The rate constant for a first‑order reaction is k_1=4.0×10^-4 s^-1 at 280 K and k_2=2.5×10^-3 s^-1 at 330 K. Using the Arrhenius equation, calculate the activation energy E_a in kJ mol^-1.

The rate constant for a first‑order reaction is $k_{1} = 4.0 \times 1 0^{- 4} s^{- 1}$ at $280 K$ and $k_{2} = 2.5 \times 1 0^{- 3} s^{- 1}$ at $330 K$ . Using the Arrhenius equation, calculate the activation energy $E_{a}$ in $kJ mo l^{- 1}$ .

medium 3 marks Class 12 · Chemistry · Chemical Kinetics

Answer

E_{a} = 28 kJ mo l^{- 1} (approx.)

The correct answer is $E_{a} = 28 kJ mo l^{- 1} (approx.)$ — Convert to kJ: $E_{a} = 28.2 kJ mo l^{- 1} \approx 28 kJ mo l^{- 1}$ .

Write the two‑point Arrhenius form: $ln \frac{k _{2}}{k _{1}} = - \frac{E _{a}}{R} (\frac{1}{T _{2}} - \frac{1}{T _{1}})$ .
Calculate the left side: $ln \frac{2.5 \times 1 0 ^{- 3}}{4.0 \times 1 0 ^{- 4}} = ln (6.25) = 1.8326$ .
Compute the temperature term: $\frac{1}{T _{2}} - \frac{1}{T _{1}} = \frac{1}{330} - \frac{1}{280} = - 0.000541 K^{- 1}$ .
Rearrange for $E_{a}$ : $E_{a} = \frac{R l n ( k _{2} / k _{1} )}{\frac{1}{T _{2}} - \frac{1}{T _{1}}}$ .
Insert $R = 8.314 J mo l^{- 1} K^{- 1}$ : $E_{a} = \frac{8.314 \times 1.8326}{0.000541} = 2.82 \times 1 0^{4} J mo l^{- 1}$ .
Convert to kJ: $E_{a} = 28.2 kJ mo l^{- 1} \approx 28 kJ mo l^{- 1}$ .

Common mistake: Using

T_{1} - T_{2}

instead of

\frac{1}{T _{2}} - \frac{1}{T _{1}}

, which gives a wrong sign and magnitude.

Concept tested: Determination of $E_{a}$ from two rate constants