# AP REVIEW QUESTIONS: THERMODYNAMICS - Fort Bend ISD

Mechanics

Oct 27, 2013 (4 years and 6 months ago)

100 views

Name: ________________________________

Period: ____

AP REVIEW QUESTIONS: THERMODYNAMICS

1.

C
6
H
5
OH
(s)

+ 7 O
2(g)

→ 6 CO
2(g)

+ 3 H
2
O
(l)

When a 2.000 grams sample of pure phenol, C
6
H
5
OH, is completely burned according to the
equation above, 64.98 kJ of heat is released. Use the information in the tabl
e below to answer the
questions that follow.

Substance

Std. Heat of formation,
ΔH
f

(kJ/mol)

Absolute Entropy, S˚,
at 25˚C (J/K∙mol)

C
graphite

0.00

5.69

CO
2(g)

-

393.5

213.6

H
2(g)

0.00

130.6

H
2
O
(l )

-

285.85

69.91

O
2(g)

0.00

205.0

C
6
H
5
OH
(s)

?

144.0

a.

C
alculate the molar heat of combustion of phenol in kilojoules per mole at 25˚C.

-
3058
kJ/mol

b.

Calculate the standard heat of formation, ΔH˚
f
, of phenol in kilojoules p
e
r mole at 25˚C.

-
160.6 kJ/mol

c.

Calculate the value of the standard free energy change, ΔG˚
, for the combustion of phenol
at 25˚C.

-
3032 kJ/mol

d.

If the volume of the combustion container is 10.0 liters, calculate the final pressure in the
container when the temperature is changed to 110.˚C. (Assume no oxygen remains
unreacted and that all produc
ts are gaseous)

0.602 atm

2.

For the gaseous equilibrium represented below, it is observed that greater amounts of PCl
3

and Cl
2

are produced as the temperature is increased.

PCl
5(g)

↔ PCl
3(g)

+ Cl
2(g)

a.

What is the sign o
f ΔS˚ for the reaction? Explain.

Po
sitive, more moles of gas on the
products side.

b.

What change, if any, will occur in ΔG˚ for the reaction as the temperature is increased?
Explain your reasoning in terms of thermodynamic principles.

H is + (endothermic since an
increase in temp favors prod
ucts

see prompt). Therefore,

G decreases ⡧oes fro洠
positive to negative⤠as te浰erature is increased.

c.

If He gas is added to the original reaction mixture at constant volume and temperature,
what will happen to the partial pressure of Cl
2
? Explain.

Noth
ing. Partial pressure of
chlorine is affected by the amount of
chlorine

present and He is an inert gas and won’t
otherwise affect the equilibrium.

d.

If the volume of the reaction mixture is decreased at constant temperature to half the
original volume, what
will happen to the number of moles of Cl
2

in the vessel? Explain.

Decrease. The increase in pressure will cause the reaction to shift to the side with fewer
moles of gas, thus converting Cl
2

into PCl
5
.

3.

Propane, C
3
H
8
, is a hydrocarbon that is commonly used

as a fuel for cooking.

a.

Write a balanced equation for the complete combustion of propane gas.

C
3
H
8

(g) + 5 O
2

(g)

3 CO
2

(g) + 4 H
2
O (l)

b.

Calculate the volume of air at 30˚C and at 1.00 atmosphere that is needed to burn
completely 10.0 grams of propane.
Assume that air is 21.0% O
2

by volume.

134 L

c.

The heat of combustion of propane is

2220.1 kJ/mole. Calculate the heat of formation,
ΔH˚
f
, of propane given that ΔH˚
f

of H
2
O
(l)

=
-

285.3 kJ/mole and ΔH˚
f

of CO
2(g)

=
-
393.5
kJ/mole.

-
101.6 kJ/mol

d.

Assumin
g that all the heat evolved in burning 30.0 grams of propane is transferred to 8.00
kilograms of water (specific heat = 4.18 J/g
-
K), calculate the increase in temperature of
water.

45.2 C

4.

BCl
3(g)

+ NH
3(g)

→ Cl
3
B:NH
3(s)

The reaction represented a
bove is a reversible reaction.

a.

Predict the sign of the entropy change, ΔS, as the reaction proceeds to the right. Explain

Negative. Two moles converted to one mole; Gases converted to solid.

b.

If the reaction spontaneously proceeds to the r
ight, predict the sign of the enthalpy change,
ΔH. Explain your prediction.

Negative. Must be a larger negative value than T

S in order
to produce a negative ⡳pontaneous⤠value of

G (

G 㴠

T

c.

The direction in which the reaction spontaneously proce
eds changes as the temperature is
increased above a specific temperature. Explain.

The T

S value beco浥m a larger negative
value than

H, so that when it is subtracted fro洠

H,

G beco浥m positive
⡮onspontaneous⤮

d.

What is the value of the equilibrium con
stant at the temperature referred to in (c); that is,
the specific temperature at which the direction of the spontaneous reaction changes?
Explain.

K = 1

5.

The tables below contain information for determining thermodynamic p
roperties of the reaction:

C
2
H
5
C
l
(g)

+ Cl
2(g)

→ C
2
H
4
Cl
2(g)

+ HCl
(g)

Standard Heats of Formation at 298 K

Substance

ΔG˚
f

298 K (kJ/mol)

C
2
H
4
Cl
2(g)

-

80.3

C
2
H
5
Cl
(g)

-

60.5

HCl
(g)

-

95.3

Cl
2(g)

0

Average Bond Dissociation Energies at 298 K

Bond

Energy (kJ/mol)

C

H

㐱4

C

㌴3

㌷3

Cl

㈴2

㐳4

a.

Calculate the ΔH˚ for the reaction above, using the table of average bond dissociation
energies.

-
151 kJ/mol

b.

Calculate the ΔS˚ for the reaction at 298 K, using data from either table as needed.

-
1
20
J/molK

c.

Calculate the value of K
eq

for the reaction at 298 K.

1.5 x 10
20

d.

What is the effect of an increase in temperature on the value of the equilibrium constant?

K will decrease because the reverse (endothermic) rxn will be
favored

with addition of heat.