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Question

Answers

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

Answer
Verified

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}}$

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. ${t_{\dfrac{1}{2}}} = \dfrac{{\ln 2}}{k}$ .