Using two temperatures T1 = 300 K and T2 = 320 K with k2/k1 = 4, estimate Ea.

• A. About 1 kJ/mol
• B. About 5.5 kJ/mol
• C. About 155 kJ/mol
• D. About 55 kJ/mol

Answer: (D)

The concept tested is how the Arrhenius relation links the temperature dependence of a rate constant to the activation energy. With k = A e^{-Ea/(RT)}, the ratio at two temperatures gives k2/k1 = exp[-Ea/R (1/T2 - 1/T1)]. Taking natural logs, ln(k2/k1) = -Ea/R (1/T2 - 1/T1). Solve for Ea: Ea = -R ln(k2/k1) / (1/T2 - 1/T1).

Plugging in k2/k1 = 4 and T1 = 300 K, T2 = 320 K:

1/T2 - 1/T1 = 1/320 - 1/300 ≈ -0.0002083.

ln(4) ≈ 1.386.

Ea ≈ - (8.314 J/mol·K)(1.386) / (-0.0002083) ≈ 55,000 J/mol ≈ 55 kJ/mol.

So the activation energy is about 55 kJ/mol.