Q40 / 150 · MHT CET 2025 (5 May Shift 2) · Chemistry

For the reaction, $2 \mathrm{~N}_2 \mathrm{O}_{5(\mathrm{~g})} \rightarrow 4 \mathrm{NO}_{2(\mathrm{~g})}+\mathrm{O}_{2(\mathrm{~g})}$ $\mathrm{N}_2 \mathrm{O}_5$ disappears at a rate of $x \mathrm{~mol~} \mathrm{dm}^{-3} \mathrm{~s}^{-1}$ Find the rate of formation of $\mathrm{O}_2$ ?

  • A
    $x \mathrm{~mol} \quad \mathrm{dm}^{-3} \mathrm{~s}^{-1}$
  • B
    $2 x \mathrm{~mol} \quad \mathrm{dm}^{-3} \mathrm{~s}^{-1}$
  • Correct answer
    C
    $\frac{x}{2} \mathrm{~mol} \quad \mathrm{dm}^{-3} \mathrm{~s}^{-1}$
  • D
    $\frac{3 x}{2} \mathrm{~mol} \quad \mathrm{dm}^{-3} \mathrm{~s}^{-1}$

Solution

From the balanced reaction $2\mathrm{N}_2\mathrm{O}_{5(\mathrm{g})} \to 4\mathrm{NO}_{2(\mathrm{g})} + \mathrm{O}_{2(\mathrm{g})}$, the reaction rate is related to concentration changes by:

$-\frac{1}{2}\frac{d[\mathrm{N}_2\mathrm{O}_5]}{dt} = +\frac{1}{4}\frac{d[\mathrm{NO}_2]}{dt} = +\frac{d[\mathrm{O}_2]}{dt}$

Given $-\frac{d[\mathrm{N}_2\mathrm{O}_5]}{dt} = x\ \mathrm{mol\ dm}^{-3}\ \mathrm{s}^{-1}$, the formation rate of $\mathrm{O}_2$ becomes:

$+\frac{d[\mathrm{O}_2]}{dt} = -\frac{1}{2}\frac{d[\mathrm{N}_2\mathrm{O}_5]}{dt} = \frac{1}{2}x\ \mathrm{mol\ dm}^{-3}\ \mathrm{s}^{-1}$

Answer: $\boxed{\text{C}}$

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

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