# AP Chemistry ConceptsTHERMODYNAMICS

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27 Οκτ 2013 (πριν από 5 χρόνια και 10 μήνες)

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AP Chemistry Concepts

-

THERMODYNAMICS

1.

H
0

rxn

= ∑ ∆
H
f

0
Products
-

∑∆
H
f
0

Reactants

= ∑ Bond Energy Reactants
-

∑ Bond energy Products

H
rxn

-

exothermic ∆
H
rxn

+ endothermic

2.

S
0

rxn

= ∑
S
f
0

Products
-

S
f
0

Reactants

S
0
rxn

-

ordered ∆
S
0
rxn

+ disordered

3.

G
0

rxn

= ∆
H
0
rxn

-

T

S

0
rxn

G
0
rxn

-

spontaneous ∆
G
0
rxn

+ nonspontaneous

4.

G
0
rxn

=
-

RT

ln
Q

Q

=
K
eq

free energy and equilibrium

5.

G
0

rxn

=
-

nF E
0

free energy and electrochemistry

F

= 96,500 coulombs / mole electrons

6.

Phase diagrams

7.

H

rxn

=
q

=
m

(
c

) ( ∆
T

)

Free Response Questions

1998 # 3

C
6
H
5
OH(s) + 7 O
2
(g) → 6 CO
2
(g) + 3H
2
O(l)

When a 2.000
-
gram sample of pure phenol, C
6
H
5
OH(s), is completely burned according to the
equation above, 64.98 kilojoules of heat is released. Use the information in the table below to

Substance

Standard Heat of Formation,
Δ
H
°
f
, at 25°C (kJ/mol)

Absolu
te Entropy,
S
°, at 25°C (J/mol
-
K)

C(graphite)

0.00

5.69

CO
2
(g)

-
395.5

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
.

Calculate the molar heat of combustion of phenol in kilojoules per mole at 25°C.

b
.

Calculate the standard heat of formation,
Δ
H
°
f
, of phenol in kilojoules per mole at 25°C.

c
.

Calculate the value of the standard free
-
energy change,
Δ
G
° for the combus
tion of
phenol at 25°C.

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

9
points

a)

1 point

Heat released per mole

Or, ΔH
comb
=
-
3058 kJ mol
-
1

1 point

Units not necessary

b)

ΔH
comb
=
-
3058 kJ mol
-
1

1 point

-
3058 kJ = [6(
-
395.5) + 3(
-
285.85)]

[ΔH
f
o

(phenol)]

1 point

ΔH
f
o

(phenol) =
-
161 kJ

1 point

One point earned for correct sign of heat of combustion, one point for correct use of
moles / coefficients, and one point for correct substitution

c)

ΔS
o

= [3(69.91) + 6(213.6)] [7(205.0) + 144.0] =

-
87.67
J
/
K

1 point

ΔG
o

= ΔH
o

-

TΔS
o

= 3058 kJ

(298 K)(
-
0.08767 kJ K
-
1
) =
-
3032 kJ

1 point

Units not necessary; no penalty if correct except for wrong ΔH
comb

for part a

d)

moles gas = 9 × [moles from part a] = 9 (0.02125 mol) = 0.1913 moles gas

1
point

Units necessary; no penalty for using Celcius temperature if also lost point in part c for same error

1996 #3

C
2
H
2
(g) + 2 H
2
(g) → C
2
H
6
(g)

Substance

S
° (J/mol K)

Δ
H
°
f

(kJ/mol)

Bond

Bond Energy (kJ/mol)

C
2
H
2
(
g)

200.9

226.7

C
-
C

347

H
2
(g)

130.7

0

C=C

611

C
2
H
6
(g)

--------

-
84.7

C
-
H

414

H
-
H

436

a
.

If the value of the standard entropy change,
Δ
S
°, for the reaction is
-
232.7 joules per mole
Kelvin, calculate the standard molar entropy,
S
°, of C
2
H
6

gas.

b
.

Calculate the value of the standard free
-
energy change,
Δ
G
°, for the reaction. What does
the sign of
Δ
G
° indicate about the reaction above?

c
.

Calculate the value of the equilibrium constant,
K
, for the reaction at 298 K.

d
.

Calculate the val
ue of the C

C bond energy in C
2
H
2

in kilojoules per mole.

(a) two points; one for line of answer

-

232.7 J/K = S° (C
2
H
6
)
-

(261.4 + 200.9) J./K

S° (C
2
H
6
) = 229.6 J/K

units ignored; 1 point earned for 98.9 J/K; 1 point lost if stoichiometry

is not implied
in process

(b) three points total; one point each portion; any value for T (e.g., 273 K or 298 K) is
allowable:

Δ
H° = (
-

84.7 kJ)
-

(226.7 kJ) =
-
311.4kJ

ΔG
°=
-

311.4kJ
-

(298 K) (
-

0.2327kJ/K)

ΔG
°=
-

311.4 kJ + 69.3 kJ

ΔG
°=
-

242.1 kJ

Ne
gative

Δ
G° therefore reaction is spontaneous, or K
eq

> 1 therefore reaction is
spontaneous, or products are favored at equilibrium.

(c) two points

ln K = 242.1 ÷ [(8.31
×

10
-
3
) (298)] = 97.7

K = 3
×

10
42

(1,2,or 3 significant figures acceptable)

(d) two points; first point earned for correct substitution and correct number of bonds,
second point earned for setting equal to
ΔH
rxn

and correct calculation of answer; no
points earned for "extrapolation" techniques to find carbon
-
carbon triple bond ene
rgy;
E* is the energy of the carbon
-
carbon triple bond.

-

311.4 kJ = [(2) (436) + E* + (2) (414)]
-

[(347) + (6) (414)]

E* = 820 kJ

1997 #7

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 of
Δ
S
° for the reaction? Explain.

b
.

What change, if any, will occur in
Δ
G
° for the reaction as the temperature is increased.
ms of thermodynamic principles.

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.

d
.

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

in the reaction vessel?
Explain.

(a)

S° is positive (or "+", or ">0") 1 point

Moles products > moles reactants 1 point

Note
; all species are gaseous, so (g) need not be

indicated. To earn credit, number of
particles (moles) must be discussed. No explanation point earned for just noting that
disorder increases, or that PCl
5

is decomposing or dissociating.

(b)

G° will decrease (or become more negative, or become smaller)
. 1 point

G° =

-

T

and since

S° is positive, T

S° is positive ( > 0). Thus increasing T will result in a larger
term being subtracted from

H°, or,

G° =
-
RT ln K and K is going up in value since T is
increasing.)

Note: full credit earned for
part (b) if:

S° < 0 in part (a) which leads to

G° is increasing because T

H°, or,

S° = 0 in part (a) which leads to no change in

(c) no change (one point)

P
He

is not part of the a) reaction (He is not involved) or, b) law of mass action or, c)
reaction quotient or, d) equilibrium constant expression; one point

hence altering P
He

has no effect on the position at equilibrium

(d) moles of Cl
2

will decrease (one
point)

The decrease in volume leads to an increase in pressure (concentration), therefore the
reaction shifts to the left because:

(one point for any of the following)

Q > K
sp

(Q > K
c
, or,

the rate of the reverse reaction increase more than the rate of t
he forward reaction, or,

the reaction shifts toward the lesser moles of gas.

Note: "LeChatelier's principle" alone is not sufficient to earn the explanation point. If
response suggests that the number of moles of Cl
2

is halved because the system is "cut"
in
half, only one point is earned.

1999 # 6

Answer the following questions in terms of thermodynamic principles and concepts of kinetic
molecular theory.

a.

Consider the reaction represented below, which is spontaneous at 298 K.

CO
2
(
g
) + 2 NH
3
(
g
)

C
O(NH
2
)
2
(
s
) + H
2
O(
l
); Δ

298

=
-
134 kJ

i.

For the reaction, indicate whether the standard entropy change,

Δ

298
, is positive, or

ii.

Which factor, the change in enthalpy,

Δ

298
, or the change in entropy, Δ

298
,
provides the principal driving force for the reaction at 298 K? Explain.

iii.

For the reaction, how is the value of the standard free energy change, Δ

, affected
by an increase in temperature? Explain.

b.

Some reactions that are predicted by their s
ign of Δ

to be spontaneous at room
temperature do not proceed at a measurable rate at room temperature.

i.

ii.

A suitable catalyst increases the rate of such a reaction. What effect does the
catalyst have on Δ

for the reaction? Explain.

(a)(i)

ΔS° is negative (−)
OR
ΔS° < 0
OR
entropy is decreasing.

1 point

3 moles of gaseous particles converted to 2 moles of solid/liquid.

1 point

• One point earned for correct identification of (−) sign of ΔS°

• One poi
nt earned for correct explanation (mention of phases is crucial for point)

• No point earned if incorrect ΔS° sign is obtained from the presumed value of
ΔG°

(ii)

ΔH° drives the reaction.

1 point

The decrease in entropy (ΔS° < 0) cannot drive the rea
ction, so the decrease in
enthalpy (ΔH° < 0) MUST drive the reaction.

OR

ΔG° = ΔH° − TΔS°; for a spontaneous reaction ΔG° < 0, and a negative value of
ΔS° causes a positive ΔG°.

1 point

• One point earned for identifying ΔH° as the principal driving
force for the
reaction

• One point earned for correct justification

• Justification point earned by mentioning the effects of changes in entropy and
enthalpy on the spontaneity of the reaction OR by a mathematical argument using
the Gibbs
-
Helmholtz equatio
n and some implication about the comparison
between the effects of ΔS° and ΔH°

(iii)

Given that ΔG° = ΔH° − TΔS° and ΔS° < 0, an increase in temperature causes an
increase in the value of ΔG° (ΔG° becomes less negative).

1 point

• One point earne
d for the description of the effect of an increase in temperature
on ΔS° and consequently on ΔG°

• No point earned for an argument based on Le Châtelier.s principle

(b)(i)

The reaction rate depends on the reaction kinetics, which is determined by the
value

of the activation energy, E
act
. If the activation energy is large, a reaction
that is thermodynamically spontaneous may proceed very slowly (if at all).

1 point

• One point earned for linking the rate of the reaction to the activation energy,
which may be explained verbally or
using

a reaction profile diagram

(ii)

The catalyst has no effect on the value of ΔG°.

1 point

The catalyst reduces the value of E
act
, inc
reasing the rate of reaction, but has no
effect on the values of ΔH° and ΔS°, so it cannot affect the thermodynamics of the
reaction.

1 point

• One point earned for indicating no change in the value of ΔG°

• One point earned for indicating (verbally, o
r with a reaction
-
profile diagram)
that the catalyst affects the activation energy

2002 # 8

Carbon (graphite), carbon dioxide, and carbon monoxide form an equilibrium mixture, as
represented by the equation

C(
s
) + CO
2
(
g
)

2CO(
g
)

a.

Predict the sign f
or the change in entropy,

S
, for the reaction. Justify your prediction.

b.

In the table below are data that show the percent of CO in the equilibrium mixture at two
different temperatures. Predict the sign for the change in enthalpy,

H
, for the
reaction.

Temperature

% CO

700
o
C

60

850
o
C

94

c.

Appropriate
ly

complete the potential energy diagram for the reaction by finishing the
curve on the graph below. Also, clearly indicate

H

for the reaction on the graph.

d.

If the initial amount of C(
s
) were doubled, what would be the effect on the percent of CO

a)

Δ
S
= +; There is more disorder in a gas than in a solid, so the product is more
disordered than the reacta
nts. The change in entropy is therefore positive.

OR

There is 1 mole of gas in the reactants and 2 moles of gas in the product.

1 point earned for indicating that Δ
S
is positive

1 point earned for explanation

b)

Δ
H
= +; More CO at the higher temperature
indicates that the reaction shifts to
the right with increasing temperature. For this to occur, the reaction must be
endothermic.

1 point earned for indicating that Δ
H
is positive

1 point earned for explanation

c)

1 point earned for completing the graph
according to the information in part (b)

1 point earned for appropriately labeling Δ
H
rxn

for the reaction as drawn

d)

An increase in the amount of C(
s
) has no effect. Solids do not appear in the
equilibrium expression, so adding more C(
s
) will not affect

the percent of CO in the
equilibrium

mixture.

1 point earned for indicating no effect

1 point earned for explanation

Note: Since the question asks about “percent of CO” a student might think of % by mass
or % by mole.

Adding carbon will not shift the equi
librium, so P(CO) and P(CO
2
) stay the same. The %
CO then decreases, because now there are more total moles in the system: % CO =
n
CO/(
n
CO +
n
CO2 +
n
C). As
n
C is raised, the denominator increases, and % CO
decreases.

2010#2

(a) Determine the change in te
mperature of the solution that results from the dissolution of the
urea.

(b) According to the data, is the dissolution of urea in water an endothermic
process or an

The process is endothermic. The decrease in temperature
indicates that the process for the dissolution of urea in
water requires energy.

One point is earned for the correct choice
with justification.

(c) Assume that the specific heat capacity of the calorimeter is negligible and that the specific heat
capacity of the solution of urea and water is 4.2 J g

1

°C

1

throughout the experiment.

(i) Calculate the heat of dissolution of the urea in
joules.

Assuming that no heat energy is lost from the calorimeter
and given that the calorimeter has a negligible heat
capacity, the sum of the heat of dissolution,
q
soln
and the
change in heat energy of the urea
-
water mixture must equal
zero.

q
soln
+
mc
Δ
T
= 0

q
soln
= −
mc
Δ
T

m
soln
= 5.13 g + 91.95 g = 97.08 g

q
soln
= −(97.08 g)(4.2 J g

1
°C

1
)(−3.2°C) = 1.3 × 10
3

J

One point is earned for the correct setup.

One point is earned for the correct numerical
result for the heat of dissolution.

(ii) Calculate the molar enthalpy of solution,
Δ
H
soln
o
, of urea in kJ mol
−1
.

Δ
H
soln
o

= q
soln

/ mol solute

molar mass of urea = 4(1.0) + 2(14.0) + 12.0 + 16.0 = 60.0 g
mol

1

One point is earned for the calculation
of moles of urea.

One point is earned for the correct
numerical result with correct algebraic

Δ
T
= 21.8 − 25.0 = −3.2 Celsius degrees

One point is earned for the correct temperature change.

moles of urea=5.13 g urea ×1 mol urea / 60.0 g urea=0.0855mol

Δ
H
soln
o

= 1.3
×

10
3

J / 0.0855 mol = 1.5
×

10
4

J mol
-
1 = 15 kJ
mol
-
1

sign.

(d) Using the information in the table below, calculate th
e value of the molar entropy of solution,
Δ
S
soln
o

Δ
G
° = Δ
H
°−
T
Δ
S
°

− 6.9 kJ mol
-
1

= 14.0 kJ mol
-
1
− (298 K)(Δ
S
°)

Δ
S
soln
o

= 0.0701 kJ mol
-
1

K
-
1

= 70.1 J mol
-
1

K
-
1

One point is earned for the correct
setup.

One point is earned for the correct
numerical result with correct units.

(e) The student repeats the experiment and this time obtains a result for
Δ
H
soln
o

of urea that is 11
percent below the accepted value.
Calculate the value of
Δ
H
soln
o

that the student obtained in this
second trial.

Error = (0.11)(14.0 kJ mol

1
) = 1.54 kJ mol

1

14.0 kJ mol

1

1.54 kJ mol

1

= 12.5 kJ mol

1

One point is earned for the correct
numerical result.

(f) The student performs a third trial of the experiment but this time adds urea that has been taken
directly from a refrigerator at 5°C. What effect, if any, would using the cold urea instead of urea
at 25°C have on the experimentally obtained value of
Δ
H
s
oln
o

There would be an increase in the obtained value for
Δ
H
soln
o

because the colder urea would have caused a larger negative
temperature change.

One point is earned for the correct prediction
with justification.

Multiple Choice Questions

1999
#
61

C
2
H
4
(g) + 3 O
2
(g)

2 CO
2
(g) + 2 H
2
O(g)

For the reaction of ethylene represented above,

H

is
-

1,323 kJ. What is the value of

H

if the
combustion produced liquid water H
2
O(l), rather than water vapor H
2
O(g)? (

H

for the phase
change H
2
O(g)

H
2
O(l) is
-
44 kJ mol
-
1
.)

A)
-
1,235 kJ

B)
-
1,279 kJ

C)
-
1,323 kJ

D)
-
1,367 kJ

E)
-
1,411 kJ

(31%)

2002
#
25

3 C
2
H
2
(g)

C
6
H
6
(g)

What is the standard enthalpy change, Δ
H
o
, for the reaction represented above? (Δ
H
o
f

of C
2
H
2
(g)
is 230 kJ mol
-
1
; Δ
H
o
f

of C
6
H
6
(g) is 83 kJ mol
-
1
.)

A)

-
607 kJ

(66%)

B)

-
147 kJ

C)

-
19 kJ

D)

+19 kJ

E)

+773 kJ

1999 #22

Of the following reactions, which involves the largest decrease in entropy?

A) CaCO
3
(
s
)

CaO(
s
) + CO
2
(
g
)

B) 2 CO(
g
) + O
2
(
g
)

2 CO
2
(
g
)

C) Pb(NO
3
)
3

+ 2 KI

PbI
2

+ 2 KNO
3

D) C
3
H
8

+ 5O
2

3 CO
2

+ 4 H
2
O(
l
)

E) 4 La + 3 O
2

㈠La
2
O
3

(68%)

2002 #41

When solid NH
4
SCN is mixed with solid Ba(OH)
2

in a closed container, the
temperature drops and a gas is produced. Which of the following indicates the correct signs for
Δ
G
, Δ
H
, and Δ
S

for the process?

Δ
G

Δ
H

Δ
S

A)

-

-

-

B)

-

+

-

C)

-

+

+

(43%)

D)

+

-

+

E)

+

-

-

2002 #73

X(s)

X(l)

Which of

the following is true for any substance undergoing the process represented above at its
normal melting point?

A)

ΔS < 0

B)

Δ
H

= 0

C)

Δ
H

=
T
Δ
G

D)

T
Δ
S

= 0

E)

Δ
H

=
T
Δ
S
(34%)