Question

The decomposition of ${{\text{N}}_{\text{2}}}{{\text{O}}_{\text{5}}}$ at ${\text{318K}}$ according to the following equation follows first order reaction:
${{\text{N}}_{\text{2}}}{{\text{O}}_{\text{5}}}{\text{(g)}} \to {\text{2N}}{{\text{O}}_{\text{2}}}{\text{(g) + }}\dfrac{1}{2}{{\text{O}}_{\text{2}}}{\text{(g)}}$
The initial concentration of ${{\text{N}}_{\text{2}}}{{\text{O}}_{\text{5}}}$ was ${\text{1}}{{.24 \times 1}}{{\text{0}}^{{\text{ - 2}}}}{\text{mol}}{{\text{L}}^{{\text{ - 1}}}}$ and after 60 minutes was ${\text{0}}{{.20 \times 1}}{{\text{0}}^{{\text{ - 2}}}}{\text{mol}}{{\text{L}}^{{\text{ - 1}}}}$ . Calculate the rate constant of the reaction at ${\text{318K}}$ .

Answer 1

If the rate constant of a reaction is dependent only upon one concentration term, then it is said to be a first order reaction.
The expression for the rate constant for a reaction of first order from concentration of the reactant after some time ‘t’ is given by the following expression:
${\text{k = }}\dfrac{{{\text{2}}{\text{.303}}}}{{\text{t}}}{\text{log}}\dfrac{{{{\left[ {\text{A}} \right]}_{\text{0}}}}}{{\left[ {\text{A}} \right]}}$
where k is the rate constant of the reaction, t is equal to the time taken for the process of decay of the reactant, $\left[ {{{\text{A}}_{\text{0}}}} \right]$ is equal to the initial concentration of the reactant and $\left[ {\text{A}} \right]$ is equal to the concentration of the reactant after time ‘t’.

Complete step by step solution:
Given that the reaction of decomposition of ${{\text{N}}_{\text{2}}}{{\text{O}}_{\text{5}}}$ is a first order reaction.
Also given that the initial concentration of ${{\text{N}}_{\text{2}}}{{\text{O}}_{\text{5}}}$ is equal to ${\text{1}}{{.24 \times 1}}{{\text{0}}^{{\text{ - 2}}}}{\text{mol}}{{\text{L}}^{{\text{ - 1}}}}$ .
And that after some time which was equal to 60 minutes, the concentration of ${{\text{N}}_{\text{2}}}{{\text{O}}_{\text{5}}}$ is ${\text{0}}{{.20 \times 1}}{{\text{0}}^{{\text{ - 2}}}}{\text{mol}}{{\text{L}}^{{\text{ - 1}}}}$ .
The temperature is given to be equal to ${\text{318K}}$ .
We need to calculate the value of the rate constant for this decomposition reaction of ${{\text{N}}_{\text{2}}}{{\text{O}}_{\text{5}}}$ .
Now, according to the question, the reactant is ${{\text{N}}_{\text{2}}}{{\text{O}}_{\text{5}}}$ . So, the initial concentration of ${{\text{N}}_{\text{2}}}{{\text{O}}_{\text{5}}}$ , $\left[ {{{\text{A}}_{\text{0}}}} \right]$ is equal to ${\text{1}}{{.24 \times 1}}{{\text{0}}^{{\text{ - 2}}}}{\text{mol}}{{\text{L}}^{{\text{ - 1}}}}$ .
The time taken for decomposition is equal to 60 minutes. So after ${\text{t = 60min}}$ , the concentration of ${{\text{N}}_{\text{2}}}{{\text{O}}_{\text{5}}}$ , $\left[ {\text{A}} \right]$ is equal to ${{0}}{{.20 \times 1}}{{\text{0}}^{{\text{ - 2}}}}{\text{mol}}{{\text{L}}^{{\text{ - 1}}}}$ .
Now put the values of $\left [ A_{0} \right ]= 1.24\times 10^{-2}molL^{-1}$ , $\left [ A \right ]= 0.20\times 10^{-2}molL^{-1}$ and $t= 60min$ in the expression for first order rate constant mentioned above. So we will now have:
$k= \dfrac{2.303}{60min}log\dfrac{1.24\times 10^{-2}molL^{-1}}{0.20\times 10^{-2}molL^{-1}}$
The units of concentration will get cancelled and we will have ${\text{mi}}{{\text{n}}^{{\text{ - 1}}}}$ as the unit of the rate constant k.
So,
$k= \dfrac{2.303}{60}log\dfrac{1.24}{0.20} \Rightarrow k= \dfrac{2.303}{60}\times 0.7924 \Rightarrow k= 0.03041min^{-1}$
Hence, the value of the rate constant is found to be ${\text{k = 0}}{\text{.03041mi}}{{\text{n}}^{{\text{ - 1}}}}$ .

Note:
The unit of k for a first order reaction depends upon only the unit of time ‘t’, it is independent of the units of the concentrations. The half-life of a reaction is the time taken by the reaction to undergo reduction to half of the initial value of the reactant. For a first order reaction, it is:
${{\text{t}}_{\dfrac{1}{2}}} = \dfrac{{0.693}}{{\text{k}}}$