johnson (dej437) – Homework 1 – Sutcliffe – (51060)
1
This
printout
should
have
16
questions.
Multiplechoice questions may continue on
the next column or page – find all choices
before answering.
Note there are a number of numerical (free
response) questions on this assignment. I rec
ommend you use the accurate Kelvin  Cel
sius conversion to help avoid rounding errors.
Watch your signs and always see if you can
predict a qualitative version of the problem
to estimate the sign and possible size of your
answer.
001
10.0 points
If Δ
G
◦
rxn
is positive, then the forward reaction
is (spontaneous / nonspontaneous) and K is
(less / greater) than one.
1.
spontaneous, less
2.
nonspontaneous; less
correct
3.
spontaneous, greater
4.
nonspontaneous; greater
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
10.0 points
Evaluate Δ
G
0
at 724
◦
C for a gas phase reac
tion for which
K
p
= 7
.
91
×
10
4
at 724
◦
C.
Correct answer:

93
.
4879 kJ
/
mol rxn.
Explanation:
K
p
= 7
.
91
×
10
20
T
= 724
◦
C + 273 = 997 K
Δ
G
0
=

RT
ln
K
p
=

(8
.
314 J
/
mol
·
K) (997 K)
×
ln(7
.
91
×
10
4
)
=

93
.
4879 kJ
/
mol rxn
003
10.0 points
Calculate the equilibrium constant at 307 K
for the reaction
HgO(s)
⇀
↽
Hg(
ℓ
) +
1
2
O
2
(g)
,
given the data
S
◦
Δ
H
◦
f
(
J
K
·
mol
)
(
kJ
mol
)
HgO(s)
70
.
29

90
.
83
Hg(
ℓ
)
76
.
02
0
O
2
(g)
205
.
14
0
Correct answer: 1
.
59349
×
10
−
10
.
Explanation:
HgO(s)
⇀
↽
Hg(
ℓ
) +
1
2
O
2
(g)
,
Δ
H
◦
r
=

[Δ
H
◦
f
,
HgO(s)
]
=

(

90
.
83) = 90
.
83 kJ
/
mol
Δ
S
◦
r
=
S
◦
Hg(
ℓ
)
+
1
2
S
◦
O
2
(g)

S
◦
HgO(s)
= 76
.
02
J
K
·
mol
+
1
2
parenleftbigg
205
.
14
J
K
·
mol
parenrightbigg

70
.
29
J
K
·
mol
= 108
.
3
J
K
·
mol
At 307 K,
Δ
G
◦
r(307 K)
= 90
.
83 K

(307 K)
×
parenleftbigg
108
.
3
J
K
·
mol
parenrightbiggparenleftbigg
1 kJ
1000 J
parenrightbigg
= 57
.
5819 kJ
/
mol
Δ
G
◦
r(307 K)
=

RT
ln
K
ln
K
=

Δ
G
◦
r(307 K)
RT
=

57
.
5819 kJ
(8
.
314 J
/
K)(307 K)
×
1000 J
1 kJ
=

22
.
5599
K
=
e
−
22
.
5599
= 1
.
59349
×
10
−
10
.
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johnson (dej437) – Homework 1 – Sutcliffe – (51060)
2
004
10.0 points
Which combination of ΔG
◦
and K is possible
at standard conditions?
1.
Δ
G
◦
=

42
.
8 J,
K
= 1
.
02
correct
2.
Δ
G
◦
=

76
.
3 J,
K
= 8
.
11
×
10
−
13
3.
Δ
G
◦
= 69
.
6 kJ,
K
= 1
.
03
4.
Δ
G
◦
=

77 kJ,
K
= 0
.
986
5.
Δ
G
◦
= 35
.
9 J,
K
= 2
.
38
×
10
13
Explanation:
If ΔG
◦
is positive, K must be less than 1;
if ΔG
◦
is negative, K must be greater than 1.
Also, if ΔG
◦
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.
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 Spring '10
 McCord
 Chemistry, Vapor pressure, Phase transition

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