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

Hint: 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}}} $