For a chemical reaction, the table below shows the values of the rate constant, kkk, at two different temperatures. Experiment 1: Temperature T1=298 KT_1 = 298\text{ K}T1=298 K, Rate constant k1=2.45×10−4 s−1k_1 = 2.45 \times 10^{-4}\text{ s}^{-1}k1=2.45×10−4 s−1. Experiment 2: Temperature T2=328 KT_2 = 328\text{ K}T2=328 K, Rate constant k2=3.91×10−3 s−1k_2 = 3.91 \times 10^{-3}\text{ s}^{-1}k2=3.91×10−3 s−1. The Arrhenius equation can be rearranged to calculate the activation energy, EaE_{\text{a}}Ea:
ln(k1k2)=EaR(1T2−1T1) \ln\left(\frac{k_1}{k_2}\right) = \frac{E_{\text{a}}}{R}\left(\frac{1}{T_2} - \frac{1}{T_1}\right) ln(k2k1)=REa(T21−T11)Calculate the value, in kJ mol−1\text{kJ mol}^{-1}kJ mol−1, of the activation energy, EaE_{\text{a}}Ea. Give your answer to 3 significant figures. (The gas constant, R=8.31 J K−1 mol−1R = 8.31\text{ J K}^{-1}\text{ mol}^{-1}R=8.31 J K−1 mol−1)