Q7 / 170 · Manipal (MET) 2015 · Chemistry

Equivalent amounts of $\mathrm{H}_2$ and $\mathrm{I}_2$ are heated in a closed vessel till equilibrium is obtained. If $80 \%$ of the hydrogen can be converted to HI, the $K_C$ at this temperature is

  • Correct answer
    A
    $64$
  • B
    $16$
  • C
    $0.25$
  • D
    $4$

Solution

$\mathrm{H}_2+\mathrm{I}_2 \rightleftharpoons 2 \mathrm{HI}$ $\begin{array}{lccc}\text { At initial } & 1 & 1 & 0 \\ \text { At equili } & 1-0.8 & 1-0.8 & 2 \times 0.8 \\ & =0.2 & =0.2 & =1.6\end{array}$ $\begin{array}{ll}\because & K_C=\frac{[\mathrm{HI}]^2}{\left[\mathrm{H}_2\right]\left[\mathrm{I}_2\right]} \\ \therefore & K_C=\frac{1.6 \times 1.6}{0.2 \times 0.2} \\ \Rightarrow & K_C=64\end{array}$

Hence, the correct answer is option: 'A'.

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