Question

In the formation of sulphur trioxide by contact process, $2S{O}_2+O_2\rightleftharpoons 2S{O}_3$, the rate of reaction was measured as $\frac{-d\left[O_2\right]}{dt}=2.5\times {10}^{-4}{\text{mol L}}^{-1}{\text{sec}}^{-1}$ the rate of reaction expressed in terms of $S{O}_3$ will be:

• A. $-1.25\times {10}^{-4}{\text{mol L}}^{-2}{\text{sec}}^{-1}$
• B. $50\times {10}^{-4}{\text{mol L}}^{-1}{\text{sec}}^{-1}$
• C. $-3.75\times {10}^{-4}{\text{mol L}}^{-1}{\text{sec}}^{-1}$
• D. $5.00\times {10}^{-4}{\text{mol L}}^{-1}{\text{sec}}^{-1}$

Answer 1

The correct option is D $5.00\times {10}^{-4}{\text{mol L}}^{-1}{\text{sec}}^{-1}$

The given reaction is:
$2S{O}_2+O_2\rightleftharpoons 2S{O}_3$
Rate of the reaction: $\frac{-d\left[O_2\right]}{dt}=\frac{1}{2}\frac{d\left[S{O}_3\right]}{dt}$
$\frac{d\left[S{O}_3\right]}{dt}=2\times 2.5\times {10}^{-4}=5\times {10}^{-4}{\text{mol L}}^{-1}{\text{sec}}^{-1}$