Derive the general expression for the equilibrium constant Kp & Ke for the reaction.
Answers
Answer:
Let the gaseous reaction is a state of equilibrium is-
aA
(g)
+bB
(g)
⇌cC
(g)
+dD
(g)
Let p
A
,p
B
,p
C
and p
D
be the partial pressure of A,B,C and D repectively.
Therefore,
K
c
=
[A]
a
[B]
b
[C]
c
[D]
d
.....(1)
K
p
=
p
A
a
p
B
b
p
C
c
p
D
d
.....(2)
For an ideal gas-
PV=nRT
⇒P=
V
n
RT=CRT
Whereas C is the concentration.
Therefore,
p
A
=[A]RT
p
B
=[B]RT
p
C
=[C]RT
p
D
=[D]RT
Substituting the values in equation (2), we have
K
p
=
[A]
a
(RT)
a
[B]
b
(RT)
b
[C]
c
(RT)
c
[D]
d
(RT)
d
⇒K
p
=
[A]
a
[B]
b
[C]
c
[D]
d
(RT)
[(c+d)−(a+b)]
⇒K
p
=K
c
(RT)
Δn
g
(From (1)]
Here,
Δn
g
= Total no. of moles of gaseous product − Total no. of moles of gaseous reactant
Hence the relation between K
p
and K
c
is-
K
p
=K
c
(RT)
Δn
g
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Answer:
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