For a reaction $A \rightarrow P$ at $T\text{ K}$, the half life $(t_{1/2})$ is plotted as a function of initial concentration $[A]_0$ of $A$ as given below. ![JEE Main 2026 Chemistry, Chemical Kinetics — question figure](https://prepforbharat.s3.ap-south-1.amazonaws.com/exam/615f0e999476412f48314daf/Chemistry/images/Chemical_Kinetics/69d8076381239f6567c4f528/question_1__q_69d8076381239f6567c4f528__cdn-question-pool.getmarks.app__05APRS2__q75_d1__5850d6901f_final_ppt_sync.jpg) The value of $x$ in the given figure is _______ s (Nearest integer)

JEE Main 2026 — Chemistry Physical Chemistry

2026integermedium

For a reaction $A \rightarrow P$ at $T\text{ K}$ , the half life $(t_{1/2})$ is plotted as a function of initial concentration $[A]_0$ of $A$ as given below.

JEE Main 2026 Chemistry, Chemical Kinetics — question figure

The value of $x$ in the given figure is _______ s (Nearest integer)

Official previous-year question

Held on 5 Apr 2026 · Verified 6 Jul 2026.

Did you get this right?

Solution

From the given graph, the plot of half-life $t_{1/2}$ versus initial concentration $[A]_0$ is a straight line passing through the origin.

This indicates that $t_{1/2}$ is directly proportional to $[A]_0$ , which means $t_{1/2} \propto [A]_0$ .

Therefore, the ratio $\dfrac{t_{1/2}}{[A]_0}$ is constant.

Using the coordinates from the graph, we get:

$\dfrac{240}{4 \times 10^{-3}} = \dfrac{x}{1.5 \times 10^{-3}}$

Solving for $x$ :

$x = \dfrac{240 \times 1.5 \times 10^{-3}}{4 \times 10^{-3}}$

$x = 60 \times 1.5 = 90$

The value of $x$ is $90$ .

Answer: $90$

Your note

JEE Main Physical Chemistry in other years

Physical Chemistry 2025 Physical Chemistry 2024 Physical Chemistry 2023 Physical Chemistry 2022 Physical Chemistry 2021 Physical Chemistry 2020 Physical Chemistry 2019 Physical Chemistry 2018 Physical Chemistry 2017 Physical Chemistry 2016 Physical Chemistry 2015 Physical Chemistry 2014

Keep practicing Physical Chemistry

Work through every JEE Main Physical Chemistry PYQ, year by year.

Browse all Chemistry topics

JEE Main 2026 — Chemistry Physical Chemistry

2026integermedium

For a reaction $A \rightarrow P$ at $T\text{ K}$ , the half life $(t_{1/2})$ is plotted as a function of initial concentration $[A]_0$ of $A$ as given below.

JEE Main 2026 Chemistry, Chemical Kinetics — question figure

The value of $x$ in the given figure is _______ s (Nearest integer)

Official previous-year question

Held on 5 Apr 2026 · Verified 6 Jul 2026.

Did you get this right?

Solution

From the given graph, the plot of half-life $t_{1/2}$ versus initial concentration $[A]_0$ is a straight line passing through the origin.

This indicates that $t_{1/2}$ is directly proportional to $[A]_0$ , which means $t_{1/2} \propto [A]_0$ .

Therefore, the ratio $\dfrac{t_{1/2}}{[A]_0}$ is constant.

Using the coordinates from the graph, we get:

$\dfrac{240}{4 \times 10^{-3}} = \dfrac{x}{1.5 \times 10^{-3}}$

Solving for $x$ :

$x = \dfrac{240 \times 1.5 \times 10^{-3}}{4 \times 10^{-3}}$

$x = 60 \times 1.5 = 90$

The value of $x$ is $90$ .

Answer: $90$

Your note

JEE Main Physical Chemistry in other years

Keep practicing Physical Chemistry

Work through every JEE Main Physical Chemistry PYQ, year by year.

Browse all Chemistry topics

JEE Main 2026 — Chemistry Physical Chemistry

Did you get this right?

Solution

Your note

JEE Main Physical Chemistry in other years

More JEE Main Physical Chemistry questions

Other Chemistry topics for JEE Main

Organic Chemistry

Inorganic Chemistry

Keep practicing Physical Chemistry

JEE Main 2026 — Chemistry Physical Chemistry

Did you get this right?

Solution

Your note

JEE Main Physical Chemistry in other years

More JEE Main Physical Chemistry questions

Other Chemistry topics for JEE Main

Organic Chemistry

Inorganic Chemistry

Keep practicing Physical Chemistry