Chem 102/105 Solutions to Problem Set 4 Dr. M. Cowie

1. Are the following processes exothermic or endothermic? a. When solid KBr is dissolved in water, the solution gets colder.

b. Natural gas is burned in a furnace. c. When concentrated H2SO4 is added to water, the solution gets very hot.

d. Water is boiled in a kettle.

a. endothermic Heat is absorbed by the system (KBr, K+, Br–) from the surroundings (solution)

b. exothermic Heat is released.

c. exothermic Heat released by the system (H2SO4) to the surrounding (solution)

d. endothermic Heat is added from the surroundings to boil the water

2. Consider a mixture of air and gasoline vapor in a cylinder with a piston. The original volume is 40.0

cm3. If the combustion of this mixture releases 950.0 J of energy to what volume will the gases expand

against a constant pressure of 650.0 torr if all the energy of combustion is converted into work to push

back the piston?

w = – PΔV

ΔV = – wP = –

– 950 J

650760 atm

= 1110.8 J

atm x 1 L atm101.3 J = 10.96 L

ΔV = V2 – V1 ⇒ V2 = ΔV + V1 = 10.96 + 0.04 = 11.00 L

3. Thermite mixtures are used for certain types of welding. The thermite reaction is highly exothermic.

Fe2O3 (s) + 2 Al (s) → Al2O3 (s) + 2 Fe (s) ΔH = –852 kJ

1.00 mol Fe2O3 and 2.00 mol Al are mixed at room temperature (25 °C), and a reaction is initiated. The

liberated heat is retained within the products, whose combined specific heat over a broad temperature

range is about 0.8 J g–1 °C–1. The melting point of iron is 1530 °C. Show that the quantity of heat

liberated is more than sufficient to raise the temperature of the products to the melting point of iron.

ΔH = – qcal = – 852 x103 J

qcal = 852 x103 J = m c ΔT

m = 1.00 mol Al2O3 101.96 g Al2O3

1 mol Al2O3 + 2.00 mol Fe

55.85 g Fe1 mol Fe = 213.66 g

ΔT = qcalmc =

852000 J213.66 g 0.8 J g–1 °C–1 = 4985 °C

ΔT = Tf – Ti ⇒ Tf = ΔT + Ti = 4985 + 25 = 5010 °C >> 1530 °C

4. The heat of solution of KI (s) in water is +20.3 kJ/mol KI. If a quantity of KI is added to sufficient

water at 23.5 °C in a Styrofoam cup to produce 150.0 mL of 2.50 M KI, what will be the final

temperature? (Assume a density of 1.30 g/mL and a specific heat of 2.7 J g–1 °C–1 for 2.50 M KI.)

(Note that density and specific heat are significantly different from those of pure H2O for a concentrated

solution).

Find qrxn

qrxn = 0.1500 L solution 2.50 mol KI1 L solution

20.3 kJ1 mol KI = 7.6125 kJ

qrxn = – qsol ⇒ qsol = – 7612.5 J

qsol = – 7612.5 J = m c DT

ΔT = qsolmc =

– 7612.5 J150.0 mL 1.30 g/mL 2.7 J g–1 °C–1 = – 14.5 °C

ΔT = Tf – Ti ⇒ Tf = ΔT + Ti = – 14.5 + 23.5 = 9 °C

5. A 9.00 g ice cube, initially at 0°C, is added to a cup of coffee at 363K which contains 120.0 g coffee

(liquid, not beans or powder). Assume the heat capacity of coffee is the same as water. The enthalpy of

fusion (melting) of ice is 6.0 kJ/mol. Calculate the temperature of the coffee after all the ice melts.

9.00 g H2O

18.0 g/mol H2O = 0.500 mol H2O 363 K = 90°C

qice → water

+ qcoffee = 0

Call the final temperature of the mixture Tf. Ice will absorb heat (ΔHfus) to convert ice at 0oC to water at 0oC, then water will warm to Tf.

qice → water

= ΔHfus + mcΔT = 0.5 mol x 6000 J/mol + 9.0 g x 4.18 J

g °C x (Tf – 0.00)°C

qcoffee = 120.0 g x 4.18

Jg °C x (Tf – 90.0)°C

3000 + 37.62 Tf + 501.6 Tf – 45144 = 0 539.22 Tf = 42144

Tf = 78.15°C

6. a. Can a system do work and give off heat at the same time?

b. Can it do it while maintaining a constant internal energy?

a. Yes

b. No ΔU = q + w If both q and w are negative, then ΔU must be negative. Contradiction

7. The internal energy of a fixed quantity of an ideal gas depends only on its temperature. If a sample is

allowed to expand at a constant temperature (isothermal expansion): (a) Does the gas do work? (b) Does

the gas exchange heat with its surroundings? (c) What happens to the temperature of the gas? (d) What

is ΔE for the gas?

a. Yes w = – P ΔV = (–)(+)(+) = –ve gas does work

b. Yes ΔU is zero (see part d) so q = – w = non zero

c. Temperature remains the same expansion is done at constant temperature (given)

d. ΔU = 0 given that ΔU only depends on T and T doesn’t change

8. The standard heat of combustion per mole of 1,3-butadiene [C4H6 (g)], butane [C4H10 (g)] and H2 (g) are –

2540.2, –2877.6 and –285.8 kJ, respectively. Use these data to calculate the heat of hydrogenation of

1,3-butadiene to butane. C4H6 (g) + 2 H2 (g) → C4H10 (g)

Write combustion reactions

Note that combustion products are gaseous CO2 and liquid H2O

C4H6 (g) + 112 O2 (g) → 4 CO2 (g) + 3 H2O (l) ΔHo

1 = – 2540.2 kJ

C4H10 (g) + 132 O2 (g) → 4 CO2 (g) + 5 H2O (l) ΔHo

2 = – 2877.6 kJ

H2 (g) + 12 O2 (g) → H2O (l) ΔHo

3 = – 285.8 kJ

Net rxn = rxn1 + 2 rxn3 – rxn2

ΔHorxn = ΔHo

1 + 2ΔHo3 – ΔHo

2 = (– 2540.2) + 2 (–285.8) – (– 2877.6) = – 234.2 kJ

9. Use data from Appendix B to calculate the standard enthalpy change for the following reaction at 25°C.

Fe2O3 (s) + 3 CO (g) → 2 Fe (s) + 3 CO2 (g)

ΔHorxn = ∑nprodΔHo

f prod – ∑nreactΔHof react

ΔHorxn = 2 ΔHo

f (Fe (s)) + 3 ΔHof (CO2 (g) – ΔHo

f (Fe2O3 (s)) – 3 ΔHof (CO (g)) =

= 2 (0) + 3 (–393.5) – (– 824.2) – 3 (–110.5) = – 24.8 kJ

10. Calculate the standard enthalpy of dimerization of NO2 (g). 2 NO2 (g) → N2O4 (g)

ΔHrxn = ΔHof (N2O4 (g)) – 2 ΔHo

f ( NO2 (g))= 9 – 2 (33) = – 57 kJ/ mol N2O4

11. Given the following data: 4 CuO (s) → 2 Cu2O (s) + O2 (g) ΔH° = 288 kJ/mol rxn

Cu2O (s) → Cu (s) + CuO (s) ΔH° = 11 kJ/mol rxn

Calculate the standard enthalpy of formation for CuO (s)

Need Cu (s) + 12 O2 (g) → CuO (s) Reverse rxn 2 + reverse half rxn 1

Cu (s) + CuO (s) + Cu2O (s) + 12 O2 (g) → Cu2O (s) + 2 CuO (s)

Net: Cu (s) + 12 O2 (g) → CuO (s)

ΔHof = – ΔH2 –

12 ΔH1 = – 11 –

12 (288) = – 155 kJ/mol

12. Use data from Table 9.2 to estimate the enthalpy change (ΔH) for the following reaction.

C2H6 (g) + Cl2 (g) → C2H5Cl (g) + HCl (g) ΔH = ?

C C

H

H H

H

H H C C

H

H H

H

H Cl + H––ClCl Cl+

Bonds Broken = C–H + Cl–Cl = 414 kJ mol–1 + 243 kJ mol–1

= 657 kJ mol–1

Bonds Formed = C–Cl + H–Cl = 339 kJ mol–1 + 431 kJ mol–1

= 770 kJ mol–1

ΔHreaction = BEbroken – BEmade = 657 kJ mol–1 + 770 kJ mol–1

= –113 kJ mol–1

13. Equations (1) and (2) can be combined to yield the equation for the formation of CH4 (g) from its

elements.

(1) C (s) → C (g) ΔH = 717 kJ

(2) C (g) + 2 H2 (g) → CH4 (g) ΔH = ?

Overall: C (s) + 2 H2 (g) → CH4 (g) ΔHof = –75 kJ mol–1

Use the above data and a bond energy of 436 kJ mol–1 for H2 to estimate the C–H bond energy in

methane (CH4).

First we need to determine ΔH for reaction (2).

Since overall equation = eqn. (1) + eqn. (2)

ΔH overall = ΔH1 + ΔH2

–75 kJ mol–1– = 717 kJ mol–1 + ΔH2

ΔH2 = –792 kJ mol–1

In reaction (2) (all compounds as gases):

Bonds broken = 2 H–H = 2(436 k mol–1) = 872 kJ mol–1

ΔH = BEbroken – BEmade

–792 kJ mol–1 = 872 kJ mol–1 – BE made

BEmade = (872 + 792) kJ mol–1 = 1664 kJ mol–1

Bonds made = 4 (C–H)

BEmade = 1664 kJ mol–1 = 4 (C–H)

∴ C–H = 416 kJ mol–1

14. a. Does the entropy increases or decreases when you separate a mixture of fine sand and charcoal?

b. Which has the greatest entropy, steam at 110°C, water at 25°C or ice at – 40°C?

a. Decrease A mixture has less order than its pure components, so separation ⇒ decrease in entropy.

b. Steam Gas phase less order ⇒ higher entropy

15. Calculate ΔS° for the reaction N2 (g) + 3 H2 (g) → 2 NH3 (g)

When is the reaction spontaneous?

ΔS° = 2 Sof (NH3 (g)) – So

f (N2 (g)) – 3 Sof (H2 (g)) = 2 x (193) – 192 – 3 x (131) = – 199 J mol–1 K–1

ΔH° = 2 ΔHof (NH3 (g)) – ΔHo

f (N2 (g)) – 3 ΔHof (H2 (g)) = 2 x (– 46) – 0 – 3 x 0 = – 92 kJ mol–1

0 > ΔH° – T ΔS°

0 > – 92000 + 199 T

T < 92000 J mol–1

199 J mol–1 K–1 = below 462.3 K

16. The very poisonous hydrogen sulfide can be removed from natural gas (mainly methane CH4) by the

reaction 2 H2S (g) + SO2 (g) 3 S (s) + 2 H2O (g)

a. Calculate the equilibrium constant for this reaction at 25°C.

b. At what temperature is the reaction spontaneous?

c. Calculate the equilibrium constant at 100°C.

a. ΔGorxn = 3 ΔGo

f (S (s)) + 2 ΔGof (H2O (g)) – 2 ΔGo

f (H2S (g)) – ΔGof (SO2 (g)) =

= 3 x (0) + 2 x (– 229) – 2 x (– 34) – (– 300) = – 90 kJ/mol

ΔGorxn = – RT ln K so ln K = –

ΔGorxn

RT = – – 90000 J mol–1

8.3145 J mol–1 K–1 x 298.15 K = 36.31

K = e36.31 = 5.85 x 1015

b. ΔHorxn = 3 ΔHo

f (S (s)) + 2 ΔHof (H2O (g)) – 2 ΔHo

f (H2S (g)) – ΔHof (SO2 (g)) =

= 3 x (0) + 2 x (– 242) – 2 x (– 21) – (– 297) = – 145 kJ/mol

ΔSorxn = 3 So

f (S (s)) + 2 Sof (H2O (g)) – 2 So

f (H2S (g)) – Sof (SO2 (g)) =

= 3 x (32) + 2 x (189) – 2 x (206) – 248 = – 186 J mol–1 K–1

0 > ΔH° – T ΔS°

0 > – 145000 + 186 T

T < 145000 J mol–1

186 J mol–1 K–1 = below 780 K

c. ln K100K25

= ΔHo

rxn

R

1

T25 –

1T100

= – 145000 J mol–1

8.3145 J mol–1 K–1 x

1

298.15 – 1

373.15 K–1 = – 11.756

K100K25

= e– 11.756

K100 = K25 x e–11.756 = 5.85 x 1015 x e–11.756 = 4.59 x 1010

17. It is quite common for a solid to change from one structure to another at a temperature below its melting

point. For example, sulfur undergoes a phase change from the rhombic crystal structure to the

monoclinic crystal at temperatures above 95°C.

a. Predict the signs of ΔH and ΔS for the process Srhombic → Smonoclinic.

b. Which form of sulfur has the more ordered crystalline structure?

a. The process is spontaneous above 95°C ⇒ 0 > ΔG = ΔH – TΔS

Four possible sign combinations

ΔH < 0 and ΔS > 0 ⇒ ΔG < 0 always spontaneous contradicts data

ΔH > 0 and ΔS < 0 ⇒ ΔG > 0 never spontaneous contradicts data

ΔH < 0 and ΔS < 0 ⇒ ΔG < 0 only at low T (the +ve TΔS is smaller than –ve ΔH term)

ΔH > 0 and ΔS > 0 ⇒ ΔG < 0 only at high T (the –ve TΔS is greater than +ve ΔH term)

b. Since ΔS > 0 and systems go from ordered structure to a less ordered structure

then Srhombic is the more ordered crystalline structure .

18. Use molar entropies from Appendix B, together with the following data, to estimate the bond

dissociation energy of the F2 molecule F2 (g) → 2 F (g) ΔG° = 123.9 kJ.

Compare your result with the value listed in Table 13 of the CDS.

ΔSorxn = 2 So

f (F (g)) – Sof (F2 (g)) = 2 (158.8) – 202.8 = 114.8 J K–1

ΔGorxn = ΔHo

rxn – T ΔSorxn

ΔHorxn = ΔGo

rxn + T ΔSorxn = 123900 J + 298.15 K 114.8 J K–1 = 158.1 kJ/mol bonds

Value in Table 10.3 is 159 kJ/mol

19. Use thermodynamic data at 298K to decide in which direction the reaction

H2 (g) + Cl2 (g) →← 2 HCl (g)

is spontaneous when the partial pressures of H2, Cl2 and HCl are all 0.5 atm.

ΔGorxn = 2 ΔGo

f (HCl (g)) – ΔGof (H2 (g)) – ΔGo

f (Cl2 (g)) = 2 (– 95.30) – 0 – 0 = – 190.6 kJ

ΔG = ΔGorxn + RT ln Q

Q = PHCl

2

PH2 PCl2

= 0.52

0.5 0.5 = 1

ΔG = ΔGorxn + RT ln Q = ΔGo

rxn + RT ln 1 = ΔGorxn = – 190.6 kJ < 0

spontaneous in the forward direction

20. The enthalpy of vaporization, ΔHovap , of PbCl2 is 104 kJ/mol. ΔSo

vap of PbCl2 is 90.6 J/(mol K). Find the

normal boiling point of PbCl2.

PbCl2 (l) PbCl2 (g)

At equilibrium ΔGorxn = 0 = ΔHo

rxn – T ΔSorxn = ΔHo

vap – T ΔSovap rxn is evaporation !

Rearrange T = ΔHo

vap

ΔSovap

= 104000 J mol–1

90.6 J mol–1 K–1 = 1147.9 K