Question

The value of equilibrium constant of the reaction, $H I(g) \rightleftharpoons \frac{1}{2} H_{2}(g)+\frac{1}{2} I_{2}(g)$ is $8.0$. The equilibrium constant of the reaction, $H _{2}(g)+I_{2}(g) \rightleftharpoons 2 H I(g)$ Will be

• A. $ \frac{1}{16} $
• B. $ \frac{1}{64} $
• C. $16$
• D. $ \frac{1}{8} $

Answer 1

Answer: (B)

For reaction,
$HI _{( g )} \rightleftharpoons \frac{1}{2} H _{2( g )}+\frac{1}{2} I _{2( g )} ; K _{ c }=8.0 $
$ K _{ c }=\frac{[ HI ]}{\left[ H _{2}\right]^{1 / 2}\left[ H _{2}\right]^{1 / 2}}$
$\Rightarrow \frac{[ HI ]}{\left[ H _{2}\right]^{1 / 2}\left[ H _{2}\right]^{1 / 2}}=8 \rightarrow \text { (i) }$
Now, for the rection,
$ H _{2( g )}+ I _{2( g )} \rightleftharpoons 2 HI _{( g )} ; K _{ c }^{\prime}=?$
$ K _{ c }^{\prime}=\frac{\left[ H _{2}\right]\left[ I _{2}\right]}{[ HI ]^{2}} $
$ K _{ c }^{\prime}=\frac{1}{\left(\frac{[ HI ]}{\left[ H _{2}\right]^{1 / 2}\left[ H _{2}\right]^{1 / 2}}\right)^{2}}$
From $e^{ n }$ (i), we have
$K _{ c }^{\prime}=\frac{1}{(8)^{2}}=\frac{1}{64}$