Question
In the reaction of formation of sulphur trioxide by contact process $$2S{O_2} + {O_2} \rightleftharpoons 2S{O_3}$$ the rate of reaction was measured as $$\frac{{d\left[ {{O_2}} \right]}}{{dt}} = - 2.5 \times {10^{ - 4}}\,mol\,{L^{ - 1}}{s^{ - 1}}.$$ The rate of reaction is terms of $$\left[ {S{O_2}} \right]$$ in $$mol\,{L^{ - 1}}{s^{ - 1}}$$ will be :
A.
$$ - 1.25 \times {10^{ - 4}}$$
B.
$$ - 2.50 \times {10^{ - 4}}$$
C.
$$ - 3.75 \times {10^{ - 4}}$$
D.
$$ - 5.00 \times {10^{ - 4}}$$
Answer :
$$ - 5.00 \times {10^{ - 4}}$$
Solution :
$$\eqalign{
& {\text{From rate law}} \cr
& - \frac{1}{2}\frac{{dS{O_2}}}{{dt}} = - \frac{{d{O_2}}}{{dt}} = \frac{1}{2}\frac{{dS{O_3}}}{{dt}} \cr
& \therefore \,\, - \frac{{dS{O_2}}}{{dt}} = - 2 \times \frac{{d{O_2}}}{{dt}} \cr
& = - 2 \times 2.5 \times {10^{ - 4}} \cr
& = - 5 \times {10^{ - 4}}\,mol\,{L^{ - 1}}{s^{ - 1}} \cr} $$