Equilibrium Ques 5
5. The value of equilibrium constant of the reaction
$HI(g) \rightleftharpoons \frac{1}{2} H_2(g)+\frac{1}{2} I_2$ is $8.0$
The equilibrium constant of the reaction $H_2(g)+I_2(g) \rightleftharpoons 2 HI(g)$ will be:
[2008]
(a) $\frac{1}{16}$
(b) $\frac{1}{64}$
(c) 16
(d) $\frac{1}{8}$
Show Answer
Answer:
Correct Answer: 5.(b)
Solution:
- (b) Given : Equilibrium constant $(K_1)$ for the reaction:
$HI(g) \stackrel{K_1}{\rightleftharpoons} \frac{1}{2} H_2(g)+\frac{1}{2} I_2(g) ; K_1=8 ;$ $\quad$ ……..(i)
To find equilibrium constant for the following reaction:
$H_2(g)+I_2(g) \rightleftharpoons 2 HI(g) ; K_2=?$ $\quad$ ……..(ii)
multiply (i) by $2$ , we get
$2 HI(g) \rightleftharpoons H_2(g)+I_2(g)$;
$K_1=8^{2}=64 $ $\quad$ ……..(iii)
Now reverse equation (iii), we get
$H_2(g)+I_2(g) \rightleftharpoons 2 HI(g) ; K=\frac{1}{64}$ $\quad$ ……..(iv)
Equation (iv) is the same as the required equation
(ii), thus $K_2$ for equation (ii) is $\frac{1}{64}$ i.e. option (b) is correct.
When the equation for an equilibrium is multiplied by a factor, the equilibrium constant must be raised to the power equal to the factor.
For a reversible reaction, the equilibrium constant of the backward reaction is inverse of the equilibrium constant for the forward reaction.
BETA
