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JEE Main 20264 AprilMorning Shift

Exam: JEE Main
Year: 2026
Date: 4 April
Shift: Morning Shift
Subject: Chemistry
Chapter: Chemical Kinetics

JEE Main 2026 · 4 April · Morning Shift · Chemistry · Chemical Kinetics

Consider the first order reaction R arrow P . The fraction of...

Consider the first order reaction

R \rightarrow P

. The fraction of molecules decomposed in the given first order reaction can be expressed as

Options

A $1 - e^{k_1 t}$
B $1 + e^{k_1 t}$
C $1 + e^{-k_1 t}$
D $1 - e^{-k_1 t}$

Correct answer

D. $1 - e^{-k_1 t}$

Step-by-step solution

For a first order reaction, the concentration of reactant at time

t

is given by

[R]_t = [R]_0 e^{-k_1 t}

The amount of reactant decomposed is

[R]_0 - [R]_t = [R]_0 - [R]_0 e^{-k_1 t} = [R]_0 (1 - e^{-k_1 t})

The fraction of molecules decomposed is

\dfrac{[R]_0 - [R]_t}{[R]_0} = \dfrac{[R]_0 (1 - e^{-k_1 t})}{[R]_0} = 1 - e^{-k_1 t}

Answer:

1 - e^{-k_1 t}

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