Question

A mixture of $H_2$ and $I_2$ in molecular proportion of 2 : 3 was heated at ${440}^{\circ }C$ till the reaction $H_2+I_2$ $\rightleftharpoons$ 2HI reached equilibrium state. Calculate the percentage of iodine converted into HI $K_{c}at440$${}^{\circ }$C is 0.02)

• A. 3.38%
• B. 4.38%
• C. 5.38%
• D. 6.38%

Answer 1

The correct option is C

5.38%

$\begin{array}{cccc}& & & \\ & H_2+& I_2\rightleftharpoons & 2HI\\ Initialmoles& 2& 3& 0\\ AtEq{b}^{\prime }& \frac{2-x}{V}& \frac{3-x}{V}& \frac{2X}{V}\end{array}$

$K_C=\frac{4x^2}{(2-x)(3-x)}=0.02$

$199x^2+5x-6=0$

x = 0.1615

Out of 3 moles , 0.1615 moles of $I_2$ is converted into HI\)

Percentage of $I_2$ converted to

$HI=\frac{0.1615\times 100}{3}=5.38$