Two rate constants of a first‑order reaction are measured: k_1 = 1.5×10^-4 s^-1 at T_1 = 280,K and k_2 = 6.0×10^-4 s^-1 at T_2 = 330,K. Using the Arrhenius equation, calculate the activation energy E_a in kJ mol^-1.

Two rate constants of a first‑order reaction are measured: $k_{1} = 1.5 \times 1 0^{- 4} s^{- 1}$ at $T_{1} = 280, K$ and $k_{2} = 6.0 \times 1 0^{- 4} s^{- 1}$ at $T_{2} = 330, K$ . Using the Arrhenius equation, calculate the activation energy $E_{a}$ in kJ mol $^{- 1}$ .

hard 5 marks Class 12 · Chemistry · Chemical Kinetics

Answer

21.3 kJ mol^{- 1}

The correct answer is $21.3 kJ mol^{- 1}$ — Convert to kJ mol $^{- 1}$ : $21.3 kJ mol^{- 1}$ .

Write the Arrhenius form: $l n \frac{k _{2}}{k _{1}} = - \frac{E _{a}}{R} (\frac{1}{T _{2}} - \frac{1}{T _{1}})$ .
Calculate the left side: $l n (6.0 \times 1 0^{- 4} /1.5 \times 1 0^{- 4}) = l n 4 = 1.3863$ .
Compute $\frac{1}{T _{2}} - \frac{1}{T _{1}} = \frac{1}{330} - \frac{1}{280} = - 5.411 \times 1 0^{- 4} K^{- 1}$ .
Insert values and solve for $E_{a}$ : $E_{a} = - R \times 1.3863/ (- 5.411 \times 1 0^{- 4}) = 8.314 \times 2561.5 J mol^{- 1} = 2.13 \times 1 0^{4} J mol^{- 1}$ .
Convert to kJ mol $^{- 1}$ : $21.3 kJ mol^{- 1}$ .

Common mistake: Using the wrong sign in the temperature term or using

R

in kJ rather than J

Concept tested: Arrhenius activation energy

Published 9 September 2026 · Last updated 16 September 2026 · How this solution was prepared and checked