Gaseous cyclobutene isomerizes to butadiene in a first order process which has a ' K ' value of 3 . 3 × 10 ^ - 4 s ^ - 1 at 153 ° C . The time in minutes it takes for the isomerization to proceed 40 % to completion at

Correct answer: 0 Kt = ln [A] _ 0 [A] _ t 3 . 3 × 10 ^ - 4 × t = ((100)/(60)) t = 1547 . 956sec t = 25 . 799min 26min

JEE Main 202124 FebruaryMorning Shift

Exam: JEE Main
Year: 2021
Date: 24 February
Shift: Morning Shift
Subject: Chemistry
Chapter: Chemical Kinetics

JEE Main 2021 · 24 February · Morning Shift · Chemistry · Chemical Kinetics

Gaseous cyclobutene isomerizes to butadiene in a first order...

Gaseous cyclobutene isomerizes to butadiene in a first order process which has a '

K

' value of

3 . 3 \times {10}^{- 4}{s}^{- 1}

153 ° C .

The time in minutes it takes for the isomerization to proceed

40 %

to completion at this temperature is_____. (Rounded off to the nearest integer) Gaseous cyclobutene isomerizes to butadiene in a first order process which has a ' K ' value of 3 . 3 × 10 ^ - 4 s ^ -

Gaseous cyclobutene isomerizes to butadiene in a first order process which has a ' K ' value of 3 . 3 × 10 ^ - 4 s ^ -

Correct answer

Step-by-step solution

Gaseous cyclobutene isomerizes to butadiene in a first order process which has a ' K ' value of 3 . 3 × 10 ^ - 4 s ^ -

$Kt = ln\dfrac{{\left[A\right]}_{0}}{{\left[A\right]}_{t}}$

$3 . 3 \times {10}^{- 4} \times t = \ln \left(\dfrac{100}{60}\right)$

$t = 1547 . 956sec$

$t = 25 . 799min$

$26min$

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