Question

A mixture of hydrogen and iodine in the mole ratio $1.5:1$ is maintained at ${450}^{\circ }C$ After the attainment of equilibrium, $H_2\left(g\right)+I_2\left(g\right),$ it is found on analysis that the mole ratio of $I_2$ to $HI$ is $1:18$. The equilibrium constant is :

• A. 54
• B. 46
• C. 64
• D. 76

Answer 1

The correct option is A 54

Initially, the number of moles of hydrogen iodine and $HI$ are 1.5, 1 and 0 respectively.
At equilibrium, the number of moles of hydrogen iodine and $HI$ are $(1.5-x),(1-x),2x$ respectively.
But the mole ratio of $I_2$ to $HI$ is $1:18$.
Thus, $\frac{I_2}{HI}=\frac{1}{18}=\frac{1-x}{2x}$ or $x=0.9$.
At equilibrium, the number of moles of hydrogen iodine and $HI$ are 0.6, 0.1 and 1.8 respectively.
The expression for the equilibrium constant is $K=\frac{[HI{]}^2}{\left[H_2\right]\left[I_2\right]}=\frac{{1.8}^2}{0.6\times 0.1}=54$.