For the reaction A → B, the rate constant k(in s⁻¹) is given by
log ₁₀k = 20.35 - (2.47 × 10³)/T
The energy of activation in kJ mol⁻¹ is ____________. (Nearest integer) [Given : R = 8.314 J K⁻¹ mol⁻¹]

Solution

Chemistry ma’am · 4 steps

Match the given line to the log form of Arrhenius: the slope is -Eₐ/2.303R.

1. Log form. Take log₁₀ of k = A e^(-Eₐ/RT). Converting ln to log₁₀ brings in the factor 2.303.

   log₁₀ k = log₁₀ A - Eₐ/(2.303 R T)

2. Compare terms. The term with 1/T must match. So Eₐ/2.303R = 2.47 × 10³ K, and the constant 20.35 is just log₁₀ A.
3. Solve for Ea. Eₐ = 2.47 × 10³ × 2.303 × 8.314 = 2470 × 19.147 = 4.73 × 10⁴ J mol⁻¹.
4. So. Eₐ ≈ 47.3 kJ mol⁻¹, which rounds to 47.

The common slip: 21. Leaving out the 2.303 and using Eₐ = 2470 × 8.314 gives about 20.5 kJ, which rounds to 21, not 47. That shortcut works only when the equation uses ln k.

The idea to keep. When log₁₀ k is given as a line in 1/T, the coefficient of 1/T is Eₐ/2.303R. Check whether the log is base 10 or base e before you multiply.

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