For a first-order reaction at 500 K, the rate constant is 4.0 × 10−5 s−1 and the Arrhenius factor (A) is 4.0 × 1013 s−1. What is the activation energy of the reaction? (R = 8.314 J K−1 mol−1; ln x = 2.303 log x)
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → 172.3 kJ mol−1
The Arrhenius equation links the rate constant with the Arrhenius factor and the activation energy. Taking the natural logarithm gives a form that can be solved for Ea:
\( \displaystyle \ln k = \ln A - \frac{E_a}{RT} \)
\( \displaystyle E_a = RT\ln\frac{A}{k} = 2.303\,RT\log\frac{A}{k} \)
\( \displaystyle \frac{A}{k} = \frac{4.0 \times 10^{13}\ \mathrm{s^{-1}}}{4.0 \times 10^{-5}\ \mathrm{s^{-1}}} = 1.0 \times 10^{18}, \quad \log\frac{A}{k} = 18 \)
\( \displaystyle E_a = 2.303 \times 8.314\ \mathrm{J\,K^{-1}\,mol^{-1}} \times 500\ \mathrm{K} \times 18 \)
\( \displaystyle E_a = 1.723 \times 10^{5}\ \mathrm{J\,mol^{-1}} = 172.3\ \mathrm{kJ\,mol^{-1}} \)
The ratio k/A = 1.0 × 10−18 is the value of \( e^{-E_a/RT} \): only this tiny fraction of the molecules has energy equal to or greater than Ea at 500 K.
NCERT (2026-27 edition):
3.4 Temperature Dependence of the Rate of a Reaction – “k = A e–Ea/RT … where A is the Arrhenius factor or the frequency factor. It is also called pre-exponential factor. It is a constant specific to a particular reaction. R is gas constant and Ea is activation energy measured in joules/mole (J mol–1).”
“In the Arrhenius equation (3.18) the factor e–Ea/RT corresponds to the fraction of molecules that have kinetic energy greater than Ea. Taking natural logarithm of both sides of equation (3.18)”
\( \displaystyle \ln k = -\frac{E_a}{RT} + \ln A \qquad (3.19) \)