Q8 / 150 · MHT CET 2025 (23 Apr Shift 1) · Chemistry

Find the rate of following reaction. $2 \mathrm{~N}_2 \mathrm{O}_{5(\mathrm{~g})} \rightarrow 4 \mathrm{NO}_{2(\mathrm{~g})}+\mathrm{O}_{2(\mathrm{~g})}$ if concentration of $\mathrm{NO}_2$ increases to $5.2 \times 10^{-3} \mathrm{M}$ in 100 seconds.

  • A
    $1.3 \times 10^{-3} \mathrm{~mol} \quad \mathrm{dm}^{-3} \mathrm{~s}^{-1}$
  • Correct answer
    B
    $1.3 \times 10^{-5} \mathrm{~mol} \quad \mathrm{dm}^{-3} \mathrm{~s}^{-1}$
  • C
    $4.0 \times 10^{-5} \mathrm{~mol} \quad \mathrm{dm}^{-3} \mathrm{~s}^{-1}$
  • D
    $5.2 \times 10^{-5} \mathrm{~mol} \quad \mathrm{dm}^{-3} \mathrm{~s}^{-1}$

Solution

The rate of formation of $\mathrm{NO}_2$ is determined from the concentration change and time interval as $\frac{5.2 \times 10^{-3} \mathrm{M}}{100 \mathrm{s}} = 5.2 \times 10^{-5} \mathrm{M/s}$.

From the stoichiometry of the reaction $2 \mathrm{N_2O_5(g)} \rightarrow 4 \mathrm{NO_2(g)} + \mathrm{O_2(g)}$, the relationship between the reaction rate and the rate of $\mathrm{NO_2}$ formation is $\text{Rate} = \frac{1}{4} \frac{d[\mathrm{NO_2}]}{dt}$.

Substituting gives $\text{Rate} = \frac{1}{4} \times (5.2 \times 10^{-5} \mathrm{M/s}) = 1.3 \times 10^{-5} \mathrm{M/s}$.

This matches option B: $1.3 \times 10^{-5} \mathrm{mol~dm^{-3}~s^{-1}}$.

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

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