Question

how to derive D = k' P

Answer 1

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