Q16 · CSIR-NET Chemistry, December 2011
Paper: CSIR-NET December 2011 · Subject: Physical Chemistry · Chapter: Chemical Kinetics · Topic: Arrhenius Activation · Marks: 2 · Difficulty: Medium
For a reaction, the rate constant k at $27^{\circ}\mathrm{C}$ was found to be $k=5.4\times10^{11}\,e^{-50}$. The activation energy of the reaction is:
(a)$50\,\mathrm{J\,mol^{-1}}$
(b)$415\,\mathrm{J\,mol^{-1}}$
(c)$15{,}000\,\mathrm{J\,mol^{-1}}$
(d)$125{,}000\,\mathrm{J\,mol^{-1}}$
Answer
Answer: D ✓ checked by 4AB · confidence high
The source book printed no answer; this one was worked out and checked — see the explanation.
Explanation
Compare k = 5.4×10¹¹ e⁻⁵⁰ with the Arrhenius equation k = A e^(−Ea/RT): Ea/RT = 50.
T = 27 °C = 300 K, so Ea = 50 × 8.314 J K⁻¹ mol⁻¹ × 300 K = 124,710 J mol⁻¹ ≈ 125,000 J mol⁻¹. Answer (d).
Study loop for Chemical Kinetics
· Browse this chapter in the app