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JEE Advanced 2019Paper 2
Exam
JEE Advanced
Year
2019
Paper
Paper 2
Subject
Chemistry
Chapter
Chemical Kinetics

JEE Advanced 2019 · Paper 2 · Chemistry · Chemical Kinetics

The decomposition reaction 2 N _ 2 O _ 5 (g) ∆ → 2 N _ 2 O _ 4 (g)...

The decomposition reaction 2N2O5(g)→∆2N2O4(g)+O2(g)2{N}_{2}{O}_{5}\left(g\right)\overset{∆}{\to}2{N}_{2}{O}_{4}\left(g\right) + {O}_{2}\left(g\right) is started in a closed cylinder under the isothermal isochoric condition at an initial pressure of 1atm1atm. After Y×103sY \times {10}^{3}s , the pressure inside the cylinder is found to be 1.45atm1.45atm . If the rate constant of the reaction is 5×10−4s−15 \times {10}^{- 4}{s}^{- 1}, assuming ideal gas behaviour, the value of YY isThe decomposition reaction 2 N _ 2 O _ 5 (g) ∆ → 2 N _ 2 O _ 4 (g) + O _ 2 (g) is started in a closed cylinder under

Correct answer

0

Step-by-step solution

The decomposition reaction 2 N _ 2 O _ 5 (g) ∆ → 2 N _ 2 O _ 4 (g) + O _ 2 (g) is started in a closed cylinder under
P0=1atm,Pt=1.45atm,P∝=1.5atm{\text{P}}_{0} = 1atm, {P}_{t} = 1 .45 atm, {P}_{∝} = 1 .5 atm
From the unit of K=5×10−4s−1,K = 5 \times {10}^{- 4}{s}^{- 1},
The reaction is first-order
t=2.3032×5×10−1log⁡(1.5−11.5−1.45)t = \dfrac{2.303}{2 \times 5 \times {10}^{- 1}}\log \left(\dfrac{1.5 - 1}{1.5 - 1.45}\right)
t=2.303×103=y×103t = 2 .303 \times {10}^{3} = y \times {10}^{3}
∴y=2.303∴ y = 2.303
After round off: y=2.3y = 2 .3

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