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Question: The decomposition of \[{N_2}{O_5}\] is a first order reaction with a rate constant of \[5 \times {10...

The decomposition of N2O5{N_2}{O_5} is a first order reaction with a rate constant of 5×10−4s−15 \times {10^{ - 4}}{s^{ - 1}} at 45∘C{45^ \circ }C i.e. 2N2O5(g)→4NO2(g)+O2(g)2{N_2}{O_5}(g) \to 4N{O_2}(g) + {O_2}(g) . If the initial concentration of N2O5{N_2}{O_5} is 0.25M0.25M , calculate its concentration after 2min⁡2\min . Also calculate half-life for decomposition of N2O5(g){N_2}{O_5}(g).

Explanation

Solution

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. N2O5(g){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:
2N2O5(g)→4NO2(g)+O2(g)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=k1×[N2O5]rate = {k_1} \times [{N_2}{O_5}]
Where, k1{k_1} is the rate constant for the first order decomposition reaction and [N2O5][{N_2}{O_5}] is the concentration of dinitrogen pentoxide.
The integrated rate law expression for the given reaction can be written as follows:
k1=2.303t×log⁡([N2O5]0[N2O5]){k_1} = \dfrac{{2.303}}{t} \times \log \left( {\dfrac{{{{[{N_2}{O_5}]}_0}}}{{[{N_2}{O_5}]}}} \right)
Where, [N2O5]0{[{N_2}{O_5}]_0} represents the initial concentration of the reactant dinitrogen pentoxide and tt represents the duration of time for which the rate measurement takes place.
According to the question, the time is t=2min⁡=2×60=120sec⁡t = 2\min = 2 \times 60 = 120\sec and the rate constant is k1=5×10−4s−1{k_1} = 5 \times {10^{ - 4}}{s^{ - 1}} and the initial concentration is [N2O5]0=0.25M{[{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,
5×10−4s−1=2.303120s×log⁡(0.25M[N2O5])5 \times {10^{ - 4}}{s^{ - 1}} = \dfrac{{2.303}}{{120s}} \times \log \left( {\dfrac{{0.25M}}{{[{N_2}{O_5}]}}} \right)
log⁡(0.25M[N2O5])=0.0261\log \left( {\dfrac{{0.25M}}{{[{N_2}{O_5}]}}} \right) = 0.0261
Taking antilog on both sides we get,
0.25M[N2O5]=100.0261\dfrac{{0.25M}}{{[{N_2}{O_5}]}} = {10^{0.0261}}
This equation can be rearranged to give,
[N2O5]=0.25M100.0261=0.235M[{N_2}{O_5}] = \dfrac{{0.25M}}{{{{10}^{0.0261}}}} = 0.235M
The formula for half-life for a first order reaction is:
t0.5=0.693k1{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:
t0.5=0.693k1=0.6935×10−4s−1=1386s{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 0.235M0.235M and the half-life is 1386s1386s.

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.