SPRING 2013, CHM 1120, SECTION 004, EXAM III, PAGE 1 of 6

CHM 1120 –EXAM III – SECTION 004 – 04/24/13

Name:____Dr. Timothy M. Trygstad – Answer Key____ Score:__100__/ 100 (please write your NAME above AND on the last page of this exam!!!)

Be sure to show all of your work clearly (partial credit) and circle your final answers.

Don’t forget to label your answers with the correct units when appropriate.

Use the following information to answer questions 1-3.

The combustion of butane produces heat according to the equation

2 C4H10(g) + 13 O2(g) 8 CO2(g) + 10 H2O(l)

= -5314 kJ/mol

(O2(g)) = 0 kJ/mol

(CO2(g)) = -393.5 kJ/mol

(H2O(l)) = -285.8 kJ/mol

1) Is the reaction exothermic or endothermic? (2 pts.) Answer: _____exothermic_____

2) Calculate the of C4H10(g) in kJ/mol. (4 pts.)

(show all of your work for full credit) Answer: -346 kJ/mol

-5314 = [(8)(-393.5) + (10)(-285.8)] – [(2)(HofC4H10) + (13)(0)]

-5314 = (-6006) – (2)(HofC4H10)

6006 – 5314 = – (2)(HofC4H10)

692 = – (2)(HofC4H10)

(692)(-2) = HofC4H10 = –346 kJ/mol

3) How many grams of butane must be burned to release 1.00 104 kJ of heat? (8 pts.)

(show all of your work for full credit) Answer: _218.758 g_

FW of C4H10 = (12.011)(4) + (1.008)(10) = 58.124 g/mol

4) Which of the following conditions is not correct for standard enthalpy of formations? (3 pts.)

a) the temperature must be 273.15 K

b) all gases must have pressures of 1.00 atm

c) the concentration of all solutions must be 1.00 M

d) all 3 conditions above are correct for standard enthalpy of formations

5) In one or two sentences, describe the First Law of Thermodynamics. (3 pts.)

Energy can be converted from one form to another, but cannot be created or destroyed.

SPRING 2013, CHM 1120, SECTION 004, EXAM III, PAGE 2 of 6

6) Which of the following processes would have an increase in entropy? (3 pts.)

a) water condensing

b) perfume vapor diffusing through the air

c) salt crystallizing from a solution

d) Na(s) + ½ Cl2(g) → NaCl(s)

7) Which of the following processes would have G < 0? (3 pts.)

a) a boulder rolling up hill

b) a soft-boiled egg becomes raw

c) a teaspoon of sugar dissolves in hot tea

d) water decomposing to H2 and O2 gases

8) Fill in the blanks in the following table: (5 pts.)

Srxn Hrxn Grxn Spontaneity

positive negative negative

spontaneous

positive

0 negative spontaneous

negative positive

positive

non-spontaneous

0

positive

positive non-spontaneous

9) Which of the following would have the highest entropy? (3 pts.)

a) a stick of butter

b) a spoonful of sugar

c) a popsicle

d) a glass of Kool-Aid

10) Which of the following has an absolute zero-point? (3 pts.)

a) enthalpy

b) entropy

c) free energy

d) some of the above

e) none of the above

11) For the generic reaction A + B → C, the value of K is 4.75x10-6

. Which of the following is true? (3 pts.)

a) G < 0

b) G > 0

c) the reaction is fast

d) a + c

e) b + c

12) In one or two sentences, describe the Second Law of Thermodynamics. (3 pts.)

The entropy of the universe increases in a spontaneous process

and remains unchanged in an equilibrium process.

13) In one or two sentences, describe the Third Law of Thermodynamics. (3 pts.)

The entropy of a pure crystalline solid is zero at absolute zero.

SPRING 2013, CHM 1120, SECTION 004, EXAM III, PAGE 3 of 6

Use the following information to answer questions 14-17.

2 POCl3(g) 2 PCl3(g) + O2(g)

= 572 kJ/mol

= 179 J/mol·K

14) Calculate the value of . (4 pts.)

Go = H

o - TS

o T = 25

oC 273.15 + 25

oC = 298.15 K 179 J/mol·K = 0.179 kJ/mol·K

Go = 572 kJ/mol – (298.15 K)(0.179 kJ/mol·K) = 572 kJ/mol – 53.36885 kJ/mol = 518.63115 kJ/mol

15) Calculate the value of K. (4 pts.)

Go = -RTlnK

518.63115 kJ/mol = -(0.008314 kJ/mol·K)(298.15 K)lnK

518.63115 kJ/mol = -(2.4788191 kJ/mol)lnK

-209.2250903 = lnK e^(-209.2250903)

= K = 1.3636 x 10-91

16) At what temperature would this reaction switch from non-spontaneous to spontaneous? (6 pts.)

Go = H

o - TS

o = 0 thus, H

o = TS

o thus, T = H

o/S

o

T = (572 kJ/mol)/(0.179 kJ/mol·K) = 3195.5 K

17) Calculate the value of G if [POCl3] = 0.250 M; [PCl3] = 0.150 M; [O2] = 0.100 M. (8 pts.)

GGo + RTlnQ

Q = ([O2][PCl3]2)/([POCl3]

2) = ((0.100)(0.150)

2)/((0.250)

2) = 0.036

G 518.63115 kJ/mol + (0.008314 kJ/mol·K)(298.15 K)ln(0.036)

G 518.63115 kJ/mol + (2.4788191 kJ/mol)ln(0.036)

G 518.63115 kJ/mol + (2.4788191 kJ/mol)(-3.324236341)

G 518.63115 kJ/mol + (-8.240180534) = 510.391 kJ/mol

SPRING 2013, CHM 1120, SECTION 004, EXAM III, PAGE 4 of 6

Use the following information to answer questions 18-23.

18) Give the oxidation number and total charge contribution for each of the atoms shown above. (12 pts.)

19) Which species is the oxidizing agent? (2 pts.)

ClO3-

20) Which species is the reducing agent? (2 pts.)

Cl-

21) Give the balanced half-reactions if run under acidic reaction conditions. (8 pts.)

Show the step-by-step process.

Cl- Cl2

2 Cl- Cl2

2 Cl- Cl2 + 2 e

-

ClO3- ClO2

ClO3- ClO2 + H2O

2 H+ + ClO3

- ClO2 + H2O

1 e- + 2 H

+ + ClO3

- ClO2 + H2O

Oxidation Half-Reaction: _______________________2 Cl- Cl2 + 2 e

-___________________________

Reduction Half-Reaction: ___________________1 e- + 2 H

+ + ClO3

- ClO2 + H2O______________

22) Give the balanced redox reaction if run under acidic reaction conditions. (4 pts.)

Show the step-by-step process.

2 Cl- Cl2 + 2 e

-

2 x (1 e- + 2 H

+ + ClO3

- ClO2 + H2O)

Redox Reaction (acidic): ______4 H+ + 2 ClO3

- + 2 Cl

- Cl2 + 2 ClO2 + 2 H2O________________

23) Give the balanced redox reaction if run under basic reaction conditions. (4 pts.)

Show the step-by-step process.

4 H+ + 2 ClO3

- + 2 Cl

- Cl2 + 2 ClO2 + 2 H2O

4 HO- + 4 H

+ + 2 ClO3

- + 2 Cl

- Cl2 + 2 ClO2 + 2 H2O + 4 HO

-

4 H2O + 2 ClO3- + 2 Cl

- Cl2 + 2 ClO2 + 2 H2O + 4 HO

-

Redox Reaction (basic): ___________2 H2O + 2 ClO3- + 2 Cl

- Cl2 + 2 ClO2 + 4 HO

-___________

SPRING 2013, CHM 1120, SECTION 004, EXAM III, PAGE 5 of 6

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SPRING 2013, CHM 1120, SECTION 004, EXAM III, PAGE 6 of 6

Name:___Dr. Timothy M. Trygstad – Answer Key__ CSS Box #:__S 3304__