The decomposition of $\mathrm{NH}_3$ on platinum surface is zero order reaction. The rates of production of $\mathrm{N}_2$ and $\mathrm{H}_2$ are respectively $\left(k=2.5 \times 10^{-4} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1}\right)$

• A

  $7.5 \times 10^{-4} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1} ; 2.5 \times 10^{-4} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1}$
• Correct answer
  B

  $2.5 \times 10^{-4} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1} ; 7.5 \times 10^{-4} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1}$
• C

  $5.0 \times 10^{-4} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1} ; 2.5 \times 10^{-4} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1}$
• D

  $2.5 \times 10^{-4} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1} ; 5.0 \times 10^{-4} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1}$

Solution

For the reaction, $2 \mathrm{NH}_3 \rightarrow \mathrm{N}_2+3 \mathrm{H}_2$

From equation (i) and (ii)
$\therefore$ Rate of production of $\mathrm{N}_2, \frac{d\left[\mathrm{~N}_2\right]}{d t}=k$ $=2.5 \times 10^{-4} \mathrm{~mol} \mathrm{~L} \mathrm{~L}^{-1} \mathrm{~s}^{-1}$ From equation (iii) Rate of production of $\mathrm{H}_2, \frac{d\left[\mathrm{H}_2\right]}{d t}=3 k$ $\begin{aligned} & =3 \times 2.5 \times 10^{-4} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1} \\ & =7.5 \times 10^{-4} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1}\end{aligned}$

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