N _ 2 O _ 5 ( g ) → 2 NO _ 2 ( g ) + (1)/(2) O _ 2 ( g ) In the above first order reaction the initial concentration of N _ 2 O _ 5 is 2 . 40 × 10 ^ - 2 mol L ^ - 1 at 318K . The concentration of N _ 2 O _ 5 after 1

Correct answer: 0 K = (2 . 303)/(t)log [ N _ 2 O _ 5 ] _ 0 [ N _ 2 O _ 5 ] _ 1 = (2 . 303)/(60)log(2 . 4)/(1 . 6) = 6 . 76 × 10 ^ - 3 ^ - 1 7 × 10 ^ - 3 ^ - 1

JEE Main 202122 JulyMorning Shift

Exam: JEE Main
Year: 2021
Date: 22 July
Shift: Morning Shift
Subject: Chemistry
Chapter: Chemical Kinetics

JEE Main 2021 · 22 July · Morning Shift · Chemistry · Chemical Kinetics

N _ 2 O _ 5 ( g ) → 2 NO _ 2 ( g ) + (1)/(2) O _ 2 ( g ) In the...

{N}_{2}{O}_{5 ( g )} \to 2{NO}_{2 ( g )} + \dfrac{1}{2}{O}_{2 ( g )}

In the above first order reaction the initial concentration of

{N}_{2}{O}_{5}

2 . 40 \times {10}^{- 2}mol{L}^{- 1}

318K .

The concentration of

{N}_{2}{O}_{5}

after

1

hour was

1 . 60

\times {10}^{- 2}mol{L}^{- 1} .

The rate constant of the reaction at

318K

\times {10}^{- 3}\min^{- 1}

(Nearest integer):

[

Given

: \log 3 = 0 . 477, \log 5 = 0 . 699 ]

Correct answer

Step-by-step solution

K = \dfrac{2 . 303}{t}log\dfrac{{\left[{N}_{2}{O}_{5}\right]}_{0}}{{\left[{N}_{2}{O}_{5}\right]}_{1}}

= \dfrac{2 . 303}{60}log\dfrac{2 . 4}{1 . 6} = 6 . 76 \times {10}^{- 3}\min^{- 1} \approx 7 \times {10}^{- 3}\min^{- 1}

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