For a reaction A rightarrow P at TK , the half life ( t 1 / 2 ) is plotted as a function of initial concentration [ A ] 0 of A as given below : The value of x in the given figure is \ \ \ \ s (Nearest integer)

Chemistry · Chemical Kinetics

JEE Main 2026 — 5 April, Evening Shift — Question 70

For a reaction $\mathrm{A} \rightarrow \mathrm{P}$ at TK , the half life $\left(\mathrm{t}_{1 / 2}\right)$ is plotted as a function of initial concentration $[\mathrm{A}]_{0}$ of A as given below : The value of x in the given figure is $\_\_\_\_$ s (Nearest integer)

Answer: 90

Numerical answer — enter this value.

Step-by-step solution

From the plot, $\mathrm{t}_{1 / 2} \propto[\mathrm{~A}]_{0}$

\propto\left[\mathrm{A}_{0}\right]^{1-\mathrm{N}}

$1-\mathrm{N}=1$ $\mathrm{N}=0$ (order) $\mathrm{t}_{1 / 2}=\frac{\mathrm{A}_{0}}{2 \mathrm{~K}}$

240=\frac{4 \times 10^{-3}}{2 \mathrm{~K}} \end{gathered}$$ $$\begin{gathered} \mathrm{x}=\frac{1.5 \times 10^{-3}}{2 \mathrm{~K}} \end{gathered}$$ Divide / $\frac{240}{x}=\frac{4}{1.5}$ $\mathrm{x}=90$

Answer key and solution verified before publishing.

Practise Chemical Kinetics

Start with this question, then two more from the same chapter — with a tutor that explains every step. Free.

Solve a similar one free

Exam: JEE Main 2026
Paper: 5 April, Evening Shift
Subject: Chemistry
Chapter: Chemical Kinetics
Topic: Basics of Rate of a Chemical Reaction