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# The rate constant for the reaction ${{\rm{N}}_2}{{\rm{O}}_5}\left( g \right) \to 2{\rm{N}}{{\rm{O}}_2}\left( g \right) + \dfrac{1}{2}{{\rm{O}}_2}$ is $2.3 \times {10^{ - 2}}{\rm{ se}}{{\rm{c}}^{ - 1}}$. Which equation given below describes the change of $\left[ {{{\rm{N}}_{\rm{2}}}{{\rm{O}}_{\rm{5}}}} \right]$ with time, ${\left[ {{{\rm{N}}_{\rm{2}}}{{\rm{O}}_{\rm{5}}}} \right]_0}$ and ${\left[ {{{\rm{N}}_{\rm{2}}}{{\rm{O}}_{\rm{5}}}} \right]_t}$ corresponds to concentration of ${{\rm{N}}_{\rm{2}}}{{\rm{O}}_{\rm{5}}}$ initially and time t respectively? This question has multiple correct options.( A ) ${\left[ {{{\rm{N}}_{\rm{2}}}{{\rm{O}}_{\rm{5}}}} \right]_0} = {\left[ {{{\rm{N}}_{\rm{2}}}{{\rm{O}}_{\rm{5}}}} \right]_t}{e^{kt}}$( B ) $\log \dfrac{{{{\left[ {{{\rm{N}}_{\rm{2}}}{{\rm{O}}_{\rm{5}}}} \right]}_0}}}{{{{\left[ {{{\rm{N}}_{\rm{2}}}{{\rm{O}}_{\rm{5}}}} \right]}_t}}} = kt$( C ) ${\left[ {{{\rm{N}}_{\rm{2}}}{{\rm{O}}_{\rm{5}}}} \right]_t} = {\left[ {{{\rm{N}}_{\rm{2}}}{{\rm{O}}_{\rm{5}}}} \right]_0} + kt$( D ) ${\left[ {{{\rm{N}}_{\rm{2}}}{{\rm{O}}_{\rm{5}}}} \right]_t} = {\left[ {{{\rm{N}}_{\rm{2}}}{{\rm{O}}_{\rm{5}}}} \right]_0} + kt$

Last updated date: 03rd Mar 2024
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Hint: Obtain the order of the reaction from the unit of the rate constant.For the first order reaction, the unit of the rate constant is the reciprocal of time.

Complete step by step answer:
The rate constant for the reaction ${{\rm{N}}_2}{{\rm{O}}_5}\left( g \right) \to 2{\rm{N}}{{\rm{O}}_2}\left( g \right) + \dfrac{1}{2}{{\rm{O}}_2}$ is $2.3 \times {10^{ - 2}}{\rm{ se}}{{\rm{c}}^{ - 1}}$.
This suggests that the reaction follows first order kinetics. For the first order reaction, the unit of the rate constant is the reciprocal of time.
For the first order reaction, the rate of the reaction is directly proportional to the first power of the concentration of the reaction.
$\begin{array}{l} {\rm{Rate}} \propto \left[ {{{\rm{N}}_{\rm{2}}}{{\rm{O}}_{\rm{5}}}} \right]\\ {\rm{Rate}} = k \times \left[ {{{\rm{N}}_{\rm{2}}}{{\rm{O}}_{\rm{5}}}} \right] \end{array}$
But the rate of the reaction is ${\rm{Rate}} = - \dfrac{{d\left[ {{{\rm{N}}_{\rm{2}}}{{\rm{O}}_{\rm{5}}}} \right]}}{{dt}}$.
Hence ${\rm{Rate}} = - \dfrac{{d\left[ {{{\rm{N}}_{\rm{2}}}{{\rm{O}}_{\rm{5}}}} \right]}}{{dt}} = k \times \left[ {{{\rm{N}}_{\rm{2}}}{{\rm{O}}_{\rm{5}}}} \right]$ … …(1)

When the above equation is integrated, the following relationship is obtained.

$\begin{array}{l} {\left[ {{{\rm{N}}_{\rm{2}}}{{\rm{O}}_{\rm{5}}}} \right]_t} = {\left[ {{{\rm{N}}_{\rm{2}}}{{\rm{O}}_{\rm{5}}}} \right]_0}{e^{ - kt}}\\ {\rm{Rearrange above expression}}\\ {\left[ {{{\rm{N}}_{\rm{2}}}{{\rm{O}}_{\rm{5}}}} \right]_0} = {\left[ {{{\rm{N}}_{\rm{2}}}{{\rm{O}}_{\rm{5}}}} \right]_t}{e^{kt}} \end{array}$

This is the same as in option A. Hence, option A is correct.
Upon integration of equation (1), the following reactions are also obtained.
$\begin{array}{l} \ln \dfrac{{{{\left[ {{{\rm{N}}_{\rm{2}}}{{\rm{O}}_{\rm{5}}}} \right]}_0}}}{{{{\left[ {{{\rm{N}}_{\rm{2}}}{{\rm{O}}_{\rm{5}}}} \right]}_t}}} = kt\\ \log \dfrac{{{{\left[ {{{\rm{N}}_{\rm{2}}}{{\rm{O}}_{\rm{5}}}} \right]}_0}}}{{{{\left[ {{{\rm{N}}_{\rm{2}}}{{\rm{O}}_{\rm{5}}}} \right]}_t}}} = \dfrac{{kt}}{{2.303}} \end{array}$

Hence, the option B) is incorrect.

Upon integration of equation (1), the following reactions are also obtained.

$\begin{array}{l} \ln {\left[ {{{\rm{N}}_{\rm{2}}}{{\rm{O}}_{\rm{5}}}} \right]_t} = \ln {\left[ {{{\rm{N}}_{\rm{2}}}{{\rm{O}}_{\rm{5}}}} \right]_0} - kt\\ 2.303{\log _{10}}{\left[ {{{\rm{N}}_{\rm{2}}}{{\rm{O}}_{\rm{5}}}} \right]_t} = 2.303{\log _{10}}{\left[ {{{\rm{N}}_{\rm{2}}}{{\rm{O}}_{\rm{5}}}} \right]_0} - kt\\ {\log _{10}}{\left[ {{{\rm{N}}_{\rm{2}}}{{\rm{O}}_{\rm{5}}}} \right]_t} = {\log _{10}}{\left[ {{{\rm{N}}_{\rm{2}}}{{\rm{O}}_{\rm{5}}}} \right]_0} - \dfrac{{kt}}{{2.303}} \end{array}$
Hence, the option C) is incorrect.

The relation ${\left[ {{{\rm{N}}_{\rm{2}}}{{\rm{O}}_{\rm{5}}}} \right]_t} = {\left[ {{{\rm{N}}_{\rm{2}}}{{\rm{O}}_{\rm{5}}}} \right]_0} + kt$ is incorrect as it shows a linear relationship between concentration and time, however, in first order reaction, the concentration shows exponential decay.

Hence, the option C) is incorrect.

Hence, only option A) is the correct option.

Note:
Please take care of positive and negative signs used in the formula. Also remember that when natural logarithm is converted to logarithm to base 10, it is multiplied with 2.303.