Question
For the reaction, $$2{N_2}{O_5} \to 4N{O_2} + {O_2}$$ rate and rate constant are $$1.02 \times {10^{ - 4}}\,mol\,{L^{ - 1}}{s^{ - 1}}$$ and $$3.4 \times {10^{ - 5}}\,{s^{ - 1}}$$ respectively. The concentration of $${N_2}{O_5}$$ in $$mol\,{L^{ - 1}}$$ will be
A.
$$3.4 \times {10^{ - 4}}$$
B.
$$3.0$$
C.
$$5.2$$
D.
$$3.2 \times {10^{ - 5}}$$
Answer :
$$3.0$$
Solution :
$$\eqalign{
& \left[ {{N_2}{O_5}} \right] = \frac{{{\text{Rate}}}}{k} \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{1.02 \times {{10}^{ - 4}}\,mol\,{L^{ - 1}}{s^{ - 1}}}}{{3.4 \times {{10}^{ - 5}}\,{s^{ - 1}}}} \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 3\,mol\,{L^{ - 1}} \cr} $$