Relationship between Homogeneous Equilibrium and the Law of Mass Action (Part C)::
Case
C- In gaseous reaction when no of moles decreases:
To understand this case, suppose the formation of ammonia by Haber Process
t=0 a b 0
t=eq (a-x) (b-3x) 2x
Suppose “x” moles of N2 are
converted to product then, 3x moles of H2 are consumed to give the
2x moles of NH3 .
Total no of moles “N” for this reaction
is (a-x)+ (b-3x)+2x = (a+b-2x)
Expression for Kc:
[N2] = (a-x)/V
[H2] = (b-3x)/V
[NH3] = 2x/V
Put the values in Kc expression
Kc
= [NH3]2
[N2][H2]3
Kc
= [2x/V]2
[(a-x)/V][ (b-3x)/V] 3
Kc = 4x2V2
(a-x)(b-3x)
If the V of the reaction mixture is increased, the x has to be decreased to keep the equilibrium constant and reaction goes towards the backward direction.
Expression for Kp:
According to Dalton theory,
pi = XiP we know that Xi = n/N
PN2
= (a-x) P
(a+b-2x)
PH2
= (b-3x) P
(a+b-2x)
PNH3
= (2x) P
(a+b-2x)
Put the values in Kp expression
Kp
= P NH32
PN2 (PH2)3
(2x)2
P2
Kp
= [(a+b-2x)]
2
(a-x) P [(b-3x) ] 3 P3
(a+b-2x) [(a+b-2x)]3
Kp = 4x2(a+b-2x)2
(a-x)(b-3x)3 P2
Expression for Kx:
XN2
= (a-x)
(a+b-2x)
XH2
= (b-3x)
(a+b-2x)
XNH3
= (2x)
(a+b-2x)
Put the values in Kx expression
Kx
= X NH32
XN2
(XH2)3
(2x)2
Kx= [(a+b-2x)] 2
(a-x) [(b-3x) ]3
(a+b-2x) [(a+b-2x)]3
Kx = 4x2(a+b-2x)2
(a-x)(b-3x)3
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