The activation energy for a particular reaction is 83.1 kJ/mol. By what factor will the rate constant increase when the temperature is increased from 550.0 degrees C to 60.0 degrees C?

• A. 2.53
• B. 1.00
• C. 0.927
• D. 0.395

Answer: (B)

The temperature dependence of a reaction rate is described by the Arrhenius equation, which relates how the rate constant changes with temperature through the activation energy. The ratio of rate constants at two temperatures is k2/k1 = exp[ -Ea/R (1/T2 − 1/T1) ], where Ea is the activation energy and R is the gas constant.

Convert the given temperatures to kelvin: T1 = 550.0°C + 273.15 = 823.15 K, T2 = 60.0°C + 273.15 = 333.15 K. Compute the difference in reciprocals: 1/T2 − 1/T1 ≈ 0.003001 − 0.001214 = 0.001787 K⁻¹. With Ea = 83.1 kJ/mol = 83,100 J/mol and R = 8.314 J/mol·K, Ea/R ≈ 10,005 K. Then the exponent is −(Ea/R)(1/T2 − 1/T1) ≈ −10,005 × 0.001787 ≈ −17.9. Therefore k2/k1 ≈ e^(−17.9) ≈ 1.8 × 10⁻⁸.

So cooling from 550.0°C to 60.0°C would decrease the rate constant by about eight orders of magnitude (a factor of roughly 1.8 × 10⁻⁸).