Question

In going from room temperature (25 C) to 10 C above room temperature, the rate of reaction doubles. Calculate the activation energy for the reaction .

Answer 1

Answer:

Ea=5.29 × 10⁴ J/mol

Explanation:

In going from 25 °C (298 K) to 35 °C (308 K), the rate of the reaction doubles. Since the rate of the reaction depends on the rate constant (k), this implies that the rate constant doubles. We can find the activation energy (Ea) using the two-point form of the Arrhenius equation.

$ln\frac{k_{2}}{k_{1}} =\frac{-Ea}{R} .(\frac{1}{T_{2}}-\frac{1}{T_{1}})\\ln\frac{2k_{1}}{k_{1}}=\frac{-Ea}{8.314J/K.mol}.(\frac{1}{308K}-\frac{1}{298K} )\\Ea=5.29 \times 10^{4} J/mol$