The reactions rate $\mathrm{N}_{2}(\mathrm{~g})+3 \mathrm{H}_{2}(\mathrm{~g}) ightarrow 2…

The reactions rate $\mathrm{N}_{2}(\mathrm{~g})+3 \mathrm{H}_{2}(\mathrm{~g}) ightarrow 2 \mathrm{NH}_{3}(\mathrm{~g})$ was measured $\frac{\mathrm{d}\left[\mathrm{NH}_{3}ight]}{\mathrm{dt}}=2 \times 10^{-4} \mathrm{~mol} \mathrm{sec}^{-1}$ The rates of reactions expressed in terms of $\mathrm{N}_{2}$ and $\mathrm{H}_{2}$ are
  1. Rate in terms of $\mathbf{N}_{2}$ = $2 \times 10^{-4}$ $\left(\mathrm{mol} \mathrm{L}^{-1} \mathrm{sec}^{-1}ight)$
    Rate in terms of $\mathbf{H}_{2}$ = $2 \times 10^{-4}$ $\left(\mathrm{mol} \mathrm{L}^{-1} \mathrm{sec}^{-1}ight)$
  2. Rate in terms of $\mathbf{N}_{2}$ = $3 \times 10^{-4}$ $\left(\mathrm{mol} \mathrm{L}^{-1} \mathrm{sec}^{-1}ight)$
    Rate in terms of $\mathbf{H}_{2}$ = $1 \times 10^{-4}$ $\left(\mathrm{mol} \mathrm{L}^{-1} \mathrm{sec}^{-1}ight)$
  3. Rate in terms of $\mathbf{N}_{2}$ = $1 \times 10^{-4}$ $\left(\mathrm{mol} \mathrm{L}^{-1} \mathrm{sec}^{-1}ight)$
    Rate in terms of $\mathbf{H}_{2}$ = $3 \times 10^{-4}$ $\left(\mathrm{mol} \mathrm{L}^{-1} \mathrm{sec}^{-1}ight)$
  4. Rate in terms of $\mathbf{N}_{2}$ = $2 \times 10^{-1}$ $\left(\mathrm{mol} \mathrm{L}^{-1} \mathrm{sec}^{-1}ight)$
    Rate in terms of $\mathbf{H}_{2}$ = $2 \times 10^{-3}$ $\left(\mathrm{mol} \mathrm{L}^{-1} \mathrm{sec}^{-1}ight)$

Solution

$\mathrm{N}_{2}+3 \mathrm{H}_{2} ightleftharpoons 2 \mathrm{NH}_{3} ;$ Rate is given by any of
the expressions
$\frac{-\mathrm{d}\left[\mathrm{N}_{2}ight]}{\mathrm{dt}}=-\frac{1 \mathrm{~d}\left[\mathrm{H}_{2}ight]}{3 \mathrm{dt}}=\frac{1}{2} \frac{\mathrm{d}\left(\mathrm{NH}_{3}ight]}{\mathrm{dt}}$
Rate of disappearance of $\mathrm{N}_{2}=\frac{1}{2}$ the rate of formation of $\mathrm{NH}_{3}=1 \times 10^{-4}$
Rate of disappearance of $\mathrm{H}_{2}=3 / 2$ the rate of formation of $\mathrm{NH}_{3}=3 \times 10^{-4}$

Asked in: JEE-TOPICTESTS-CHEMISTRY

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