$\mathrm{N}_{2} \mathrm{O}_{5(\mathrm{~g})} \rightarrow 2 \mathrm{NO}_{2(\mathrm{~g})}+\frac{1}{2} \mathrm{O}_{2(\mathrm{~g})}$ In the above first order reaction the initial concentration of $\mathrm{N}_{2} \mathrm{O}_{5}$ is $2.40 \times 10^{-2} \mathrm{~mol} \mathrm{L}^{-1}$ at 318 K. The concentration of $\mathrm{N}_{2} \mathrm{O}_{5}$ after 1 hours was $1.60 \times 10^{-2} \mathrm{~mol} \mathrm{L}^{-1}$. The rate constant of the reaction at 318 K is _____ $\times 10^{-3} \mathrm{~min}^{-1}$. (Nearest integer) [Given: $\log 3=0.477, \log 5=0.699$]

JEE Main · 2021 · 22 Jul 2021 · Shift 2 · Chemical Kinetics

N_{2} O_{5 (g)} \to 2 NO_{2 (g)} + \frac{1}{2} O_{2 (g)}

In the above first order reaction the initial concentration of

N_{2} O_{5}

2.40 \times 1 0^{- 2} mol L^{- 1}

at 318 K. The concentration of

N_{2} O_{5}

after 1 hours was

1.60 \times 1 0^{- 2} mol L^{- 1}

. The rate constant of the reaction at 318 K is _____

\times 1 0^{- 3} min^{- 1}

. (Nearest integer)
[Given:

lo g 3 = 0.477, lo g 5 = 0.699

]

Answer: 7

For a first order reaction

k = \frac{2.303}{t} lo g \frac{[ A ] _{0}}{[ A ]}

, with

t = 1

hour

= 60

min. Using

lo g 2 = 1 - lo g 5 = 0.301

lo g \frac{2.40 \times 1 0 ^{- 2}}{1.60 \times 1 0 ^{- 2}} = lo g 1.5 = lo g 3 - lo g 2 = 0.477 - 0.301 = 0.176

k = \frac{2.303 \times 0.176}{60} = 6.76 \times 1 0^{- 3} mi n^{- 1}

The rate constant is

6.76 \times 1 0^{- 3} mi n^{- 1}

, which to the nearest integer is

7 \times 1 0^{- 3} mi n^{- 1}

JEE Main · 2021 · 22 Jul 2021 · Shift 2 · Integer answer