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

Answer: (A)

$\frac{1}{64}$

$HI\left(g\right)\rightleftharpoons \frac{1}{2}{H}_2\left(g\right)+\frac{1}{2}{I}_2$ $K_c=8.0$

When we reverse the reaction the new reaction constant is

$K_c=\frac{1}{K_c}=\frac{1}{8}$

$\frac{1}{2}{H}_2\left(g\right)+\frac{1}{2}{I}_2\left(g\right)\rightleftharpoons HI\left(g\right)$
$K_c=\frac{1}{K_c}$

When we multiply a
reaction by a factor n

$K_c=(K_c{)}^n$

Multiply by $2$

$H_2\left(g\right)+I_2\left(g\right)\rightleftharpoons 2HI\left(g\right)$

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