Q23 / 120 · VITEEE 2018 · Chemistry

For the reaction $2 \mathrm{~N}_{2} \mathrm{O}_{5} \rightarrow 4 \mathrm{NO}_{2}+\mathrm{O}_{2}$, rate and rate constant are $1.02 \times 10^{-4} \mathrm{~mol} \mathrm{lit}^{-1} \mathrm{sec}^{-1}$ and $3.4 \times 10^{-5} \mathrm{sec}^{-1}$ respectively then concentration of $\mathrm{N}_{2} \mathrm{O}_{5}$ at that time will be

  • A
    $1.732 \mathrm{M}$
  • Correct answer
    B
    $3 \mathrm{M}$
  • C
    $3.4 \times 10^{5} \mathrm{M}$
  • D
    $1.02 \times 10^{-4} \mathrm{M}$

Solution

$2 \mathrm{~N}_{2} \mathrm{O}_{5} \rightarrow 4 \mathrm{NO}_{2}+\mathrm{O}_{2}$ from the unit of rate constant it is clear that the reaction follow first order kinetics. Hence by rate law equation, $\mathrm{r}=k\left[\mathrm{~N}_{2} \mathrm{O}_{5}\right]$ where $\mathrm{r}=1.02 \times 10^{-4}, k=3.4 \times 10^{-5}$ $1.02 \times 10^{-4}=3.4 \times 10^{-5}\left[\mathrm{~N}_{2} \mathrm{O}_{5}\right]$ $\left[\mathrm{N}_{2} \mathrm{O}_{5}\right]=3 \mathrm{M}$

Hence, the correct answer is option: 'B'.

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