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?
Answer & explanation
Correct answer: option 1
The correct answer is option 1. \(239.3\, \ kJmol^{-1}\)
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)
\(\frac{E_a}{2.303 × R} = 1.25 × 10^4\)
\(∴ E_a = 1.25 × 10^4 ×2.303 × R\)
\(⇒ E_a = 1.25 ×10^4 × 2.303 × 8.314\)
\(⇒ E_a = 23.93 × 10^4\)
\(⇒ E_a = 239.3 \text{ kJ mol}^{-1}\)