Q122 · CSIR-NET Chemistry, December 2019

Paper: CSIR-NET December 2019 · Subject: Physical Chemistry · Chapter: Chemical Kinetics · Topic: Chemical Kinetics – General · Marks: 2 · Difficulty: Medium

For a reaction, raising the temperature from 200 K to 300 K results in the increase of rate constant by a factor of 2. The activation energy for this reaction in $\mathrm{kJ\,mol^{-1}}$ is closest to $(\ln 2=0.69)$ Options:
(a)7.0
(b)3.5
(c)14.0
(d)0.83
Answer
Answer: B ✓ checked by 4AB · confidence high

The source book printed no answer; this one was worked out and checked — see the explanation.

Explanation
ln 2 = (Ea/R)(1/200 − 1/300) = Ea/(600R) ⇒ Ea = 0.69 × 8.314 × 600 = 3440 J ≈ 3.5 kJ mol⁻¹.

Study loop for Chemical Kinetics

1. Practise the PYQsPrevious-year questions, with answers

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