Question

Question 7.16 What is the equilibrium concentration of each of the substances in the equilibrium when the initial concentration of ICl was 0.78 M?

2 ICl(g)⇌I2(g) + Cl2(g) ; KC = 0.14

Class XI Equilibrium Page 225

Answer 1

2ICl (g) <=> I2(g) + Cl2(g) ; Kc = 0.14
at t = 0
conc. of ICl = 0.78 M
conc. of I2 = 0 M
conc. of Cl2 = O M

at equilibrium ,
conc. of ICl = (0.78 - 2x ) M
conc. of I2 = x M
conc. of Cl2 = x M

now, Kc = [I2][Cl2]/[ICl]²
Kc = x.x/(0.78 - 2x )²
0.14 = x²/(0.78 - 2x )²
take square root both sides,
√0.14 = x/(0.78 - 2x )
0.374 = x/(0.78 - 2x )
x = 0.29172 - 0.748x
1.748x = 0.21972
x = 0.29172/1.748 = 0.1668 ≈ 0.17

hence, conc. of ICl at equilibrium = 0.78 - 0.34 = 0.44 M
conc. of I2 = x = 0.17 M
conc. of Cl2 = x = 0.17 M

Answer 2

Answer: Equilibrium concentration of $I_2$ is 0.17

Equilibrium concentration of $Cl_2$ is 0.17

Equilibrium concentration of $ICl$ is 0.44

Explanation:

The given balanced equilibrium reaction is,

                   $2ICl(g)\rightleftharpoons I_2(g)+Cl_2(g)$

  Initial conc.         0.78 M          0     0  

At eqm. conc.    (0.78-2x) M   (x) M   (x) M

Given : $K_c=0.14$

The expression for equilibrium constant for this reaction will be:

$K_c=\frac{[I_2]^1[Cl_2]^1}{[ICl]^2}$

Now put all the given values in this expression, we get :

$0.14=\frac{x\times x}{(0.78-2x)^2}$

$x=0.17$

Thus the equilibrium concentration of $I_2$ is 0.17, equilibrium concentration of $Cl_2$ is 0.17 and equilibrium concentration of $ICl$ is $(0.78-2\times 0.17)=0.44$