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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

Enthalpy versus entropy

Select one: Question 4

Which of the following statements is INCORRECT?

Entropy

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.)

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

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.

Entropy 2nd calculation Explanation

Did you remember to use J and K?

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

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

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

ΔG= ΔH-TΔS

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

Question 12

Check the correct statements for the above reaction.

Select one or more:   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

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

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

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

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

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

Detailed delta S Calculation 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

Detailed delta G calculation 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

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?

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 Answer 1 ΔGorxn < 0 Answer 2 ΔGorxn > 0 Answer 3 ΔGorxn >> 0 Answer 4 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.