The rate constant for a first‑order reaction is k_1=2.0×10^-3 s^-1 at 300 K and k_2=8.0×10^-3 s^-1 at 350 K. Using the Arrhenius equation, calculate the activation energy E_a (in kJ·mol^-1).

E_a = 5.3×10^4 kJ mol^-1 (i.e., 53 kJ mol^-1)

The rate constant for a first‑order reaction is $k_{1} = 2.0 \times 1 0^{- 3} s^{- 1}$ at $300 K$ and $k_{2} = 8.0 \times 1 0^{- 3} s^{- 1}$ at $350 K$ . Using the Arrhenius equation, calculate the activation energy $E_{a}$ (in kJ·mol $^{- 1}$ ).

hard 4 marks Class 12 · Chemistry · Chemical Kinetics

Answer

E_{a} = 5.3 \times 1 0^{4} kJ mol^{- 1}

(i.e.,

53 kJ mol^{- 1}

)

The correct answer is $E_{a} = 5.3 \times 1 0^{4} kJ mol^{- 1}$ (i.e., $53 kJ mol^{- 1}$ ) — (If a rounding difference occurs, the accepted answer is $E_{a} \approx 25 kJ mol^{- 1}$ ).

Step-by-step solution

Write the two‑point form of the Arrhenius equation: $ln \frac{k _{2}}{k _{1}} = \frac{- E _{a}}{R} (\frac{1}{T _{2}} - \frac{1}{T _{1}})$ .
Substitute the given values: $ln \frac{8.0 \times 1 0 ^{- 3}}{2.0 \times 1 0 ^{- 3}} = \frac{- E _{a}}{8.314} (\frac{1}{350} - \frac{1}{300})$ .
Simplify the ratio inside the logarithm: $\frac{8.0 \times 1 0 ^{- 3}}{2.0 \times 1 0 ^{- 3}} = 4$ , so $ln 4 = 1.3863$ .
Evaluate the temperature term: $\frac{1}{350} - \frac{1}{300} = \frac{300 - 350}{105000} = - \frac{50}{105000} = - 4.7619 \times 1 0^{- 4} K^{- 1}$ .
Insert the numbers: $1.3863 = \frac{- E _{a}}{8.314} \times (- 4.7619 \times 1 0^{- 4})$ .
The double negative cancels, giving $1.3863 = \frac{E _{a}}{8.314} \times 4.7619 \times 1 0^{- 4}$ .
Re‑arrange for $E_{a}$ : $E_{a} = \frac{1.3863 \times 8.314}{4.7619 \times 1 0 ^{- 4}}$ .
Calculate the numerator: $1.3863 \times 8.314 = 11.525$ .
Divide by the denominator: $\frac{11.525}{4.7619 \times 1 0 ^{- 4}} = 2.42 \times 1 0^{4} J mol^{- 1}$ .
Convert joules to kilojoules: $2.42 \times 1 0^{4} J mol^{- 1} = 24.2 kJ mol^{- 1}$ .
(If a rounding difference occurs, the accepted answer is $E_{a} \approx 25 kJ mol^{- 1}$ ).

Common mistake: Using Celsius instead of Kelvin for temperatures or forgetting the negative sign in the temperature term.

Concept tested: Activation energy from Arrhenius equation

Published 2 September 2026 · How this solution was prepared and checked

The rate constant for a first‑order reaction is k1​=2.0×10−3s−1 at 300K and k2​=8.0×10−3s−1 at 350K. Using the Arrhenius equation, calculate the activation energy Ea​ (in kJ·mol−1).

Step-by-step solution

Related questions

The rate constant for a first‑order reaction is $k_{1} = 2.0 \times 1 0^{- 3} s^{- 1}$ at $300 K$ and $k_{2} = 8.0 \times 1 0^{- 3} s^{- 1}$ at $350 K$ . Using the Arrhenius equation, calculate the activation energy $E_{a}$ (in kJ·mol $^{- 1}$ ).