Q75 / 197 · NEET 2009 (Mains) · Chemistry

For the reaction, $\mathrm{N}_2+3 \mathrm{H}_2 \rightarrow 2 \mathrm{NH}_3$, If $\frac{\mathrm{d}\left[\mathrm{NH}_3\right]}{\mathrm{dt}}=2 \times 10^{-4} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1}$, The value of $\frac{-\mathrm{d}\left[\mathrm{H}_2\right]}{\mathrm{dt}}$ would be -

  • A
    $1 \times 10^{-4} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1}$
  • Correct answer
    B
    $3 \times 10^{-4} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1}$
  • C
    $4 \times 10^{-4} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1}$
  • D
    $6 \times 10^{-4} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1}$

Solution

$$ \begin{aligned} & \frac{1}{3} \frac{-\mathrm{d}\left[\mathrm{H}_2\right]}{\mathrm{dt}}=\frac{1}{2} \frac{\left[\mathrm{NH}_3\right]}{\mathrm{dt}} \\ & \frac{-\mathrm{d}\left[\mathrm{H}_2\right]}{\mathrm{dt}}=\frac{3}{2} \frac{\mathrm{d}\left[\mathrm{NH}_3\right]}{\mathrm{dt}} \\ & \frac{-\mathrm{d}\left[\mathrm{H}_2\right]}{\mathrm{dt}}=\frac{3}{2} \times 2 \times 10^{-4} \\ & =3 \times 10^{-4} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1} \\ & \end{aligned} $$

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

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