The rate constant for a first order reaction is given by the following equation:
ln k = 33.24 - (2.0 × 10⁴ K)/T
The activation energy for the reaction is given by ____________ kJ mol⁻¹. (In nearest integer) (Given : R = 8.3 J K⁻¹ mol⁻¹)

Solution

Chemistry ma’am · 4 steps

Line up the given equation with the Arrhenius form; the number beside 1/T is Ea/R.

1. Arrhenius in log form. Take the natural log of k = A e^(-Eₐ/RT). It becomes a straight line in 1/T.

   ln k = ln A - Eₐ/R·1/T

2. Match the terms. Compare with ln k = 33.24 - (2.0 × 10⁴ K)/T. The constant term is ln A = 33.24, and the coefficient of 1/T gives Eₐ/R = 2.0 × 10⁴ K.
3. Solve for Ea. Eₐ = 2.0 × 10⁴ K × 8.3 J K⁻¹mol⁻¹ = 1.66 × 10⁵ J mol⁻¹.
4. So. Eₐ = 166 kJ mol⁻¹, so the answer is 166.

The common slip: 382. Bringing in 2.303 gives Eₐ = 2.303 × 8.3 × 2.0 × 10⁴ J, about 382 kJ. The equation already uses ln, so 2.303 does not belong here.

The idea to keep. When a question gives ln k as a function of 1/T, do not solve anything. Just read the slope: it is -Eₐ/R. With log₁₀ k, the slope is -Eₐ/2.303R.

Answer from the NTA final answer key. Solution by PrepWiser.