Question

The equilibrium constant at 717 K for the reaction :
$H_{2_{\left(g\right)}}+I_{2_{\left(g\right)}}\rightleftharpoons {2HI}_{\left(g\right)}$ is 50.

The equilibrium constant for the reaction :
${2HI}_{\left(g\right)}\rightleftharpoons H_{2_{\left(g\right)}}+I_{2_{\left(g\right)}}$ at same temperature is:

• A. $0.50$
• B. $2\times {10}^{-2}$
• C. $4.0$
• D. $1\times {10}^{-1}$

Answer 1

The correct option is B $2\times {10}^{-2}$

Here, $K_{equilibrium}$ = 50

If $K^1$eq = $\frac{1}{Keq}$

K= $\frac{\left[product\right]}{\left[reactant\right]}$

$K_{eq}=\frac{[HI{]}^2}{\left[H_2\right]\left[I_2\right]}$

$K_{eq}^1$=$\frac{\left[H_2\right]\left[I_2\right]}{[HI{]}^2}$

$K_{eq}^1$ = $\frac{1}{50}$

= 2 X${10}^{-}$${}^2$

Option B is correct.