr. = ─ ΔH
T Practice B: = ─ 6.03 kJ •
1
= ─ 0.0221 kJ/mol·K = ─ 22.1 J/mol·K
mol
273 K Learn the rules and equations above, then try these problems. 1. To change 22.0 grams of dry ice (solid CO2) to its gas phase at ─78ºC and 1 atm
pressure, 7.65 kJ of heat is added.
a. Is the ΔH for this reaction positive or negative?
b. What is the ΔH value for this sublimation of dry ice, in kJ/mol?
c. What is the ΔSsurr. for this process?
2. For an endothermic process,
a. Will the sign of ΔSsurr. be positive or negative? Explain your reasoning.
b. How will ΔSsurr. change at higher temperature?
3. If x = ─ energy/time, what are the SI units of x ? Predicting Spontaneity From the Change to the System
In equation form, the second law of thermodynamics is stated as
ΔSuniverse = ΔSsystem + ΔSsurroundings = Positive for spontaneous change (Eq. 1) If we substitute equation 2 into equation 1, we can write
ΔSuniverse = ΔSsystem ─ ΔHsystem = Positive for spontaneous change
T (Eq. 3) In this form, all of the changes needed to predict spontaneity are measured from the
perspective of the system: the particles involved in the process.
Note from Equation 3 that as T increases, the value of ΔH/T decreases. This means that
As temperature increases, ΔH becomes relatively less important, and ΔS becomes
relatively more important, in determining whether a process will be spontaneous. © 2011 ChemReview.Net v1r Page 1134 Module 36 — Thermodynamics Using equation 3, write answers to the following question.
Q1. If a reaction is exothermic, and the products are more random,
a. Will ΔH/T be positive or negative? b. Will the reaction be spontaneous? *****
Part a. If the reaction is exothermic, ΔH must be a negative number.
T is absolute temperature which is always a positive number.
A negative ΔH over a positive T must equal a negative number. If needed, adjust your work and complete part b.
*****
Part b. Since ΔH/T in part a is a negative number, ─ ΔH/T must be a positive number.
If the products of a reaction are more random, the system has gained entropy,
and the change in the system’s entropy (ΔS) is a positive number.
Since ΔS is positive and ─ ΔH/T is positive, their sum must be positive,
which means that ΔSuniverse is positive and the process is spontaneous. Apply the logic of equation 3 to answer this question. Write your reasoning.
Q2. If a reaction is endothermic, and the products are more random, will the reaction be
spontaneous?
*****
If the reaction is endothermic, ΔH must be positive. T is always positive.
A positive ΔH over a positive T must be positive, so ─ ΔH/T must be negative.
If the system becomes more random, it gains entropy, so the change in the system’s
entropy (ΔS) is positive.
If ΔS is positive and ─ ΔH / T is negative, whether the sum is positive or negative will
depend on which is the larger number: the positive entropy change or the number after
the minus sign in ─ ΔH / T. The answer is: whether the reaction is spontaneous will
depend on the values of ΔS, ΔH, and T.
By applying this “logic of signs” to the other two possibilities of signs for ΔH and ΔS, we
can summarize with these general rules:
Systems tend to go to lower enthalpy and higher entropy.
• If a process takes a system to both lower enthalpy (is exothermic) and higher
entropy, the process is always spontaneous (always goes toward the products).
• If a process takes a system to higher enthalpy (is endothermic) and lower entropy,
the reactants are favored, and the process will not go.
• If there is a conflict between the tendencies to lower enthalpy and higher entropy,
the magnitude of ΔH, ΔS, and T will determine whether the reaction goes. © 2011 ChemReview.Net v1r Page 1135 Module 36 — Thermodynamics Summary: Rules Involving Entropy
From the following, be able to apply the rules and write of the equations from memory.
Rules
1. Entropy (S): Comparing two systems, in the system with higher entropy, the particles
• are in arrangements that are more probable; mixed, random, or chaotic; • are less well-ordered; • have more freedom of motion; • have more options in terms of positions for the particles to occupy. • Gas particles of a substance have higher entropy than solid particles. • Other variables being equal, a gas has more entropy at lower pressure and higher
volume. 2. The second law of thermodynamics: A process will be spontaneous if it increases the
entropy of the universe.
3. The units of ΔS, per mole, are joules per (mol·K), written as J/mol·K or J·mol─1·K─1.
4. Systems tend to go to lower enthalpy and higher entropy.
• If a process takes a system to both lower enthalpy (is exothermic) and higher
entropy, the process is spontaneous (goes toward the products). • If a process takes a system to higher enthalpy and lower entropy, the process will
not go forward, but instead goes backwards. • If there is a conflict between the tendencies to lower enthalpy and higher entropy,
the magnitude of ΔH, ΔS, and T will determine whether the process goes. Equations
Eq. 1. ΔSuniverse = ΔSsystem + ΔSsurroundings = Positive for spontaneous change Eq. 2. ΔSsurroundings = ─ ΔHsys.
T Eq.3. ΔSuniverse = ΔSsystem ─ ΔHsys. = Positive for spontaneous change
T Practice C: when P and T are constant, and T is absolute. (for reaction in which P and T are constant) Write your answer and reasoning. 1. As you play solitaire with a deck of shuffled cards, what is the sign of ΔS for the cards?
What is the sign of ΔH for your body?
2. If a reaction at constant temperature and pressure is endothermic, and the products are
less random,
a. will the reaction be spontaneous?
b. Will the reverse reaction be spontaneous? © 2011 ChemReview.Net v1r Page 1136 Module 36 — Thermodynamics 3. If a reaction at constant temperature and pressure is endothermic and the products have
higher entropy, will the reaction go? ANSWERS
Practice A
1a. liquid water: the gas phase of a substance has the highest PE. During a phase change, temperature is
constant, and ΔH is approximately equal to the change in potential energy.
1b. Solid. Potential energy must be added to change a solid to a gas.
2a. liquid water: the liquid phase has less structure than the solid.
2b. gas: The molecules in the gas phase have more freedom of motion.
2c. 1 atm. 2d. 2 liters 2e. The shuffled deck has less order. 2f. 2 heads and 2 tails is more probable. 2g. The right side, with twice as many particles, has more possible arrangements.
2h. The left side. Entropy is usually higher on the side with more gas molecules.
2i. For molecules with similar composition, the larger molecule (C4H10) will generally have higher entropy. Practice B
1a. If energy is added to the system, as is always true to change a solid substance to its gas phase, ΔH must
be positive.
1b. WANTED:
DATA: ΔH in kJ/mol (use kJ to be consistent with the unit in the DATA) 7.65 kJ added = 22.0 g CO2 sublimated
44.0 g CO2 = 1 mol CO2 SOLVE:
1c. WANTED:
DATA: (2 measures of same process)
(g CO2 in data = grams prompt) kJ =
7.65 kJ • 44.0 g CO2 = + 15.3 kJ
= ΔH
mol
22.0 g CO2
1 mol CO2
mol CO2
ΔSsurr.
ΔH = + 15.3 kJ/mol ─78ºC = 195 K T The equation that relates these symbols, when P and T are constant, is
ΔSsurr. = ─ ΔH
T = ─ 15.3 kJ • 1
= ─ 0.0785 kJ/mol·K = ─ 78.5 J/mol·K
mol 195 K 2a. For an endothermic process, ΔH is positive, and T can only be positive, so ─ ΔH/T must be negative, and
that equals ΔSsurr. .
2b. As T increases, the magnitude of ΔSsurr. (the value of the number after its sign, whether positive or
negative) decreases. For an endothermic process, ΔSsurr. is negative, but as temperature increases it will
be less negative.
3. The units must be the same on both sides of an equation. Since the SI unit for energy is joules and time is
seconds, the units of x must be joules/second. © 2011 ChemReview.Net v1r Page 1137 Module 36 — Thermodynamics Practice C
1. In solitaire, you take a random deck and put it into order. The entropy of the cards decreases, so ΔS is
negative. As you order the cards, your body must burn energy. Burning is an exothermic reaction with a
negative ΔH.
2a. If the products are less random, ΔSsystem is a negative number. If the reaction is endothermic, ΔHsystem
is positive, (ΔHsystem/T) must be positive, and ΔSsystem ─ (ΔHsystem/T) is negative minus a positive,
which must be more negative. Since the right side is negative, ΔSuniverse is negative, and the process is
not spontaneous.
2b. When the process goes backwards, all of the Δsystem signs reverse. For reverse process, ΔSuniverse is
positive and the process is spontaneous.
3. If the reaction is endothermic, ΔHsystem is positive, and (ΔHsystem/T) must be positive. If the products
are more random they have higher entropy, and ΔSsystem is a positive number. ΔSsystem ─
(ΔHsystem/T) is then a positive minus a positive. Whether the result is positive or negative depends on
which of the positive terms is larger. Without more information, you can’t say whether the reaction will go.
***** Lesson 36C: Free Energy
Defining Free Energy
Our second law equation:
ΔSuniverse = ΔSsystem + ΔSsurroundings = Positive for spontaneous change
is true for all changes. When measuring chemical reactions and processes, we generally try
to hold temperature and pressure constant. If we do, the equation
ΔSuniverse = ΔSsystem ─ ΔHsystem = Positive for spontaneous change
T (Eq. 3) allows us to predict which processes will be spontaneous based on measurements of the
system alone.
A third way to predict spontaneity is to define a function termed free energy (symbol G)
that combines enthalpy, entropy, and temperature:
G = H ─ TS (Equation 4) Free energy is also called Gibbs free energy, after the 19th century Yale physicist J. Willard
Gibbs whose work laid the foundation for much of modern thermodynamics.
In chemical processes, the quantity of interest is the change in the free energy of a system,
defined as
ΔG = ΔH ─ TΔS at constant temperature. (Equation 5) ΔG is the energy of the system that is free (available) to do work. Each Δ symbol is
measured from the perspective of the particles (the system), so by convention the subscripts
are omitted.
© 2011 ChemReview.Net v1r Page 1138 Module 36 — Thermodynamics To find the SI units of ΔG, apply the laws of dimensional homogeneity:
• Terms to be added or subtracted must have the same dimensions (units); and • On both sides of an equation, the units must be the same. Based on the equation above, the units of the ΔH, ΔG, and ─ TΔS terms must be the same.
In energy relationships, coefficients and amounts are understood to be moles.
These rules mean
The SI units attached to ΔH and ΔG may always be joules, with “per moles shown”
understood.
When referring to one mole of one substance, the units of ΔH and ΔG may be stated as
either joules or joules/mol . Using ΔG To Predict Direction
ΔG values quickly indicate whether a chemical process is spontaneous. To explore why,
begin by solving ΔG = ΔH ─ TΔS for ΔS, then check your answer below. *****
There are several ways to do the algebra. One is
ΔG = ΔH ─ TΔS
ΔG ─ ΔH = ─ TΔS
ΔS sys. = ΔG ─ ΔH
─T = ─ (ΔG ─ ΔH) = ΔH ─ ΔG
─(─T)
T = ΔH ─ ΔG
T
T Now substitute the final terms on the right above in place of ΔSsys. in Eq. 3, then simplify.
*****
ΔSuniverse = ΔSsys. ─ ΔHsys. = Positive for spontaneous change
Eq. 3:
T
when P and T are constant
ΔSuniverse = ΔH ─ ΔG
T
T
Or ─ ΔHsys. = ─ ΔG = Positive for spontaneous change
T T when P and T are constant ΔSuniverse = ─ ΔG = Positive for spontaneous change
T
when P and T are constant (Eq. 6) For a process at constant T and P to be spontaneous, what must be the sign of ΔG ?
*****
T is always positive. ─ ΔG / T must be positive for spontaneous change, and that will
happen only if ΔG is negative. © 2011 ChemReview.Net v1r Page 1139 Module 36 — Thermodynamics Based on this “logic of signs,” we can write three general rules:
Using ΔG to Predict Direction: For a process at constant temperature and pressure,
if ΔG is negative, the process will be spontaneous (go toward the products).
• if ΔG is positive, the process will not go or go backward (favors the reactants).
if ΔG is zero, no net change will occur (the process is at equilibrium). Using ΔG = ΔH ─ TΔS , the temperature at which a process will be at equilibrium can be
calculated if ΔH and ΔS are known. Try the following example.
Q. For the melting of ice at 1 atm pressure, ΔH = +6.03 kJ/mol and ΔS = + 22.1 J/mol·K .
Calculate the temperature at which this phase change will be at equilibrium.
*****
Answer: To solve with equations:
1. List the WANTED unit and its symbol, list the DATA, label each item of DATA with
symbols based on its units, then solve using an equation that includes both the
WANTED and DATA symbols.
2. In the DATA table, convert to consistent units.
In thermodynamics problems, the supplied DATA units are often not consistent. If you
write a DATA table and convert to consistent units in the table before you start to solve, it
will simplify your work.
*****
WANTED: temperature at equilibrium DATA: T ΔH = +6.03 kJ/mol
ΔS = + 22.1 J/mol·K = + 0.0221 kJ/mol·K As the consistent energy unit, you may convert to either J or kJ. Converting to the
larger unit (kJ in this case) will simplify significant figures.
RULE: The system is at equilibrium when ΔG = 0 The equation that uses those four symbols is ΔG = ΔH ─ TΔS When the value of a variable is zero, a quick way to solve is to substitute the zero
into the fundamental equation, solve for the WANTED symbol in symbols, then
substitute the data.
0 = ΔH ─ TΔS
ΔH = TΔS
? = T = ΔH
ΔS = 6.03 kJ • mol·K
= 273 K
mol
0.0221 kJ This answer means that if you have ice and water at 273 K and 1 atm pressure, the
molecules will neither freeze nor melt. © 2011 ChemReview.Net v1r Page 1140 Module 36 — Thermodynamics Within the data uncertainty, this answer is consistent with what we know about
phase changes. Since 273.15 K (0ºC) is the temperature at 1 atm pressure where
solid and liquid water coexist, if the surrounding temperature is also kept at 273.15
K, heat will neither enter nor leave the water molecules, and a mixture of ice and
liquid water will remain at equilibrium.
In general,
For a substance in a closed system, at the T and P of a phase change, equilibrium exists
between the phases, and ΔG = 0 . Practice A: Learn the rules above, then apply the rules to these problems from memory. 1. For the boiling of water at 1 atm pressure and a temperature of 100ºC,
a. If the heat of vaporization is 2260 J/g, what is ΔH for the reaction, in kJ/mol ?
b. If ΔS for the vaporization is 109 J·mol─1·K─1, find ΔG for the reaction.
2. Complete the following table. At constant P and T:
If ΔH is and ΔS is Positive (+) (+) (─) (─) (─) And process goes
fwd, backward, or ? Negative (─) (+) then ΔG is
(+ or ─ or ? ) (+) Shifting the Direction of a Reaction
In chemistry, one of our goals is to be able to cause a reversible reaction to go in the
direction we would like. The form of the equation ΔG = ΔH ─ TΔS helps us to adjust conditions to favor one side of a reaction equation or the other.
According to ΔG = ΔH ─ TΔS , as we lower the temperature of a reaction toward
absolute zero, the term TΔS becomes smaller. If we increase the temperature at which a
reaction is run, the term TΔS becomes larger. Let’s summarize as:
When trying to shift a reaction direction, to make the entropy change more important,
raise the temperature.
This rule is most important in a reaction where the tendencies of enthalpy and entropy
oppose each other.
For example, for the reaction of melting ice: H2O(s) H2O(l) ΔH = +6.03 kJ/mol and ΔS = + 0.0221 kJ/mol·K © 2011 ChemReview.Net v1r Page 1141 Module 36 — Thermodynamics Enthalpy, being positive, opposes the reaction. Entropy, being positive, favors the
reaction. To get the reaction to go, should temperature be increased or decreased?
*****
Since entropy favors what we want, to make entropy more important, raise the
temperature. This matches our experience with this reaction, that to get ice to melt,
we raise its temperature.
To measure how temperature makes a difference, try this calculation.
Q. Use the data above to calculate ΔG for the melting of ice, first at ─10ºC, then at +10ºC.
*****
Answer
at ─10ºC, ΔG = ΔH ─ TΔS = + 6.03 kJ/mol ─ (263 K)(+ 0.0221 kJ·mol─1·K─1) = + 6.03 kJ/mol ─ 5.81 kJ·mol─1 = + 0.22 kJ/mol
at +10ºC, ΔG = ΔH ─ TΔS (positive ΔG = no go) = + 6.03 kJ/mol ─ (283 K)(+ 0.0221 kJ·mol─1·K─1) = + 6.03 kJ/mol ─ 6.25 kJ·mol─1 = ─ 0.22 kJ/mol (negative ΔG = reaction goes) This, too, matches our experience: ice does not melt at ─10ºC, but does melt at +10ºC. Summary: Rules Involving Free Energy
1. G is free energy. ΔG is the energy of the system that is available to do work.
2. G = H ─ TS and ΔG = ΔH ─ TΔS at constant temperature. 3. When referencing a single substance, the unit of ΔG may be joules or joules/mole.
When referencing more than one substance, the unit of ΔG is joules .
4. For a process at constant temperature and pressure:
if ΔG is negative, the process will be spontaneous (go toward the products).
• if ΔG is positive, the process will go backward or not go (favors the reactants).
if ΔG is zero, no net change occurs (the process is at equilibrium). 5. For a substance in a closed system, at the T and P of a phase change, equilibrium exists
between the phases, and ΔG = 0 .
6. To shift a reaction direction by making the entropy change more important, increase the
temperature. Practice B
1. To favor the products of a reaction, would you raise or lower the temperature if
a. ΔH is negative and ΔS is negative?
b. ΔH is positive and ΔS is negative? © 2011 ChemReview.Net v1r Page 1142 Module 36 — Thermodynamics 2. To boil water: H2O(l) H2O(g) a. What must be the sign of ΔH? b. What must be the sign of ΔS? c. Will increasing temperature make ΔG more positive or more negative?
3. To boil water at 1.0 atm pressure, at what temperature does ΔG change from positive to
negative? ANSWERS
Practice A
1a. WANTED: ΔH in kJ/mol
Heat of vaporization is a ΔH, but this ΔH is in J/g. Convert to the WANTED unit.
2,260 J = 1 g H2O sublimated DATA: (2 measures of same process) 18.0 g H2O = 1 mol H2O (g H2O in data = grams prompt) • 18.0 g H2O • 1 kJ = + 40.7 kJ
1 mol H2O
mol H2O
103 J
If energy from the surroundings must be added, as is always true to change a liquid to its gas phase, ΔH
must be given a positive sign.
kJ =
mol SOLVE: 1b. WANTED: 2,260 J
1 g H2O ΔG in kJ/mol DATA: (the units of ΔG and ΔH are the same) + 40.7 kJ/mol = ΔH
100ºC = 373 K = T
109 J·mol─1·K─1 = ΔS = 0.109 kJ·mol─1·K─1 (make units consistent in DATA table) SOLVE: The equation that uses all four of those symbols is
ΔG = ΔH ─ TΔS
ΔG = + 40.7 kJ/mol ─ (373 K)(+ 0.109 kJ·mol─1·K─1)
= + 40.7 kJ/mol ─ (40.7 kJ·mol─1)
= 0 kJ/mol = ΔG Is the answer reasonable? At the boiling point of a liquid, in a closed system, equilibrium exists between
the liquid and gas phases. At equilibrium, ΔG = 0 .
2. The equation that relates these variable is ΔG = ΔH ─ TΔS
If ΔH is and ΔS is then ΔG is
( + or ─ o r ? ) And process goes
fwd, backward, or ?
backward (─) +
?
? (+) ─ forward Positive (+) Negative (─) (+) (+) (─)
(─) © 2011 ChemReview.Net v1r , and T must be positive. ?
? Page 1143 Module 36 — Thermodynamics Practice B
1. To encourage a reaction to go, you want ΔG to be negative. ΔG = ΔH ─ TΔS a. When ΔH helps and ΔS hurts, decrease the importance of ΔS by lowering T.
b. If ΔH is positive and ΔS is negative, the (─ TΔS) term must be positive. Since both terms are
positive, and you need the result to be negative, the reaction does not go. Lowering the temperature
will make ΔG a smaller positive number, and may shift a mixture toward the products a bit, but the
mixture will favor the reactants.
2a. To boil water, heat must be added. The reaction is therefore endothermic, and ΔH must be positive.
2b. When a substance goes from a liquid to a gas, entropy increases. ΔS must be positive.
2c. From experience, we know that adding heat tends to make boiling go. Increasing temperature therefore
tends to make the ΔG of boiling negative.
From the equation ΔG = ΔH ─ TΔS
likely to be negative is to increase T. , since ΔH and ΔS are both positive, the way to make the ΔG more 3. At 1.0 atm, water starts to boil at 100ºC. As temperature changes, for a chemical equation that is a phase
change, ΔG will switch between negative and positive (have a value of zero) at the temperature of the
phase change (such as the melting point or boiling point). At that temperature, if the system is closed (no
heat enters or leaves), the two phases are at equilibrium.
***** Lesson 36D: Standard State Values
The rules for standard states were covered in Lesson 22D. Let us briefly review. Standard States
For thermodynamic measurements, for substances to be in their standard state,
• elements must be at 25ºC and 1 atm pressure and (in most cases) in the phase and/or
solid structure that is most stable. • Compounds that are gases must be at one atmosphere pressure. • Substances in solutions must have a concentration of one mol/L. • Solid and liquid compounds are in their standard state under nearly all conditions if
they are in the form that is most stable at 25ºC and 1 atm pressure. Standard conditions in thermodynamic calculations should assume a temperature of 25ºC
unless otherwise noted. Standard State Enthalpy Values
Each substance has a characteristic value for its enthalpy of formation, symbolized as ΔHºf .
The º symbol means that the values apply to a substance in its standard state.
Absolute enthalpy cannot be measured, but we can measure changes in enthalpy (∆H). To
design a scale to calculate ∆H, each element in its standard state is arbitrarily assigned a
ΔHºf value of zero kJ/mol. This choice for zero is arbitrary but convenient: in calculations,
we can assign a ∆Hºf value of zero to every element by inspection.
© 2011 ChemReview.Net v1r Page 1144 Module 36 — Thermodynamics The standard state enthalpy (∆Hºf) of each compound is then defined as the ∆H of its
formation reaction: the reaction in which one mole of the compound is formed from its
elements, with all substances in their standard states.
Example: The formation equation for carbon dioxide gas is
C(s) + O2(g) CO2(g) ∆H = ΔHºf = ― 393.5 kJ/mol The standard state enthalpy (∆Hºf) of a compound is the difference between its enthalpy and
the ∆Hºf = zero of its elements. A ∆Hºf value does not change with changes in temperature
and pressure as long as the substance remains in the same phase.
∆Hºf has a different value if the substance is in a different phase: its gas phase has a higher
enthalpy than its liquid phase, which has a higher enthalpy than its solid phase.
The symbol ΔH may be attached to any reaction. An enthalpy change may be given the
symbol ΔHº if all substances involved are in their standard states. The symbol ΔHºf
applies only to the heat of formation of a substance: the enthalpy change of its formation
reaction. Standard State Entropy Values
Substances can also be assigned standard entropy (Sº) values. To do so, we start from the
Third Law of Thermodynamics
The entropy (S) of a perfect crystal of a substance is zero at absolute zero (0 K).
This definition of a zero point for entropy is logical. In a crystal with no deformities, the
particles of a substance are in their most ordered state possible.
In addition, at absolute zero, the motion of particles (their translation, vibration, and
rotation) is zero.
Unlike ∆H values, entropy values change with temperature. Compared to its perfect
crystal at absolute zero, the disorder of a substance increases as its temperature increases
because the motion of its particles increases. Motion makes it possible for the particles to
move to more positions, which increases their entropy.
Since any substance in a real situation will be warmer than absolute zero (which cannot be
reached), it will always have more motion than at absolute zero. Entropy values for a
substance in a real situation must therefore be higher than zero: positive numbers.
Every substance can be assigned a standard entropy value Sº that represents the entropy of
the substance at 25ºC and 1 atm pressure. Though Sº values will always be positive, the
change in S (ΔS) during a reaction or process may be positive or negative.
Tables of the thermodynamic characteristics of substances list ∆Hºf and Sº values for
compounds. The use of the Δ in front of the Hº but not the Sº is an indication that the
enthalpy values are relative to an arbitrary zero, while entropy values are relative to a
logical zero. © 2011 ChemReview.Net v1r Page 1145 Module 36 — Thermodynamics Since entropy varies with temperature, it is important to distinguish between generic
entropy S and standard state entropy Sº. However, ∆H and a ∆Hº values for a process will
usually be the same, since ∆H values are not temperature dependent. Calculating Thermodynamic Changes During Reactions
If table values for ΔHºf and Sº for substances are known, the ΔH and ΔSº for chemical
reactions can be calculated. Two fundamental equations are the state functions:
ΔH = ΔHfinal ─ ΔHinitial and (Equation 7) ΔSº = Sºfinal ─ Sºinitial (Equation 8) In state functions, only the initial and final values matter.
When applying the equations above to chemical reactions,
• ΔHfinal = sum of (coefficient times ΔHºf product) terms = ∑cproducts ΔHºf products • Sºinitial = the sum of the (coefficient times Sºreactant) terms = ∑creactants Sºreactants To find the overall ΔH and ΔSº changes in a reaction, we can use
ΔHreaction = ∑cproducts ΔHºf products ─ ∑creactants ΔHºf reactants (Equation 9) ΔSºreaction (Equation 10) = ∑cproducts Sºproducts ─ ∑creactants Sºreactants The meaning of these equations will be more clear when applied to an example.
Q. For the reaction
2 CO(g) + O2(g) (kJ/mol) Substance
2 CO2(g) ΔHºf Sº
(J/mol·K) carried out at 25ºC and 1 atm pressure,
based on the table at the right, calculate CO(g) ─ 110.5 198 a. ΔHreaction O2(g) 0 205 b. ΔSºreaction CO2(g) ─ 393.5 214 c. ΔGºreaction Use equations 9 and 10 above. If you need a hint, read a part of the answer below, then
complete the problem.
*****
a. WANTED: ΔHreaction
ΔHreaction = ∑ cproducts ΔHºf products ─ ∑ creactants ΔHºf reactants
= (2)( ΔHºf of CO2(g) ) ─ [ (2)(ΔHºf of CO(g)) + (1)(ΔHºf of O2(g)) ]
= (2 mol)( ─ 393.5 kJ/mol) ─ [ (2)( ─ 110.5) + (1)(0) ]
= ( ─ 787.0 kJ) ─ [( ─ 221.0) ] = ─ 566.0 kJ For reactions, coefficients are in moles and ΔH values are in J or kJ . © 2011 ChemReview.Net v1r Page 1146 Module 36 — Thermodynamics b. WANTED: ΔSºreaction
ΔSºreaction = ∑cproducts Sºproducts ─ ∑creactants Sºreactants = (2 mol )( 214 J/mol·K) ─ [ (2)(198) + (1)(205) ]
= ( 428 ) ─ [ 601 ] = ─ 173 J/K
For reactions, ΔS values are in J/K (or kJ/K) , with per moles shown understood.
Does the negative value for the entropy change makes sense? The rule is that the side with
more gas molecules has more entropy. In this reaction, we go from three gas molecules to
two. Going to fewer gas molecules, entropy should decrease, and it does.
c. WANTED: ΔGºreaction
Strategy: We know ΔH and ΔS for the reaction from parts a and b. What
known equation relates ΔG, ΔH, and ΔS? *****
ΔG = ΔH ─ TΔS
ΔG = ΔH ─ TΔS is true for all values, whether in standard states or not.
= ─ 566.0 kJ ─ (298 K)( ─ 173 J/K)
= ─ 566.0 kJ ─ (─ 51,600 J)
= ─ 566.0 kJ + 51,600 J These last two numbers cannot be added directly. If you leave out the units, you may miss
that the units are not consistent, and you may add them by mistake.
To add or subtract two values, they must have the same units (see Lesson 2D). In most
thermodynamic calculations, the arithmetic and significant figures will be easier if you
• include the units, watch the units, and • early in each calculation, convert all J to kJ (by moving the decimal 3 places left). For relatively simple equations, you do not need to write a DATA table with every
equation, but if you do not, you should convert to consistent units by inspection as early as
possible in the calculation. Whenever unit manipulations become complex, it is best to
write out a DATA table under each equation, then convert to consistent units in the table
before attempting to solve.
In the last addition line above, converting J to kJ by inspection, then finish.
*****
ΔG = ─ 566.0 kJ + 51,600 J
= ─ 566.0 kJ + 51.6 kJ = ─ 514.4 kJ = ΔGº Since all of the calculations in parts a, b, and c above are based on ΔHº and Sº values, the
calculated values of ΔG ΔH, ΔS are also ΔHº, ΔSº and ΔGº values, the values for these
variables at 1 atm pressure and 25ºC. © 2011 ChemReview.Net v1r Page 1147 Module 36 — Thermodynamics Summary: Rules For Standard State ΔH and ΔS Calculations
1. The entropy (S) of a perfect crystal of a substance is zero at absolute zero (0 K). 2. ΔHreaction = ∑ cproducts ΔHºf products ─ ∑ creactants ΔHºf reactants
3. ΔSºreaction Practice:
1. = ∑ cproducts Sºproducts ─ ∑ creactants Sºreactants Learn the rules above, then try the problem. ─ 26.0 kJ + 29.0 J = 2. For the Haber Process reaction
N2(g) + 3 H2(g) 2 NH3(g) (kJ/mol) Substance carried out at 25ºC and 1 atm pressure, ΔHºf Sº
(J/mol·K) a. Based on the equation above, predict
whether the change in entropy will
be positive or negative. Explain
your reasoning. N2(g) 0 192 H2(g) 0 131 b. Use the table values at the right to
calculate NH3(g) ─ 46 193 i. ΔHreaction ii. ΔSºreaction iii. ΔGºreaction c. Was your prediction for the sign of the change in entropy correct? ANSWERS
1. ─ 26.3 kJ + 29.0 J = ─ 26.3 kJ + 0.0290 kJ = ─ 26.271 kJ = ─ 26.3 kJ
(When adding, round to the highest place with doubt.) 2. a. The reaction goes from 4 moles of gas to 2 moles. Entropy should decrease; the change in
entropy (ΔS) should be a negative number. bi. WANTED: ΔHreaction ΔHreaction = ∑ cproduct ΔHºf product ─ ∑ creactant ΔHºf reactant
= (2 mol)( ─ 46 kJ/mol) ─ [ (1)(0) + (3)( 0) ]
= ( ─ 92 kJ) ─ [ 0 ] = ─ 92 kJ For reactions that make more than one mole of one product, coefficients are in moles and ΔH
values are in J or kJ .
bii. WANTED: ΔSºreaction S values are temperature dependent, but since this reaction is run at standard thermodynamic
conditions, and the values in the table are Sº values, the table values may be used to solve. © 2011 ChemReview.Net v1r Page 1148 Module 36 — Thermodynamics ΔSºreaction = ∑ cproduct Sºproduct ─ ∑ creactans Sºreactant
= (2 mol)( 193 J/mol·K) ─ [ (1)(192) + (3)( 131) ]
= ( 386 J/K ) ─ [ 192 + 393 ] = 386 ─ 585 = ─ 199 J/K For reactions that make more than one mole of one product, coefficients are in moles and ΔS
values are in J/K , per moles shown.
biii. WANTED: ΔGºreaction Strategy: We know ΔH and ΔS for the reaction from parts a and b. What known equation
relates ΔG, ΔH, and ΔS? ΔG = ΔH ─ TΔS = ─ 92 kJ ─ (298 K)( ─ 199 J/K)
= ─ 92 kJ ─ (─ 59,300 J)
= ─ 92 kJ + 59.3 kJ
ΔG = ─ 33 kJ = ΔGº The reaction made 2 moles of ammonia. Per mole, this ΔG would be ─16 kJ/mol .
Since all values were for substances in their standard states, this ΔG is also a ΔGº.
c. ΔSreaction was predicted to be negative, and it is. * * ** * Lesson 36E: Adding ∆G° Equations
Prerequisites: If you have any difficulties with this lesson, review Lesson 22C on Hess’s
Law and 22D on formation reactions.
*****
ΔGº Versus ΔH
ΔGº values for chemical reactions were calculated in the previous lesson using table values
for ΔHº and Sº. A second way to calculate ΔGº values for reactions is to add the ΔGºf
values for substances.
Both ΔH and ΔG are state functions: dependent on only the characteristic values of the
initial state (reactants) and final state (products). This means that many of the rules that
apply to ΔH calculations can be applied to ΔG calculations.
If the ΔGº of a reaction is known, ΔGºreaction values
• have the same units as ΔH; • are written after the reaction with a positive or negative sign; • change sign when the reaction is written backwards; • multiply when coefficients are multiplied; • can be added in the same way that ΔH values were added using Hess’s law, and © 2011 ChemReview.Net v1r Page 1149 Module 36 — Thermodynamics • can be calculated by the summation method using
ΔGºreaction = ∑cproduct ΔGºf product ─ ∑creactant ΔGºf reactant (Eq. 11) Also similar to ΔH,
• the ΔGºf for elements in their standard state (at 25ºC and 1 atm pressure) is defined
as zero kJ/mol , and • the ΔGºf for compounds, which can be found in tables, is equal to the change in
free energy that occurs in the formation reaction: in which one mole of the
compound is formed at 25ºC and 1 atm from elements in their standard state. As in ΔH calculations, the SI units of ΔG that are attached to chemical reactions:
• Assume that coefficients are in moles; • Are written as J or kJ , with “per moles shown” understood; and • May also be written as J/mol or as kJ/mol if referencing one mole of one substance. Unlike ΔH values, ΔG values change with temperature. However, since ΔGº values must
be measured at 25ºC and 1 atm pressure, temperature is a factor in ΔG calculations, but is
not a factor in ΔGº calculations.
The above rules can be summarized by this general rule:
For calculations reversing, multiplying, or adding reaction equations with ΔGº
values attached, follow the rules for ΔH.
Try this problem. If you need a hint, read a part of the answer below, then finish.
Q. If 2 SO2 (g) + O2(g) 2 SO3(g) ΔGº = ─ 142 kJ and ΔGºf for SO2(g) = ─ 300. kJ , calculate ΔGºf for SO3(g) .
*****
Hint 1: When given a mixture of ΔGº equations and ΔGºf values, use the Hess’s law method to solve.
*****
Hint 2: Write out the formation equations and attach their ΔGºf values. *****
Hint 3: WANT: 1/8 S8(s) + 3/2 O2(g) 1 SO3(g) 1/8 S8(s) + 3/2 O2(g) SO3(g) ΔGº = ? (formation rxn.) *****
Hint 4: WANT: ΔGº = ? -------------------------1/8 S8(s) + 1 O2(g) 1 SO2(g) ΔGº = ─ 300. kJ ***** © 2011 ChemReview.Net v1r Page 1150 Module 36 — Thermodynamics WANT: 1/8 S8(s) + 3/2 O2(g) 1 SO3(g) ΔGº = ? -------------------------1/8 S8(s) + 1 O2(g)
1 SO2(g) (formation rxn.) ΔGº = ─ 300. kJ (formation rxn.) 1 SO2 (s) + 1/2 O2(g) 1 SO3(g) ΔGº = ─ 71 kJ
(1/2 of given rxn.)
_________________________________________________
1 SO3(g)
ΔGº = ─ 371 kJ/mol = ΔGºf
1/8 S8(s) + 3/2 O2(g)
Since the WANTED reaction makes a single mole of product, the unit may be kJ or kJ/mol.
Note that in combustion reactions, if we do not worry about O2 until the final adding of the
reactions, the final O2 coefficient serves as a check that the equations are added correctly. Practice:
1. If Learn the rules above, then try these problems. N2(g) + 4 H2(g) + Cl2(g) 2 NH4Cl(s) ΔGº = ─ 406 kJ , find ΔGºf for NH4Cl(s) .
2. Use the values in the table to find ΔGº for
2 NO(g) + O2(g) Substance 2 NO2(g) ΔGºf (kJ/mol) NO(g) + 87 NO2(g) + 52 3. Given this ΔGº for the burning of methyl alcohol,
CH3OH(l) + 3/2 O2(g) 1 CO2(g) + 2 H2O(g) plus the free energies of formation for the
compounds on the right, find the free energy of
formation of the methyl alcohol. ΔGº = ─ 686 kJ
Substance ΔGºf (kJ/mol) H2O(g) ─ 229 CO2(g) ─ 394 ANSWERS
1. The given reaction makes 2 moles of NH4Cl(s). A ΔGºf will be for the formation reaction of NH4Cl(s),
which is the given reaction multiplied by 1/2. ΔGºf = 1/2 x ─ 406 kJ = ─ 203 kJ/mol
2. As with ΔH , there are two ways to find ΔGº : adding the equations by the Hess’s law method, or using the
summation equation. If you know the ΔGºf values (here you do), summation is generally faster.
A key is: elements in their standard state, such as O2(g), have ΔGºf = 0 .
ΔGºreaction = ∑ cproduct ΔGºf product ─ ∑ creactant ΔGºf reactant = (2 mol)( + 52 kJ/mol ) ─ [ (2)( + 87 ) + (1)( 0) ]
= © 2011 ChemReview.Net v1r ( +104 kJ ) ─ [ + 174 ] = ─ 70. kJ Page 1151 Module 36 — Thermodynamics 3. The substance being burned (reacted with O2) is CH3OH .
When given a mixture of ΔGº equations and ΔGºf values, use the Hess’s law method to solve.
WANT: 1 C(s) + 2 H2(g) + 1/2 O2(g)
1 CH3OH(l)
-------------------------1 C(s) + 1 O2(g)
1 CO2(g) + 2 H2O(g) ΔGº = ? (formation rxn.) 1 CO2(g) ΔGº = ─ 394 kJ (formation rxn.) CH3OH(l) + 3/2 O2(g) ΔGº = + 686 kJ (given reversed) ΔGº = ─ 458 kJ
(formation x 2)
2 H2O(g)
2 H2 (g) + 1 O2(g)
_____________________________________________________________
ΔGº = ─ 166 kJ/mol = ΔGºf
1 CH3OH(l)
1 C(s) + 2 H2(g) + 1/2 O2(g)
***** Lesson 36F: Free Energy at Non-Standard Conditions
Prerequisites: You may want to review Lesson 27D on natural logs (ln) to complete the
calculations in this lesson.
***** The Non-Standard Free Energy of Reactions
For a reaction involving either gases that are not at standard (1 atm) pressure or solutions
that are not at 1 mol/L concentration, the free energy change can be calculated using
(Equation 12) ΔGreaction = ΔGºreaction + RT ln(Q) In this equation, Q is the reaction quotient: the value obtained when concentrations or partial
pressures are substituted into the K expression for the reaction (see Lesson 28H).
Let’s learn how to use this equation by example.
Q. Using Equation 12 and the table values at the right,
find the ΔGrxn. at 25ºC for
2 CO(g) + O2(g) 2 CO2(g) with gas partial pressures = 2.0 atm for each
reactant and 10.0 atm for CO2. Substance ΔGºf (kJ/mol) CO2(g) ─ 394 CO(g) ─ 137 Solving with equation 12, especially the first time, let us use our methodical steps.
1. Write the WANTED symbol and unit.
2. Write the equation that uses the WANTED and DATA symbols. Below the equation,
list in a data table each symbol in the equation.
3. Fill in the data table. If needed, either in the data table if it fits, or below the table, solve
to find the value of each symbol in the table. Put a ? after the WANTED symbol.
Do those steps, then check your answer below.
***** © 2011 ChemReview.Net v1r Page 1152 Module 36 — Thermodynamics WANT: ΔGrxn. in kJ (Since the table data is in kJ, pick a consistent WANTED unit) ΔGrxn. = ΔGºrxn. + RT ln(Q) DATA: is the one equation we know using ΔGrxn. A condition for the use of this equation: it solves based the coefficients of the balanced
equation in moles. Though the units used during the calculation may be per mole, for a
reaction, the final units will be J or kJ, with per moles shown by the coefficients as understood.
ΔGrxn. in kJ = ? = WANTED
ΔGºrxn. =
R=
T=
ln(Q) =
Below the DATA table, do any long calculations needed to fill in the DATA.
a. to find ΔG°rxn., use the summation equation. Do that step, then check below.
*****
ΔGºreaction = ∑ cproduct ΔGºf product ─ ∑ creactants Gºf reactant = (2 mol )( ─ 394 kJ/mol) ─ [ (1)(0) + (2)( ─ 137) ]
= (─ 788 kJ ) ─ [─ 274 ] = ─ 514 kJ
The unit for ΔGº for a reaction can be J or kJ , with per moles shown as understood.
b. R = ? We need a value for R. We have used several in the past, including R = 0.0821 L·atm/mol·K = 8.31 kPa·L/mol·K = 62.4 L·torr/mol·K
Which is the best R term to start from in this problem so that units are consistent?
*****
Since the problem has pressure units in atm, start from the R that uses atm.
R = 0.08206 L·atm/mol·K
(To help with rounding, we will use a more precise R here)
However, by the end of a ΔGrxn. problem, to have consistent units, we will need to
convert the unit liter·atm in R to joules. To do so, use the PV-work conversion
1 liter • atmosphere = 101.3 joules
We will need this new R value in every ΔGrxn. problem. Try the conversion now.
Q. Convert 0.08206 L·atm/mol·K to J/mol·K . *****
WANTED = ? J
mol·K © 2011 ChemReview.Net v1r = 0.08206 L·atm •
mol·K 101.3 J = 8.314 J/mol·K 1 L·atm Page 1153 Module 36 — Thermodynamics The rule will be: in ΔGrxn. or other problems that mix joules and R, use
R = 8.31 J/mol·K or 0.00831 kJ/mol·K (Equation 13) In each problem, choose the R value that has units consistent with other DATA.
Which value should be entered in the DATA table for this problem?
*****
R = 0.00831 kJ/mol·K is consistent with the unit supplied. Continue to fill in the DATA table for the equation,
*****
T in K = 25ºC + 273 = 298 K
c. ln(Q) = ? An important rule is: When K or Q are calculated:
• Gas pressures must be measured in atmospheres.
• Concentrations must be measured in moles/liter.
• Concentrations and pressures for solids, solvents, and pure liquids are given
values of 1.
• Values are substituted into K and Q expressions without units.
This rule assures that Q calculations are based on units that are consistent with the
defined standard conditions of one atmosphere pressure and one mole/liter concentration.
Calculate a value for Q in this problem, then check your answer below.
*****
To find Q, substitute the gas partial pressures into the K expression for the reaction.
*****
Q= 2
( P CO2 )
(P CO)2 • (P O2) = *****
Q= 2
( P CO2 )
(P CO)2 • (P O2) = 2
(10.0)
=
2 (2.0)
(2.0) 100.
8 = 12.5 Now finish the data table.
*****
ln(Q) = ln(12.5) = + 2.526
Especially in logarithmic calculations, it is a good idea to carry an extra significant
figure until the final step of a sequential calculation.
With the DATA table now complete, solve for the WANTED symbol.
***** © 2011 ChemReview.Net v1r Page 1154 Module 36 — Thermodynamics ΔGrxn. = ΔGºrxn. + RT ln(Q)
= ─ 514 kJ/mol + (0.00831 kJ/mol·K)(298 K)( + 2.526 ) = ─ 514 kJ/mol + 6.26 kJ/mol = ─ 508 kJ (for reactions, units include per moles shown as understood) Done!
Let’s summarize.
If ΔGrxn. = ΔGºrxn. + RT ln(Q) is the equation, use this DATA table: ΔGrxn. in kJ (per moles shown) =
(find using ∑ with table values or ΔGº = ΔHº ─ TΔSº ) ΔGºrxn. in kJ = R = 8.31 J/mol·K or 0.00831 kJ/mol·K (choose the one with consistent units) T in K = ºC + 273 =
ln(Q) = (use gas P converted to atm, [ ] to M, solid = liquid = 1 , and no units) Practice: Learn the rules in the summary above, then apply the rules to these problems
from memory.
1. Using the table values at the right,
a. find ΔGºrxn. for N2O4(g) Substance
2 NO2(g) b. find the ΔGrxn. at 25ºC, with partial pressures of
101 kPa for N2O4 and 505 kPa for NO2 .
2. In the equation ΔGrxn. = ΔGºrxn. + RT ln(Q) ΔGºf (kJ/mol) N2O4(g) + 98 NO2(g) + 51 as the [products] in a reaction mixture becomes very large compared to the [reactants], state whether the following
will increase or decrease, and state your reasoning.
a. The value of Q
d. The value of ΔGrxn b. The value of ln(Q) c. The value of RT ln(Q) e. The tendency of the reaction to form more products. 3. Under standard-state conditions,
a. all terms substituted into a K expression to calculate Q have what value and units?
b. Under those standard-state conditions, what is the calculated value for Q ?
c. Using the Q value in part b, simplify ΔGrxn. = ΔGºrxn. + RT ln(Q) © 2011 ChemReview.Net v1r . Page 1155 Module 36 — Thermodynamics ANSWERS
1a. To find ΔGº rxn. from table values, use the summation equation.
ΔGºrxn. = ∑ cproducts ΔGºf products ─ ∑ creactants ΔGºf reactants
= (2 mol )( + 51 kJ/mol) ─ [ (1)( + 98)) ]
= (+ 102) ─ [ 98 ] = + 4 kJ 1b. If ΔG rxn. and ΔGº rxn. are in the data, the equation is likely to be
ΔGrxn. = ΔGºrxn. + RT ln(Q) If that is the equation, use this data table: ΔGrxn. in kJ/mol = ?
ΔGºrxn. in kJ/mol = + 4 kJ
R from part a = 0.00831 kJ/mol·K (Choose the R with consistent units) T in K = 298 K
ln(Q) = ? Q = ? = ( PNO2 )2
( PN2O4 ) (substitute gas pressures converted to atm, concentrations to M, solid = liquid = 1 )
? P N O in atm = 101 kPa = 1.00 atm by definition
24 ? P NO2 in atm = 505 kPa • 1 atm = 5.00 atm 101 kPa
K = ( PNO2 )2 = Q =
( PN2O4) (5.00)2 = 25.0
(1.0) ln(Q) = ln(25) = + 3.22
ΔGrxn. = ΔGºrxn. + RT ln(Q)
= + 4 kJ/mol + 0.00831 kJ/mol·K(298 K)( + 3.22)
= + 4 kJ/mol + 7.97 kJ/mol
= + 12 kJ/moles shown (adding; doubt in ones place) 2a. Since Q = [products] / [reactants] , Q gets larger. 2b. As Q increases, ln(Q) increases. 2c. Since R and T are positive, and ln(Q) is positive if Q>1, which it is if [products]>[reactants}, RT ln(Q)
must be positive and get larger as Q increases. 2d. As RT ln(Q) gets to be a larger positive value, the value of ΔG rxn. must increase. It may be a
negative value, but it will shift in a more positive direction. 2e. As ΔG rxn. shifts in a more positive direction, the tendency to shift toward making more products
decreases. All reactions go toward equilibrium: a mix of reactants and products. When [products]
becomes higher than it would be at equilibrium, the direction of the reactant shifts toward reactants. © 2011 ChemReview.Net v1r Page 1156 Module 36 — Thermodynamics 3a. Under standard-state conditions, solution concentrations = 1 M and gas pressures = 1 atm . All solid and
liquid concentrations and pressures are assigned a value of 1 in the Q equation. Since units are omitted
when calculating K and Q , the value for all terms under standard conditions = 1 without units. 3b. Under standard-state conditions, Q is a ratio in which all of the values are 1, so Q = 1 . 3c. ln(Q) = ln(1) = ln(e0) = 0 , so ΔGrxn.= ΔGºrxn. + RT ln(Q) = ΔGrxn.= ΔGºrxn.+ RT(0) = ΔGºrxn. Under standard-state conditions, ΔGrxn. = ΔGºrxn.
***** Lesson 36G: Free Energy and K
Prerequisites: You may want to review Lesson 27D on natural logs (both e and ln
functions) for the calculations in this lesson.
***** ΔG and Equilibrium
For reactions with a very large K, equilibrium so strongly favors the products that a
limiting reactant is close to 100% used up. Reaction calculations can then be handled by the
relatively simple steps of stoichiometry. Stoichiometry solves reaction calculations quickly
when compared to the WRECK steps and/or rice tables needed in K calculations.
However, even for reactions that strongly favor the reactants or products, a very large or
very small K value can be written.
In a closed system,
• all reaction mixtures go to equilibrium, • at equilibrium, ΔG = 0 , and • in a reaction mixture that is at equilibrium, Q = K. Therefore, since ΔGrxn. = ΔGºrxn. + RT ln(Q) at equilibrium, this equation becomes
Or, to simplify: ΔGrxn. = 0 = ΔGºrxn. + RT ln(K) ΔGºrxn. = ─ RT ln(K) (Equation 14) In calculations using gas partial pressures, the K is a Kp. For calculations based on
concentrations, , the K is a Kc. If the K type is unspecified, Kc is understood.
Equation 14 means that knowing values for any two of the three variables in the equation:
ΔGºrxn., T, or K, we can calculate the third.
Try this example.
Q. For a reaction at 25ºC and standard pressure, ΔGºrxn. = ─20.0 kJ. What is the K
value for the reaction?
***** © 2011 ChemReview.Net v1r Page 1157 Module 36 — Thermodynamics WANT: K
The equation that links ΔGºrxn. and K is ΔGºrxn. = ─ RT ln(K)
DATA: ΔGºrxn. = ─20.0 kJ
R = 0.00831 kJ/mol·K
T in K = 25ºC + 273 = 298 K
ln(K) = ?
SOLVE:
ln(K) = ΔGºrxn. = ΔGºrxn. • 1 • 1 = ─20.0 kJ •
mol·K
•1
= + 8.076
─ RT
─R
T
mol
─ 0.00831 kJ 298 K
Separating the terms in the formula helps with unit cancellation (see Lesson 17C).
K = eln(K) = e(8.076) = 3,220 = Since x = eln(x) , 3.22 x 103 = K Finding K from ΔGºf Table Values
Small values for concentrations and pressures can be difficult to determine experimentally.
As a result, values for equilibrium constants can be difficult to determine directly.
Equation 14 is especially important because it allows us to calculate equilibrium constants
from values that can be calculated by other means. If ΔGºf values are known (and many
can be looked up in tables), a ΔGºreaction value can be found using the summation
equation. From ΔGºreaction, a K value can be calculated.
Try this example.
Q. For the Haber Process reaction,
a. find ΔGºrxn. for
N2(g) + 3 H2(g) 2 NH3(g) Substance ΔGºf (kJ/mol) NH3(g) ─ 16.7 b. find the value K value for the reaction at standard temperature.
*****
a. Given free energies of formation, either use the summation reaction.
ΔGºrxn. = ∑cproducts ΔGºf product ─ ∑creactant ΔGºf reactant = (2 mol )( ─ 16.7 kJ/mol) ─ [ (1)(0) + (3)( 0) ]
= (─ 33.4 kJ) ─ [0] = ─ 33.4 kJ per stated moles
or, since this reaction is double the formation equation, double ΔGºf .
b. The equation that links ΔGºrxn. and K is ΔGºrxn. = ─ RT ln(K)
DATA: ΔGºrxn. = ─ 33.4 kJ from part a R = 8.31 J/mol·K = 0.00831 kJ/mol·K © 2011 ChemReview.Net v1r Page 1158 Module 36 — Thermodynamics T in K = 25ºC + 273 = 298 K
ln(K) = ?
*****
SOLVE:
mol·K • 1
=
ln(K) = ΔGºrxn. = ─33.4 kJ •
─ RT
mol
─ 0.00831 kJ 298 K
WANTED = K = eln(K) = e(13.49) = + 13.49 = ln(K) 7.22 x 106 = K To summarize:
If ΔGºrxn. = ─ RT ln(K)
ΔGºrxn. in kJ = is the equation, use this data table:
(find using ∑ with table values or ΔGº = ΔHº ─ TΔSº ) R = 8.31 J/mol·K or 0.00831 kJ/mol·K (use the one with consistent units) T in K = ºC + 273 =
ln(K) = Practice: To find K, first find ln(K), then use K = eln(K) Commit to memory equation 14, then try these problems. 1. If K = 1 for a reaction, calculate ΔGºrxn.. (You should not need a calculator).
2. In mammals, internal temperature is generally 38ºC. At this temperature, ΔGºrxn. for
H2O(l)
H+(aq) + OH─(aq) is 80.5 kJ . Find the value of K for this reaction at
mammalian normal temperature.
3. For Fe(s) + 3/2 O2(g)
Find the value of Sºrxn. Fe2O3(s) , at 25ºC, ΔHºrxn. = ─ 826 kJ and K = 2.6 x 10130 . 4. Complete this problem if you are majoring in the physical sciences or engineering.
If 1 pascal (Pa) = 1 newton/meter2, 1 newton = 1 kg · m · s─2, 1 joule = 1 kg · m2 · s─2,
and one atm = 101 kPa, use those equalities to calculate the number of joules in
one liter · atm. ANSWERS
1. WANT:
Know: ΔGºrxn.
T, K The equation that relates those variables is © 2011 ChemReview.Net v1r ΔGºrxn. = ─ RT ln(K) Page 1159 Module 36 — Thermodynamics DATA:
ΔGºrxn. in kJ = ?
R = 8.31 J/mol·K = 0.00831 kJ/mol·K (pick consistent units) T in K = 25ºC + 273 = 298 K
ln(K) = ln(1) = ln(e0) = 0 ,
This simplifies ΔGºrxn. = ─ RT ln(K) to ΔGºrxn. = ─ RT (0) = 0 = ΔGºrxn. At standard conditions, all terms in a Q equation have a value of 1, so Q = 1. If the value of K is also one,
then when conditions are standard, the system is at equilibrium, and ΔG = ΔGº. = 0 .
2. WANT:
Know: K
T, ΔGºrxn. The equation that relates those variables is ΔGºrxn. = ─ RT ln(K) DATA:
ΔGºrxn. in kJ = + 80.5 kJ
R = 8.31 J/mol·K = 0.00831 kJ/mol·K (pick consistent units) T in K = 38ºC + 273 = 311 K
ln(K) = ΔGºrxn. =
─ RT + 80.5 kJ •
mol·K
•
1 = ─ 31.14
mol
─ 0.00831 kJ
311 K SOLVE: WANTED = K = eln(K) = e(─31.14) = 2.97 x 10─14 = K = Kw At 25ºC, Kw = 1.0 x 10─14 . At higher temperatures, more HOH bonds break, [ions] are higher, and Kw
is higher.
3. WANT: Sºrxn. Know: T, K, ΔHº The equations that relates those variables are ΔGº = ΔHº ─ TΔSº and ΔGºrxn. = ─ RT ln(K) You can combine the equations, but if you choose to solve them separately, the math will be more familiar.
First solve the equation that has one unknown value. Which equation is that?
*****
Since you know R, T, and K, use them find ΔGºrxn.
DATA: ΔGºrxn. = ─ RT ln(K)
ΔGºrxn. in kJ = ?
R = 8.31 J/mol·K or 0.00831 kJ/mol·K
T in K = 25ºC + 273 = 298 K
ln(K) = ln( 2.6 x 10130 ) = 300.
ΔGºrxn. = ─ RT ln(K) © 2011 ChemReview.Net v1r Page 1160 Module 36 — Thermodynamics = ─ 0.00831 kJ/mol·K(298 K)( + 300.)
= ─ 0.00831 kJ/mol·K(298 K)( + 300.)
=
Now solve ─ 743 kJ/mol = ΔGºrxn. ΔGº = ΔHº ─ TΔSº for ΔSº . ΔSº = ΔGº ─ ΔHº = ─ 743 kJ ─ (─826 kJ) = ─ 0.279 kJ/K
─T
─298 K
ΔSº = ─ 279 joules/K , or since one mole of product is formed, ─ 279 joules/mol·K
Either J or kJ can be used in the answer unit.
4. Suggested strategy: Convert all units to SI base units: kg, m, s; cancel units that are not those units.
*****
WANT: ? joules = = ? kg · m2 · s─2
DATA: 1 pascal (Pa) = 1 newton/meter2
1 newton = 1 kg · m · s─2
1 joule = 1 kg · m2 · s─2
one atm = 101 kPa
Hint: Add to the data table: 1 liter = ? dm3, 1 atm = ? Pa *****
SOLVE:
***** ? joule = ? kg · m2 · s─2 = 1 liter · atm 3·
? joules = ? kg · m2 · s─2 = 1 liter · atm ·· 1 dm3 ·
1m
1L
10 dm
*****
? kg · m2 · s─2 = 1 L· atm · 1 dm3 · 1 m3 · 101 x 103 Pa · 1 N · m─2 · 1 kg · m · s─2 =
1L
1 atm
1 Pa
1N
103 dm3 ( ) = 101 kg · m2 · s─2 = 101 joules
This calculation verifies that 1 L· atm = 101 joules . ***** Summary: Thermodynamics
Energy and Enthalpy
1. Energy is the capacity to do work.
2. The first law of thermodynamics: Energy can neither be created nor destroyed.
In any physical or chemical process, © 2011 ChemReview.Net v1r ΔEuniverse = Efinal ─ Einitial = 0 Page 1161 Module 36 — Thermodynamics 3. Energy (E), heat (q), and work (w) in the SI system are measured in joules and (in
chemistry) from the perspective of the system.
4. In any physical or chemical change:
5. In the case of PV work by a gas, ∆Esystem = q + w wsystem = ─ Pexternal ∆Vsystem From the perspective of the system, if a system does work on its surroundings, w is
negative.
6. In converting from PV work units to energy units: 1 liter • atm = 101 joules
7. If no subscript is given after a thermodynamic symbol, assume the subscript is system.
8. The definition of enthalpy (H):
9. Hfinal ─ Hinitial = H = E + PV ΔH = ΔE + Δ(PV) if work is limited to PV work. 10. If external P is held constant and work is limited to PV work, ∆H measures heat flow
into or out of the system. ΔH = q . Entropy
1. Entropy (S): Comparing two systems, in the system with higher entropy, the particles
• are in arrangements that are more probable; mixed, random, or chaotic; • are less well ordered; • have more freedom of motion or positions for the particles to occupy. • Gas particles of a substance have higher entropy than solid particles. • Other variables being equal, a gas has more entropy at lower pressure and higher
volume. • In a chemical reaction, the side of the balanced equation that has more gas molecules
(determined by adding the gas particle coefficients) generally has higher entropy. 2. The second law of thermodynamics: A process will be spontaneous if it increases the
entropy of the universe.
ΔSuniverse = ΔSsystem + ΔSsurroundings = Positive for spontaneous change
3. The units of ΔS, per mole, are joules per (mol·K), written as J/mol·K or J·mol─1·K─1.
4. Systems tend to go to lower enthalpy and higher entropy.
= ─ ΔHsys.
T 5. ΔSsurroundings 6. ΔSuniverse = ΔSsystem ─ ΔHsys.
T © 2011 ChemReview.Net v1r when P and T are constant, and T is absolute. = Positive for spontaneous change
(applies if P and T are held constant) Page 1162 Module 36 — Thermodynamics Standard State ΔH and ΔS Calculations
1. The third law of thermodynamics: Entropy (S) of a perfect crystal of a substance at
absolute zero is zero.
2. ΔHreaction = ∑ cproduct ΔHºf product ─ ∑ creactant ΔHºf reactant
3. ΔSºreaction = ∑ cproduct Sºproduct ─ ∑ creactant Sºreactant Free Energy
1. G is the symbol for free energy, also known as Gibbs free energy.
ΔG is the energy of the system that is available to do work.
2. G = H ─ TS and at constant temperature, ΔG = ΔH ─ TΔS . 3. When referencing a single substance, the unit of ΔG may be joules or joules/mole.
When referencing more than one substance, the unit of ΔG is joules .
4. For a process at constant temperature and pressure:
if ΔG is negative, the process will be spontaneous (go toward the products).
• if ΔG is positive, the process will go backward or not go (favors the reactants).
if ΔG is zero, no net change occurs (the process is at equilibrium). 6. At the T and P of a phase change in a closed system, equilibrium exists between the
phases, and ΔG = 0 .
7. When trying to shift a reaction direction, to make the entropy change more important,
raise the temperature.
8. For calculations reversing, multiplying, or adding reaction equations with ΔGº values
attached, follow the rules for ΔH.
9. ΔGºrxn can be calculated by the summation method using
ΔGºrxn = ∑ cproduct ΔGºf product ─ ∑ creactant ΔGºf reactant Free Energy, Q, and K
1. ΔGrxn. = ΔGºrxn. + RT ln(Q) 2. In problems that mix joules and R, use R = 8.31 J/mol·K or 0.00831 kJ/mol·K 3. When K or Q are calculated, gas pressures must be converted to atmospheres,
concentrations must be in moles/liter, and units are omitted. For solids, solvents, and
pure liquids, pressures and concentrations = 1
4. ΔGºrxn. = ─ RT ln(K)
##### © 2011 ChemReview.Net v1r Page 1163 ...

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