MCQ
$40\%$ of a mixture of $0.2\,mol$ of $N_2$ and $0.6\,mol$ of $H_2$ reacts to give $NH_3$ according to the equation.

${N_2}(g)\, + \,3{H_2}(g)\,\rightleftharpoons  \,2N{H_3}\,(g)$

at constant temperature and pressure. Then the ratio of the final volume to the initial volume of gases are.

  • $4:5$
  • B
    $5:4$
  • C
    $7:10$
  • D
    $8:5$

Answer

Correct option: A.
$4:5$
a
${N_2}(g) + 3{H_2}(g) \rightleftharpoons 2N{H_3}(g)$

Moles $N_2$ $H_2$ $NH_3$
Initial $0.2$ $0.6$ $0$
At equilibrium $0.2-x$ $0.6-3x$ $2x$

Also,  $0.4=\frac{\mathrm{x}}{0.2} \Rightarrow \mathrm{x}=0.08$

Ratio  $=\frac{\mathrm{V}_{\mathrm{f}}}{\mathrm{V}_{\mathrm{i}}}=\frac{\left(\mathrm{n}_{\text {Toth }}\right)_{\mathrm{f}}}{\left(\mathrm{n}_{\text {Toth }}\right)_{\mathrm{i}}}=\frac{0.8-2 \mathrm{x}}{0.8}$

$=1-\frac{\mathrm{x}}{0.4} $

$=1-\frac{0.08}{0.40}=\frac{4}{5} $

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