Question
In a reaction, $$2A \to $$ products, the concentration
of $$A$$ decreases from $$0.50\,M$$ to $$0.38\,M$$ in
$$10\,\min .$$ What is the rate of reaction $$\left( {{\text{in}}\,M\,{s^{ - 1}}} \right)$$ during this interval ?
A.
$$0.012$$
B.
$$0.024$$
C.
$$2 \times {10^{ - 3}}$$
D.
$$2 \times {10^{ - 4}}$$
Answer :
$$2 \times {10^{ - 4}}$$
Solution :
$$\eqalign{
& {\text{Rate of reaction}} = \frac{{d\left[ A \right]}}{{dt}} \cr
& {\text{Given,}}{\left[ A \right]_{initial}} = 0.50\,M \cr
& {\left[ A \right]_{final}} = 0.38\,M \cr
& dt = 10\,\min = 600\,\sec \cr
& d\left[ A \right] = 0.12 \cr
& {\text{Rate}} = \frac{{0.12}}{{600}} = 2 \times {10^{ - 4}}M\,{s^{ - 1}}. \cr} $$