A.
Calculate the equilibrium constant K for the reaction: 2NO(g) + O
2
(g) = 2NO
2
(g) at P
= 1 atm, T = 298K using the following data:
compound
∆
G
f
0
, kJ/mol
∆
H
f
0
, kJ/mol
NO
86.606
90.291
NO
2
51.258
33.095
∆
G = 2
51.258 – 2
86.606 = 70.696 kJ/mol
K = exp(
∆
G/RT) = 2.502 10
12
B.
Will the equilibrium constant increase or decrease if temperature is increased? Explain
your answer.
2
ln
/
/
K
T
H
RT
∂
∂
= ∆
∆
H = 2 33.095 – 2 86.606 = 114.392 < 0 (the reaction is exothermic)
Therefore
ln
/
0
K
T
∂
∂
<
: the equilibrium constant will decrease.
C.
Calculate the equilibrium constant for P=100 atm and T = 298K. Assume that the
reactants and the product are ideal gases.
ln[K(P,T)/K(P
0
,T)] =
0
(
ν
)ln(
/
)
i
P P
−
∑
ν
2
1
2
1
i
=
− −
= −
∑
ln[K(P,T)/K(P
0
,T)]
=
0
ln(
/
)
P
P
or K(P,T)/K(P
0
,T) =
0
/
P
P
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K(P,T) =K(P
0
,T)
0
/
P
P
= 100 K(P
0
,T) = 2.502 10
14
D. Suppose you have initially 1 mole of NO, 1 mole of O
2
and 1 mole of NO
2
.
At a later
moment, the amount of NO
2
was found to be 1.2 mol. What are the amounts of NO and
O
2
at the same time?
n(NO
2
)=1+2
ξ=1.2
mol
so
ξ=0.1
n(NO) = 1 – 2
ξ
=0.9 mol
n(O
2
)=1
ξ=0.9
mol
2.
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 Spring '08
 LIM
 Physical chemistry, Thermodynamics, Equilibrium, pH, Steady State, Daewoo K3, steady state approximation

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