The rate constant of a reaction is given in \(log_{10}k = 14.34 – \frac{1.25 × 10^4}{T}\) What will be the energy of activation?

239.3 kJ mol^-1

Given,

\(log_{10}k = 14.34 – \frac{1.25 × 10^4}{T}\) ------(1)

From Arrhenius equation, we can write

\(log_{10}k = log_{10}A – \frac{E_a}{2.303R × T}\)-------(2)

By comparing equations (1) and (2)

\(\frac{E_a}{2.303 × R} = 1.25 × 10^4\)

\(∴ E_a = 1.25 × 10^4 ×2.303 × R\)

\(⇒ E_a = 1.25 ×10^4 × 2.303 × 8.314\)

\(⇒ E_a = 23.93 × 10^4\)

\(⇒ E_a = 239.3 \text{ kJ mol}^{-1}\)