Question

In the formation of sulphur trioxide by the contact process $2S{O}_2\left(g\right)+O_2\left(g\right)\to 2S{O}_3\left(g\right)$
The rate of
reaction is expressed as $-\frac{d\left(O_2\right)}{dt}=2.5\times {10}^{-4}mol$
$L^{-1}se{c}^{-1}$ The rate of disappearance of $\left(S{O}_2\right)$ will be:

• A. $5\times {10}^{-4}mol{L}^{-1}{S}^{-1}$
• B. $-1.25\times {10}^{-4}mol{L}^{-1}{S}^{-1}$
• C. $3.75\times {10}^{-4}mol{L}^{-1}{S}^{-1}$
• D. $50.0\times {10}^{-4}mol{L}^{-1}{S}^{-1}$

Answer 1

The correct option is A $5\times {10}^{-4}mol{L}^{-1}{S}^{-1}$

$2S{O}_{2\left(g\right)}+O_{2\left(g\right)}\to 2S{O}_{3\left(g\right)}$

$\therefore \frac{-1}{2}\frac{d\left[S{O}_2\right]}{dt}=-\frac{d\left[O_2\right]}{dt}=\frac{1}{2}\frac{d\left[S{O}_3\right]}{dt}$

$\Rightarrow -\frac{d\left[S{O}_2\right]}{dt}=5\times {10}^{-4}$

$\Rightarrow$ Rate of dissapearance = $5\times {10}^{-4}$ of $S{O}_2$