Question

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

• A. $2gmi{n}^{-1}$
• B. $20gmi{n}^{-1}$
• C. $200gmi{n}^{-1}$
• D. $50gmi{n}^{-1}$

Answer 1

The correct option is B $20gmi{n}^{-1}$

For the given reaction.

$2S{O}_2+O_2⟶2SO3$

$\frac{1}{2}\frac{-d\left[SO2\right]}{dt}=\frac{-d\left[O_2\right]}{dt}=\frac{1}{2}\frac{d\left[S{O}_3\right]}{dt}$

$\frac{d\left[S{O}_3\right]}{dt}=100gmi{n}^{-1}=\frac{100}{80}molmi{n}^{-1}$

$\frac{-d\left[O_2\right]}{dt}=\frac{1}{2}\left(\frac{100}{80}\right)molmi{n}^{-1}$

$=\frac{1}{2}\times \frac{100}{80}\times 32gmi{n}^{-1}=20gmi{n}^{-1}$

Hence. option $B$ is correct.