NVAEasyJEE Main 2026 · 22 January, Shift 1Arrhenius Equation & Activation Energy

Chemistry Question from JEE Main 2026 · 22 January, Shift 1

The temperature at which the rate constants of the given below two gaseous reactions become equal is _____ K (Nearest integer).

XY,k1=106e30000TX \longrightarrow Y, \qquad k_1 = 10^{6} e^{-\frac{30000}{T}} PQ,k2=104e24000TP \longrightarrow Q, \qquad k_2 = 10^{4} e^{-\frac{24000}{T}}

Given: ln10=2.303\ln 10 = 2.303

Answer & step-by-step solution

Sign in to reveal the correct answer, the full step-by-step solution, and the common mistakes for this question.

Practice more Arrhenius Equation & Activation Energy questions

Get unlimited AI-adaptive practice, mastery tracking, and an AI tutor that explains every step - free to start.

Related questions