Question

initial pressure of A 10 atm and at equilibrium partial pressure is 6 atm calculate KP for the reaction 2A equilibrium B+C​

Answer 1

Answer:

Explanation:

Given,

$2A\rightleftharpoons B+C$

Initial Pressure of A = 10atm

Partial Pressure of A = 6atm

To Find :-

K_p for the reaction.

Solution :-

                   $\sf 2A\rightleftharpoons B+C$

For '2A' :-

Initial pressure(I) :- 10atm

Change in Pressure(C) :- - 2x

Equilibrium Pressure(E) :- 10 - 2x  = 6atm[Given]

10-6 = 2x

4 = 2x

x = 4/2 = 2.

For 'B' :-

Initial pressure(I) :- 0atm

Change in Pressure(C) :- x = 2atm

Equilibrium Pressure(E) :- 0+2atm = 2atm

For 'C' :-

Initial pressure(I) :- 0atm

Change in Pressure(C) :- x = 2atm

Equilibrium Pressure(E) :- 0+2atm = 2atm.

$K_{p}(at\:equilibrium)=\dfrac{[B][C]}{[A]^2}$

$=\dfrac{2\times 2}{6^2}$

$=\dfrac{4}{36}$

$=\dfrac{1}{9}$