In the reaction of formation of sulphur trioxide by contact process $2 \mathrm{SO}_2+\mathrm{O}_2…

In the reaction of formation of sulphur trioxide by contact process $2 \mathrm{SO}_2+\mathrm{O}_2 \rightleftharpoons 2 \mathrm{SO}_3$ the rate of reaction was measured as $\frac{\mathrm{d}\left[\mathrm{O}_2\right]}{\mathrm{dt}}=-2.5 \times 10^{-4} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1}$. The rate of reaction is terms of $\left[\mathrm{SO}_2\right]$ in $\mathrm{mol} \mathrm{L}^{-1} \mathrm{~s}^{-1}$ will be:
  1. $-1.25 \times 10^{-4}$
  2. $-2.50 \times 10^{-4}$
  3. $-3.75 \times 10^{-4}$
  4. $-5.00 \times 10^{-4}$

Solution

From rate law $\begin{aligned} &-\frac{1}{2} \frac{\mathrm{dSO}_2}{\mathrm{dt}}=-\frac{\mathrm{dO}_2}{\mathrm{dt}}=\frac{1}{2} \frac{\mathrm{dSO}_3}{\mathrm{dt}} \\ &\therefore-\frac{\mathrm{dSO}_2}{\mathrm{dt}}=-2 \times \frac{\mathrm{dO}_2}{\mathrm{dt}} \\ &=-2 \times 2.5 \times 10^{-4} \\ &=-5 \times 10^{-4} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1} \end{aligned}$

Asked in: MHT CET Full Test 10

Practice more Chemical Kinetics questions on Aicharya