tran (pt4954) – Homework 1 – sutcliffe – (51045)
1
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printout
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have
20
questions.
Multiplechoice questions may continue on
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before answering.
001
10.0points
If Δ
G
◦
rxn
is positive, then the forward reaction
is (spontaneous / nonspontaneous) and K is
(less / greater) than one.
1.
nonspontaneous; less
correct
2.
nonspontaneous; greater
3.
spontaneous, greater
4.
spontaneous, less
5.
None of these; Δ
G
is not directly related
to
K
.
Explanation:
A positive Δ
G
◦
rxn
(standard reaction free
energy)
denotes
an
endothermic
reation,
which is nonspontaneous.
Also, Δ
G
◦
rxn
=

RT
ln
K
, so a positive Δ
G
◦
rxn
would result
in a
K
that is between the values of zero and
one.
002
10.0points
Which combination of Δ
G
◦
and K is im
possible?
1.
Δ
G
◦
= 16 J,
K
= 635
correct
2.
Δ
G
◦
=

37
.
4 kJ,
K
= 3
.
57
×
10
6
3.
Δ
G
◦
= 18
.
5 J,
K
= 0
.
993
4.
Δ
G
◦
= 86
.
4 kJ,
K
= 7
.
31
×
10
−
16
5.
Δ
G
◦
=

72 J,
K
= 1
.
03
Explanation:
If Δ
G
◦
is positive, K must be less than 1;
if Δ
G
◦
is negative, K must be greater than 1.
Also, if Δ
G
◦
is small (compared to RT), then
K will be close to 1, and if Δ
G
◦
is large, K
will be much smaller or greater than 1.
003
10.0points
When Δ
G
0
is calculated for a proposed reac
tion, the process being described is
1.
any mixture of reactants converted to
equilibrium mixture.
2.
100% reactants converted to equilibrium
mixture.
3.
any mixture of reactants and products
converted to equilibrium mixture.
4.
100% reactants converted to 100% prod
ucts.
correct
Explanation:
Δ
G
0
= Δ
G
0
final

Δ
G
0
initial
= Δ
G
0
products

Δ
G
0
reactants
004
10.0points
Note: The ’reaction progress’ axis reads from
all reactants (far L) to all products (far R)
The figure represents a reaction at 298 K.
G
A
B
C
D
E
rxn progress
Based on the figure, the standard reaction is
1.
nonspontaneous.
2.
spontaneous.
correct
Explanation:
Δ
G
◦
is negative (point E is lower free en
ergy than point A), so the standard reaction
is spontaneous.
005
10.0points
A gasphase reaction has
K
p
= 3
.
99
×
10
20
at
25
◦
C. Calculate Δ
G
0
for this reaction.
Correct answer:

117
.
549 kJ
/
mol rxn.
Explanation:
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tran (pt4954) – Homework 1 – sutcliffe – (51045)
2
K
p
= 3
.
99
×
10
20
T
= 25
◦
C + 273 = 298 K
Δ
G
0
=

RT
ln
K
p
=

(8
.
314 J
/
mol
·
K) (298 K)
×
ln
(
3
.
99
×
10
20
)
=

117
.
549 kJ
/
mol rxn
006
10.0points
Consider the reaction
2 Fe
2
O
3
(s) + 3 C(s)
→
4 Fe(s) + 3 CO
2
(g)
,
Δ
H
◦
= 462 kJ
,
Δ
S
◦
= 558 J
·
K
−
1
.
Calcu
late the equilibrium constant for this reaction
at 525
◦
C.
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 Spring '11
 campsisi
 Critical Point, Vapor pressure, Tran

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