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# The decomposition of ${N_2}{O_5}$ according to the equation $2{N_2}{O_5}\left( g \right)\xrightarrow{{}}4N{O_2}\left( g \right) + {O_2}\left( g \right)$ is a first order reaction. After $30\min .$ from the start of the decomposition in a closed vessel, the total pressure developed is found to be $284.5mm$ of $Hg$. On compete decomposition, the total pressure is $584.5mm$ of $Hg$. The rate constant of the reaction is:A: $5.21 \times {10^{ - 3}}{\min ^{ - 1}}$ B: $5.12 \times {10^{ - 3}}{\min ^{ - 1}}$C: $5.89 \times {10^{ - 3}}{\min ^{ - 1}}$D: None of these

Hint: Decomposition reaction is a type of reaction in which one compound breaks down in two or more compounds. A first-order reaction is a reaction in which reaction proceeds at a rate that depends linearly on only one reactant concentration.
Formula used: $k = \dfrac{{2.303}}{t}\log \dfrac{a}{{a - x}}$ where
k is rate constant
t is time taken
a is initial pressure
x is pressure of reactant decomposed till time t

For reaction $2{N_2}{O_5}\left( g \right)\xrightarrow{{}}4N{O_2}\left( g \right) + {O_2}\left( g \right)$
It can also be written as ${N_2}{O_5}\left( g \right)\xrightarrow{{}}2N{O_2}\left( g \right) + \dfrac{1}{2}{O_2}\left( g \right)$
Let initial pressure of ${N_2}{O_5}$ be $a$ before reaction and that of $N{O_2}$ and ${O_2}$ be $0$
After time t let say $x$ amount of pressure is decomposed in reaction and by the above reaction we can say that
Pressure of $N{O_2}$ at time t is$= 2x$
Pressure of ${O_2}$ at time t is$= \dfrac{x}{2}$
Pressure of ${N_2}{O_5}$ at time t is$= a - x$
After complete dissociation pressure will be as following
Pressure of $N{O_2}$ after complete dissociation is$= 2a$
Pressure of ${O_2}$ after complete dissociation is$= \dfrac{a}{2}$
Pressure of ${N_2}{O_5}$ after complete dissociation is$= 0$
Total number of moles at time t would be $a - x + 2x + \dfrac{x}{2} = a + \dfrac{3}{2}x$
Total number of moles after complete dissociation would be $2a + \dfrac{a}{2} + 0 = \dfrac{5}{2}a$
$\dfrac{5}{2}a \propto 584.5mm$
$a + \dfrac{3}{2}x \propto 284.5mm$
Solving these equations we get $a \propto 233.8mm$ and $x \propto 33.8mm$
Using formula
$k = \dfrac{{2.303}}{t}\log \dfrac{a}{{a - x}} = \dfrac{{2.303}}{{30}}\log \dfrac{{233.8}}{{233.8 - 33.8}}$
$= 5.21 \times {10^{ - 3}}{\min ^{ - 1}}$
So, our answer to this question is option B that is $5.12 \times {10^{ - 3}}{\min ^{ - 1}}$.

Note:
Concentration time equation of first order reaction is $\ln \left[ A \right] = \ln \left[ {{A_ \circ }} \right] - kt$ and half life of first order reaction is given when $A$ becomes ${A_ \circ }$ that is reaction is half completed or reactants are half used/decomposed. ${t_{\dfrac{1}{2}}} = \dfrac{{\ln 2}}{k}$ .