Q77 · CSIR NET Chemistry, December 2016
Paper: CSIR NET December 2016 · Subject: Physical Chemistry · Chapter: Chemical Kinetics · Topic: Arrhenius Activation · Marks: 2 · Difficulty: Medium
For a reaction with an activation energy of $49.8 \mathrm{kJ} \mathrm{mol}^{-1}$, the ratio of the rate constants at 600 K and $300 \mathrm{K},\left(\mathrm{k}_{600} / \mathrm{k}_{300}\right)$, is approximately $\left(\mathrm{R}=8.3 \mathrm{J} \mathrm{mol}^{-1} \mathrm{K}^{-1}\right)$
(a)$\ln 10$
(b)10
(c)10+e
(d)$\mathrm{e}^{10}$
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
ln(k₆₀₀/k₃₀₀) = (Ea/R)(1/300 − 1/600) = (49800/8.3)(1/600) = 10 ⇒ ratio = e¹⁰.
Study loop for Chemical Kinetics