In the following reaction; xA → yB _ 10 [- (d[A])/(dt)] = _ 10 [- (d[B])/(dt)] + 0.3010 ‘A’ and ‘B’ respectively can be:

Correct answer: A. C _ 2 H _ 4 and C _ 4 H _ 8 From the rate law: (- d[A])/(dt) = log(d[B])/(dt) + 0.3010 = log (- dA)/(dt) = log (dB)/(dt) + 2 (- d[A])/(dt) = 2 × (d[B])/(dt) 2A → B and we know 2 C _ 2 H _ 4 → C _ 4 H _ 8

JEE Main 201912 AprilMorning Shift

Exam: JEE Main
Year: 2019
Date: 12 April
Shift: Morning Shift
Subject: Chemistry
Chapter: Chemical Kinetics

JEE Main 2019 · 12 April · Morning Shift · Chemistry · Chemical Kinetics

In the following reaction; xA → yB _ 10 [- (d[A])/(dt)] = _ 10 [-...

In the following reaction; $xA \to yB$

$\log_{10} \left[- \dfrac{d\left[A\right]}{dt}\right] = \log_{10} \left[- \dfrac{d\left[B\right]}{dt}\right] + 0.3010$

‘A’ and ‘B’ respectively can be:

Options

A ${C}_{2}{H}_{4}$ and ${C}_{4}{H}_{8}$
B ${C}_{2}{H}_{2}$ and ${C}_{6}{H}_{6}$
Cn-Butane and Iso-butane
D ${N}_{2}{O}_{4}$ and $N{O}_{2}$

Correct answer

A. ${C}_{2}{H}_{4}$ and ${C}_{4}{H}_{8}$

Step-by-step solution

From the rate law:

$\text\log\dfrac{- d\left[A\right]}{dt} = log\dfrac{d\left[B\right]}{dt} + 0.3010 = log \dfrac{- dA}{dt} = log \dfrac{dB}{dt} + \log 2$

$\dfrac{- d\left[A\right]}{dt} = 2 \times \dfrac{d\left[B\right]}{dt}$

$2A \to B$ and we know $2{C}_{2}{H}_{4} \to {C}_{4}{H}_{8}$

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