For the reaction $2N_2O_5→4NO_2 + O_2$ at a given instant of time, the rate and rate constant are $1.02 × 10^{-4}\, mol\, L^{-1}\, s^{-1}$ and $3.4 × 10^{-5}\, s^{-1}$, respectively. The concentration of $N_2O_5$ at that time will be:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → $3.0\, mol\, L^{-1}$
The unit of the rate constant (k) is $\text{s}^{-1}$. This indicates a first-order reaction.
For a first-order reaction, the rate law is: Rate =$k[A]$
For the reaction $2\text{N}_2\text{O}_5\rightarrow4\text{NO}_2+\text{O}_2$, the rate law is: Rate = $k[\text{N}_2\text{O}_5]$
Given:
$\text{Rate}=1.02\times10^{-4}\,\text{mol L}^{-1}\text{s}^{-1}$
$k=3.4\times10^{-5}\,\text{s}^{-1}$
$[\text{N}_2\text{O}_5]=\frac{\text{Rate}}{k}$
$[\text{N}_2\text{O}_5]=\frac{1.02\times10^{-4}\,\text{mol L}^{-1}\text{s}^{-1}}{3.4\times10^{-5}\,\text{s}^{-1}}$
$[\text{N}_2\text{O}_5]=\left(\frac{1.02}{3.4}\right)\times\left(\frac{10^{-4}}{10^{-5}}\right)\,\text{mol L}^{-1}$
$[\text{N}_2\text{O}_5]=0.3\times10^1\,\text{mol L}^{-1}$
$[\text{N}_2\text{O}_5]=3.0\,\text{mol L}^{-1}$