Question

The given reaction is a first order reaction in terms of the concentration of ${{{N}}_2}{{{O}}_{2\left( {{g}} \right)}}$.
${{{N}}_2}{{{O}}_{2\left( {{g}} \right)}} \to 2{{NO}}$
Which of the following is valid, $\left[ {{{{N}}_2}{{{O}}_2}} \right]$ being constant?
A. $\left[ {{{NO}}} \right] = {\left[ {{{{N}}_2}{{{O}}_2}} \right]_0}{{{e}}^{ - {{kt}}}}$
B. $\left[ {{{NO}}} \right] = {\left[ {{{{N}}_2}{{{O}}_2}} \right]_0}\left( {{{1 - }}{{{e}}^{{{kt}}}}} \right)$
C. $\left[ {{{NO}}} \right] = {\left[ {{{{N}}_2}{{{O}}_2}} \right]_0}\left( {{{{e}}^{ - {{kt}}}} - 1} \right)$
D. $\left[ {{{NO}}} \right] = {\left[ {{{{N}}_2}{{{O}}_2}} \right]_0}\left( {{{1 - }}{{{e}}^{ - {{kt}}}}} \right)$

Answer 1

Hint:Order of a reaction is that number which we obtain when the power of concentration of reactants are added. It can be first order, second order, third order or pseudo order reaction. In first order reactions, rate is only dependent on the reactant concentration.

Complete answer:
The concentration changes with time when the reaction proceeds. This is called the rate of reaction. Its unit is generally expressed in ${{M}}{{{s}}^{ - 1}}$. It is necessary to remember that the power of concentrations of reactants are not the stoichiometric coefficients which we write in a chemical equation.
The chemical reaction is given below:
${{{N}}_2}{{{O}}_{2\left( {{g}} \right)}} \to 2{{NO}}$
Let’s assume that the initial concentration of ${{{N}}_2}{{{O}}_{2\left( {{g}} \right)}}$ is ${\left[ {{{{N}}_2}{{{O}}_2}} \right]_0}$
At initial condition, ${{{N}}_2}{{{O}}_{2\left( {{g}} \right)}}$ has not been reacted. So only ${{{N}}_2}{{{O}}_{2\left( {{g}} \right)}}$ is present, but no ${{NO}}$ molecule is formed. And at time ${{t}}$, the concentration of ${{{N}}_2}{{{O}}_{2\left( {{g}} \right)}}$ is $\left[ {{{{N}}_2}{{{O}}_2}} \right]$ and ${{NO}}$ is $\left[ {{{NO}}} \right]$.
Thus we can say that the initial concentration of ${{{N}}_2}{{{O}}_{2\left( {{g}} \right)}}$ has a total concentration of ${{{N}}_2}{{{O}}_{2\left( {{g}} \right)}}$ and ${{NO}}$, i.e. ${\left[ {{{{N}}_2}{{{O}}_2}} \right]_0} = \left[ {{{{N}}_2}{{{O}}_2}} \right] + \left[ {{{NO}}} \right] \to (1)$
For the first order reaction, rate can be expressed as ${{r}} = - \dfrac{{d\left[ {{{{N}}_2}{{{O}}_2}} \right]}}{{d{{t}}}}{{ = k}}\left[ {{{{N}}_2}{{{O}}_2}} \right]$
Integrating in terms of $d\left[ {{{{N}}_2}{{{O}}_2}} \right]$ and $d{{t}}$, we get
$\int_{{{\left[ {{{{N}}_2}{{{O}}_2}} \right]}_0}}^{\left[ {{{{N}}_2}{{{O}}_2}} \right]} {\dfrac{{d\left[ {{{{N}}_2}{{{O}}_2}} \right]}}{{\left[ {{{{N}}_2}{{{O}}_2}} \right]}}} = - k\int_0^{{t}} {d{{t}}} \Leftrightarrow \ln \left( {\dfrac{{\left[ {{{{N}}_2}{{{O}}_2}} \right]}}{{{{\left[ {{{{N}}_2}{{{O}}_2}} \right]}_0}}}} \right) = - {{kt}}$
Taking exponent, we get
$\left( {\dfrac{{\left[ {{{{N}}_2}{{{O}}_2}} \right]}}{{{{\left[ {{{{N}}_2}{{{O}}_2}} \right]}_0}}}} \right) = {{{e}}^{ - {{kt}}}}$ or $\left[ {{{{N}}_2}{{{O}}_2}} \right] = {\left[ {{{{N}}_2}{{{O}}_2}} \right]_0}{{{e}}^{ - {{kt}}}} \to (2)$
Substituting $(2)$ in $(1)$, we get
${\left[ {{{{N}}_2}{{{O}}_2}} \right]_0} = {\left[ {{{{N}}_2}{{{O}}_2}} \right]_0}{{{e}}^{ - {{kt}}}} + \left[ {{{NO}}} \right]$
On rearranging, we get the formula for $\left[ {{{NO}}} \right]$.
i.e. $\left[ {{{NO}}} \right] = {\left[ {{{{N}}_2}{{{O}}_2}} \right]_0} - {\left[ {{{{N}}_2}{{{O}}_2}} \right]_0}{{{e}}^{ - {{kt}}}}$
Or $\left[ {{{NO}}} \right] = {\left[ {{{{N}}_2}{{{O}}_2}} \right]_0}\left( {{{1 - }}{{{e}}^{-{{kt}}}}} \right)$

Thus, the correct option is D.

Note:
From the given answer, we can conclude that in a first order reaction, the concentration of the reactant reduces from the initial concentration exponentially with time. Or we can tell that whenever the concentration of the reactant is reduced exponentially when we plot a graph between reactant concentration and time, then it is a first order reaction. It can be used to identify the order of the reaction.