Question

The rate constant $k$, for the reaction
$N_2{O}_5\left(g\right)⟶2{NO}_2\left(g\right)+\frac{1}{2}{O}_2\left(g\right)$
is $1.3\times {10}^{-2}{s}^{-1}$. Which equation given below describes the change of $\left[N_2{O}_5\right]$ with time?
${\left[N_2{O}_5\right]}_0$ and ${\left[N_2{O}_5\right]}_t$ correspond to concentration of $N_2{O}_5$ initially and at time $t$.

• A. ${\left[N_2{O}_5\right]}_t={\left[N_2{O}_5\right]}_0+kt$
• B. ${\left[N_2{O}_5\right]}_0={\left[N_2{O}_5\right]}_t{e}^{kt}$
• C. $\log {\left[N_2{O}_5\right]}_t=\log {\left[N_2{O}_5\right]}_0+kt$
• D. $\ln \frac{{\left[N_2{O}_5\right]}_0}{{\left[N_2{O}_5\right]}_t}=kt$

Answer 1

The correct option is D $\ln \frac{{\left[N_2{O}_5\right]}_0}{{\left[N_2{O}_5\right]}_t}=kt$

As the unit of rate constant is ${sec}^{-1}$, the reaction is first order reaction.

$N_2{O}_5\left(g\right)⟶2{NO}_2\left(g\right)+\frac{1}{2}{O}_2\left(g\right)$
$kt=\ln \frac{a}{(a-x)}$

$kt=\ln \frac{{\left[N_2{O}_5\right]}_0}{{\left[N_2{O}_5\right]}_t}$