Question

The rate of formation of $S{O}_3$ in the reaction $2S{O}_2+O_2\to 2S{O}_3$ $100g{min}^{-1}.$ Then, the rate of disappearance of $O_2$ is:

• A. $50g{min}^{-1}$
• B. $100g{min}^{-1}$
• C. $20g{min}^{-1}$
• D. $40g{min}^{-1}$

Answer 1

Answer: (A)

$50g{min}^{-1}$

Here reaction is: $2S{O}_2+O_2\to 2S{O}_3$

By rate law:

$=\frac{-1}{2}\frac{d\left(O_2\right)}{dt}=-\frac{d\left(O_2\right)}{dt}=\frac{+1}{2}\frac{d\left(S{O}_3\right)}{dt}.....\left(1\right)$

Given the rate of formation of $S{O}_3=\frac{+d\left(S{O}_3\right)}{dt}=100g{m}^{-1}$

From (1)

$=\frac{-d\left(O_2\right)}{dt}=\frac{+1}{2}\frac{d\left(S{O}_3\right)}{dt}$

Rate of disappearance of $O_2$ $\Rightarrow \frac{-d\left(O_2\right)}{dt}=\frac{1}{2}\times 100=50g{min}^{-1}$