### CHAPTER 17 BASIC HOMEWORK:

### THERMODYNAMICS II

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**Question 1**

Which of the following processes is NOT spontaneous?

Examples of spontaneous processes

Select one:

**Question 2**

Which of the following statements is INCORRECT?

Spontaneous processes and change in energy

Select one:

**Question 3**

Which of the following statements is INCORRECT?

Spontaneous processes and change in entropy

Select one:

**Question 4**

Which of the following statements is INCORRECT?

Select one:

**Question 5**

For which of the following reactions is there an increase in entropy (entropy of products > entropy of reactants or ΔS > 0)?

Determining the change in entropy 1

Select one or more:

**Question 6**

Check those pairs of molecules where the one on the right has a larger entropy. (Each pair of molecules has the same conditions. For example, both temperatures are the same.)

Determining the change in entropy 2 Select one or more:

**Question 7**

Calculate the increase of entropy in J/K when 1.3 mole of C2H5OH melts at -114oC and 1 atm. ΔHfus = 5.02 kJ/mol.

(Remember to use J and K.)

Entropy definition and calculation

**Explanation**

### ΔS= ΔH/T = 6526 J/ (-114+273) = 6526/159 = 41.04

### 5020 J/ mol * 1.3 mol = 6526 J

**Question 8**

Calculate the increase of entropy in J/K when 3.3 mole of C2H5OH vaporization at 78oC (the boiling point at 1 atm). ΔHvap = 39.30 kJ/mol.

**Explanation**

**Explanation**

#### Did you remember to use J and K?

39300 J/mol * 3.3 mol = 129,690 J

### ΔS= ΔH/T = 129,690/351= 369.49 J/K

**Question 9**

Consider the reaction: A(g) + 2B(g) ↔ 3C(g) + 3D(s). Keq = 1,500.

[A] = 0.10 M, [B] =0.15 M, [C] = 1.10 M, and some D is present. What is Q?

**Explanation**

**Explanation**

Q= [C]^3/ [A][B]^2 = (1.10)^3/(0.1)(0.15)^2 = 1.331/0/00225 = 591.56

**Question 10**

For the above value of Q

Select one:

**Question 11**

Under certain condition for the reaction A → B, ΔH = -22.2 kJ and ΔS = -50 J/K. What is ΔG in J at 205 K?

**Explanation**

**Explanation**

ΔG= ΔH-TΔS

=-22200-(205)(-50)=-11,950

**Question 12**

Check the correct statements for the above reaction.

Reaction direction

Select one or more:

**Explanation**

**Explanation**

All answers must be correct to get credit.

**Question 13**

For the reaction A → B, ΔH = -17.8 kJ and ΔS = -35 J/K. What is ΔG in J at 977 K?

ΔG = ΔH – TΔS

**Explanation**

**Explanation**

ΔG= ΔH-TΔS

=-17800-977(-35)=16395

**Question 14**

Check the correct statements for the above reaction.

Select one or more:

**Question 15**

Under certain condition for the reaction A → B, ΔH = 13.7 kJ and ΔS = 23 J/K. At what temperature in K is the reaction at equilibrium?

Hint given in feedback.

**Explanation**

**Explanation**

Solve ΔG = ΔH – TΔS for T.

Hint, what is ΔG at equilibrium?

ΔG= ΔH-TΔS

0=13700-T(23)

T=595.65

**Question 16**

Check the correct statements for the above reaction.

Select one or more:

**Explanation**

**Explanation**

All answers must be correct to get credit.

**Question 17**

Under certain condition for the reaction A → B, ΔH = -17.8 kJ and ΔS = -32 J/K. At what temperature in K is the reaction at equilibrium?

Hint repeated for this problem in feedback.

**Explanation**

**Explanation**

Solve ΔG = ΔH – TΔS for T.

Hint, what is ΔG at equilibrium?

ΔG= ΔH-TΔS

0=-17800-T(-32)

T=556.25

**Question 18**

Check the correct statements for the above reaction.Select one or more:

**Question 19**

Check those statements that are true.

Standard State and thermodynamic calculations

Select one or more:

**Explanation**

**Explanation**

Mercury is a liquid at room temperature (standard temperature).

**Question 20**

Calculate ΔSo in J/K for the following reaction:

2Fe2O3(s) + 3C(s) → 4Fe(s) + 3CO2(g)

So(Fe2O3(s)) = 87.4 J/K/mol

So(C(s)) = 5.7 J/K/mol

So(Fe(s)) = 27.3 J/K/mol

So(CO2(g)) = 213.7 J/K/mol

**Explanation**

**Explanation**

Remember there are 3 moles of carbon dioxide, etc.

[4(27.3)+3(213.7)]-[2(87.4)+3(5.7)]

-(109.2+641.1)-[(174.8+17.1)]

=750.3-191.9

=558.4

**Question 21**

Calculate ΔGo in kJ for the following reaction:

2Fe2O3(s) + 3C(s) → 4Fe(s) + 3CO2(g)

ΔGfo(Fe2O3(s)) = -742.2 kJ/mol

ΔGfo(C(s)) = ?

ΔGfo(Fe(s)) = ?

ΔGfo(CO2(g)) = -394.4 kJ/mol

**Explanation**

**Explanation**

Remember there are 3 moles of carbon dioxide, etc.

3(-394.4) – 2(-742.2.)

-1183.2+1484.4=301.2

**Question 22**

Check the correct statement for the reduction of iron III oxide to iron.

Reminder given in general feedback.

Select one:

**Explanation**

**Explanation**

If ΔG. = 0 the process is at equilibrium.

If ΔG. < 0 the process is spontaneous in the forward direction or to the right.

**Question 23**

For a certain reaction ΔGorxn = 18.4 kJ/mol. What is the equilibrium constant at 392C?

Answer: 0.0358

**Explanation**

**Explanation**

ΔGrxn = -RTln(Keq)

Keq= e^-ΔGrxn/RT

ΔGrxn = 18.4KJ*1000= 18400J

392C=665K

R=8.31

Keq=e^((-18400)/(665*8.31))

Keq=0.0358

**Question 24**

Correctly match the following conditions.

Hint given in general feedback.

ΔGorxn << 0

ΔGorxn < 0

ΔGorxn > 0

ΔGorxn >> 0

**Explanation**

**Explanation**

The help for the previous problem tells you the answer.

However, you can check for yourself using the Gibb’s energy equation. Try different values for the Gibb’s energy or Keq at some temperature, say 300 K, and see what happens.