Chemistry, asked by nagarajan19734041, 5 months ago

Derive the general expression for the equilibrium constant Kp & Ke for the reaction. ​

Answers

Answered by s02371joshuaprince47
0

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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Answered by deveshreem
1

Answer:

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