The decomposition reaction 2 N 2 O 5 ( g ) xrightarrow longrightarrow rightarrow 2 N 2 O 4 ( g )+ O 2 ( g ) is started in a closed cylinder under isothermal isochoric condition at an initial pressure of 1 atm . After Y × 10 3 s , the pressure inside the cylinder is found to be 1.45 atm . If the rate constant of the reaction is 5 × 10 -4 s -1 , assuming ideal gas behaviour, the value of Y is \ \ \ \ -

Chemistry · Chemical Kinetics

JEE Advanced 2019 — Paper 2 — Question 20

The decomposition reaction $2 \mathrm{~N}_{2} \mathrm{O}_{5}(\mathrm{~g}) \xrightarrow{\longrightarrow} \rightarrow 2 \mathrm{~N}_{2} \mathrm{O}_{4}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g})$ is started in a closed cylinder

under isothermal isochoric condition at an initial pressure of 1 atm . After $\mathrm{Y} \times 10^{3} \mathrm{~s}$ , the pressure inside the

cylinder is found to be 1.45 atm . If the rate constant of the reaction is $5 \times 10^{-4} \mathrm{~s}^{-1}$ , assuming ideal gas

behaviour, the value of $\mathbf{Y}$ is $\_\_\_\_$ -

Answer: 2.3

Numerical answer — enter this value.

Step-by-step solution

\text{Unit of K represent it is first order reaction.}

2\text{N}_2\text{O}_5 \longrightarrow 2\text{N}_2\text{O}_4 + \text{O}_2

& 2\text{N}_2\text{O}_5 & \longrightarrow & 2\text{N}_2\text{O}_4 & + & \text{O}_2 \\ t=0 & 1 & & 0 & & 0 \\ t=t & 1-P & & P & & P/2 \end{array}$$

1 - P + P + \frac{P}{2} = 1.45

\frac{P}{2} = 0.45, P = 0.9

t = \frac{2.303}{2 \times 5 \times 10^{-4}} \log \frac{1}{1 - P}

y \times 10^{-3} = \frac{2.303}{2 \times 5 \times 10^{-4}} \log \frac{1}{1 - 0.9} = \frac{2.303}{2 \times 5 \times 10^{-4}} \log 10

y \times 10^{-3} = \frac{2.303}{10^{-3}}

Y = 2.30

Answer key and solution verified before publishing.

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Exam: JEE Advanced 2019
Paper: Paper 2
Subject: Chemistry
Chapter: Chemical Kinetics
Topic: Integrated Rate Laws