Question
For a reaction $$\frac{1}{2}A \to 2B,$$ rate of disappearance of $$‘A’$$ is related to the rate of appearance of $$'B’$$ by the expression
A.
$$ - \frac{{d\left[ A \right]}}{{dt}} = \frac{1}{2}\frac{{d\left[ B \right]}}{{dt}}$$
B.
$$ - \frac{{d\left[ A \right]}}{{dt}} = \frac{1}{4}\frac{{d\left[ B \right]}}{{dt}}$$
C.
$$ - \frac{{d\left[ A \right]}}{{dt}} = \frac{{d\left[ B \right]}}{{dt}}$$
D.
$$ - \frac{{d\left[ A \right]}}{{dt}} = 4\frac{{d\left[ B \right]}}{{dt}}$$
Answer :
$$ - \frac{{d\left[ A \right]}}{{dt}} = \frac{1}{4}\frac{{d\left[ B \right]}}{{dt}}$$
Solution :
$$\eqalign{
& {\text{The rates of reactions for the reaction}} \cr
& \frac{1}{2}A \to 2B \cr
& {\text{can be written either as}} \cr
& - 2\frac{d}{{dt}}\left[ A \right]\,\,\,{\text{with respect to }}'A' \cr
& or\,\,\frac{1}{2}\frac{d}{{dt}}\left[ B \right]\,\,\,{\text{with respect to }}'B' \cr
& {\text{From the above, we have}} \cr
& - 2\frac{d}{{dt}}\left[ A \right] = \frac{1}{2}\frac{d}{{dt}}\left[ B \right] \cr
& or\,\, - \frac{d}{{dt}}\left[ A \right] = \frac{1}{4}\frac{d}{{dt}}\left[ B \right] \cr
& {\text{i}}{\text{.e}}{\text{., correct answer is (B)}} \cr} $$