Question

The value of equilibrium constant of the reaction $HI\left(g\right)\rightleftharpoons \frac{1}{2}{H}_2\left(g\right)+\frac{1}{2}{I}_2$ is $8.0$
The equilibrium constant of the reaction
$H_2\left(g\right)+I_2\left(g\right)\rightleftharpoons 2HI\left(g\right)$ will be:

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

Answer 1

The correct option is A $\frac{1}{64}$

For reaction,

${HI}_{\left(g\right)}\rightleftharpoons \frac{1}{2}{H_2}_{\left(g\right)}+\frac{1}{2}{I_2}_{\left(g\right)};K_c=8.0$

$K_c=\frac{\left[HI\right]}{{\left[H_2\right]}^{1/2}{\left[H_2\right]}^{1/2}}$

$\Rightarrow \frac{\left[HI\right]}{{\left[H_2\right]}^{1/2}{\left[H_2\right]}^{1/2}}=8⟶\left(i\right)$

Now, for the rection,

${H_2}_{\left(g\right)}+{I_2}_{\left(g\right)}\rightleftharpoons 2{HI}_{\left(g\right)};{K_c}^{\prime }=?$

${K_c}^{\prime }=\frac{\left[H_2\right]\left[I_2\right]}{{\left[HI\right]}^2}$

${K_c}^{\prime }=\frac{1}{{\left(\frac{\left[HI\right]}{{\left[H_2\right]}^{1/2}{\left[H_2\right]}^{1/2}}\right)}^2}$

From ${eq}^n\left(i\right)$, we have

$K_c^{\prime }=\frac{1}{{\left(8\right)}^2}=\frac{1}{64}$