Equilibrium - Result Question 12
####13. The value of equilibrium constant of the reaction
$HI(g) \rightarrow \frac{1}{2} H_2(g)+\frac{1}{2} I_2$ is 8.0
[2008]
The equilibrium constant of the reaction $H_2(g)+I_2(g) \rightarrow 2 HI(g)$ will be:
(a) $\frac{1}{16}$
(b) $\frac{1}{64}$
(c) 16
(d) $\frac{1}{8}$
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Solution:
- (b) Given : Equilibrium constant $(K_1)$ for the reaction:
$HI(g) \stackrel{K_1}{\rightarrow} \frac{1}{2} H_2(g)+\frac{1}{2} I_2(g) ; K_1=8 ;$
To find equilibrium constant for the following reaction:
$H_2(g)+I_2(g) \rightarrow 2 HI(g) ; K_2=?$
multiply (i) by 2 , we get
$2 HI(g) \rightarrow H_2(g)+I_2(g)$;
$ \begin{equation*} K_1=8^{2}=64 . \tag{iii} \end{equation*} $
Now reverse equation (iii), we get
$H_2(g)+I_2(g) \rightarrow 2 HI(g) ; K=\frac{1}{64}$
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.
