The temperature at which the rate constants of the given below two gaseous reactions become equal is \ \ \ \ K. (Nearest integer). X longrightarrow Y k 1 =10 6 e frac -30000 T P longrightarrow Q k 2 =10 4 e frac -24000 T Given : ln 10=2.303

Chemistry · Chemical Kinetics

JEE Main 2026 — 22 January, Morning Shift — Question 68

The temperature at which the rate constants of the given below two gaseous reactions become equal is $\_\_\_\_$ K. (Nearest integer). $\mathrm{X} \longrightarrow \mathrm{Y} \mathrm{k}_{1}=10^{6} \mathrm{e}^{\frac{-30000}{\mathrm{~T}}}$ $\mathrm{P} \longrightarrow \mathrm{Q} \mathrm{k}_{2}=10^{4} \mathrm{e}^{\frac{-24000}{\mathrm{~T}}}$ Given : $\ln 10=2.303$

Answer: 1303

Numerical answer — enter this value.

Step-by-step solution

$10^{4} \mathrm{e}^{\frac{-24000}{\mathrm{~T}}}=10^{6} \mathrm{e}^{\frac{-30000}{\mathrm{~T}}}$ $\mathrm{e}^{\frac{6000}{\mathrm{~T}}}=100$ $\frac{6000}{T}=2 \ln 10$ $\mathrm{T}=\frac{6000}{2 \times 2.303}$ $\mathrm{T}=1302.64 \mathrm{~K}$ $\mathrm{T} \approx 1303 \mathrm{~K}$

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: 22 January, Morning Shift
Subject: Chemistry
Chapter: Chemical Kinetics
Topic: Arrhenius Equation

A → products (First order reaction). Three sets of experiment were performed for a reaction under similar experimental conditions. Run…

Answer: 1303

Step-by-step solution

Practise Chemical Kinetics

More Chemical Kinetics questions from this paper