Question

The decomposition of \[{N_2}{O_5}\] is a first order reaction with a rate constant of \[5 \times {10^{ - 4}}{s^{ - 1}}\] at \[{45^ \circ }C\] i.e. \[2{N_2}{O_5}(g) \to 4N{O_2}(g) + {O_2}(g)\] . If the initial concentration of \[{N_2}{O_5}\] is \[0.25M\] , calculate its concentration after \[2\min \]. Also calculate half-life for decomposition of \[{N_2}{O_5}(g)\].

Answer 1

Hint: Dinitrogen pentoxide present in its gaseous form has a tendency to decompose to give nitrogen dioxide and oxygen as the products. Since, this reaction follows the first order kinetics, the rate of the reaction solely depends on the concentration of the single reactant i.e. \[{N_2}{O_5}(g)\]

Complete answer:
A first order reaction shows a direct rate dependence on the concentration of a single reactant. The reaction involving the decomposition of dinitrogen pentoxide is given as follows:
\[2{N_2}{O_5}(g) \to 4N{O_2}(g) + {O_2}(g)\]
The rate law expression for the given reaction can be written as follows:
\[rate = {k_1} \times [{N_2}{O_5}]\]
Where, \[{k_1}\] is the rate constant for the first order decomposition reaction and \[[{N_2}{O_5}]\] is the concentration of dinitrogen pentoxide.
The integrated rate law expression for the given reaction can be written as follows:
\[{k_1} = \dfrac{{2.303}}{t} \times \log \left( {\dfrac{{{{[{N_2}{O_5}]}_0}}}{{[{N_2}{O_5}]}}} \right)\]
Where, \[{[{N_2}{O_5}]_0}\] represents the initial concentration of the reactant dinitrogen pentoxide and \[t\] represents the duration of time for which the rate measurement takes place.
According to the question, the time is \[t = 2\min = 2 \times 60 = 120\sec \] and the rate constant is \[{k_1} = 5 \times {10^{ - 4}}{s^{ - 1}}\] and the initial concentration is \[{[{N_2}{O_5}]_0} = 0.25M\] . On inserting all these values in the integrated rate law expression we get and solving for the concentration of dinitrogen pentoxide we get,
\[5 \times {10^{ - 4}}{s^{ - 1}} = \dfrac{{2.303}}{{120s}} \times \log \left( {\dfrac{{0.25M}}{{[{N_2}{O_5}]}}} \right)\]
\[\log \left( {\dfrac{{0.25M}}{{[{N_2}{O_5}]}}} \right) = 0.0261\]
Taking antilog on both sides we get,
\[\dfrac{{0.25M}}{{[{N_2}{O_5}]}} = {10^{0.0261}}\]
This equation can be rearranged to give,
\[[{N_2}{O_5}] = \dfrac{{0.25M}}{{{{10}^{0.0261}}}} = 0.235M\]
The formula for half-life for a first order reaction is:
\[{t_{0.5}} = \dfrac{{0.693}}{{{k_1}}}\]
Using the formula of half-life and given value of rate constant, the half-life can be calculated:
\[{t_{0.5}} = \dfrac{{0.693}}{{{k_1}}} = \dfrac{{0.693}}{{5 \times {{10}^{ - 4}}{s^{ - 1}}}} = 1386s\]
Hence, the concentration of dinitrogen pentoxide is \[0.235M\] and the half-life is \[1386s\].

Note:
The half-life of a first order reaction is the time taken for the reaction to consume half of its reactant and is always independent of the initial or final concentration of the reactant. Thus, the calculation for half-life can be done before or after the calculation of final concentration of dinitrogen pentoxide.