MCQ
Equivalent amounts of $H_2$ and $I_2$ are heated in a closed vessel till equilibrium is obtained. If $80\%$ of the hydrogen is converted to $HI$, the $K_c$ at this temperature is
- ✓$64$
- B$16$
- C$0.25$
- D$14$
✓
Answer
Correct option: A.
$64$
a
$\mathrm{H}_{2}+\mathrm{I}_{2} \rightleftharpoons 2 \mathrm{HI}$
$\mathrm{H}_{2}+\mathrm{I}_{2} \rightleftharpoons 2 \mathrm{HI}$
At initial $\quad 1 \quad 1 \quad 0$
At equal $(1-0.8)(1-0.8) \quad 2 \times 0.8$
$\quad=0.2=0.2 \quad=1.6$
$\therefore \mathrm{K}_{\mathrm{c}}=\frac{[\mathrm{HI}]^{2}}{\left[\mathrm{H}_{2}\right]\left[\mathrm{I}_{2}\right]}$
$\therefore \mathrm{K}_{\mathrm{c}}=\frac{1.6 \times 1.6}{0.2 \times 0.2}$
$\Rightarrow \mathrm{K}_{\mathrm{c}}=64$
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