Question
In a reaction, $$2X \to Y,$$ the concentration of $$X$$ decreases from $$0.50\,M$$ to $$0.38\,M$$ in $$10\,\min .$$ What is the rate of reaction in $$M\,{s^{ - 1}}$$ during this interval?
A.
$$1 \times {10^{ - 4}}$$
B.
$$4 \times {10^{ - 2}}$$
C.
$$2 \times {10^{ - 2}}$$
D.
$$1 \times {10^{ - 2}}$$
Answer :
$$1 \times {10^{ - 4}}$$
Solution :
$$\eqalign{
& {\text{Rate of reaction}} = - \frac{1}{2}\frac{{\Delta \left[ X \right]}}{{\Delta t}} \cr
& \Delta \left[ X \right] = 0.38 - 0.50 = 0.12\,M \cr
& {\text{Rate}} = \frac{1}{2}\frac{{0.12}}{{10 \times 60}} \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{2 \times {{10}^{ - 4}}}}{2} \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\, = 1 \times {10^{ - 4}}\,M\,{s^{ - 1}} \cr} $$