A graph is plotted of ln(k)\ln(k)ln(k) against 1/T1/T1/T (where k k\,k is the rate constant and T T\,T is the temperature in K\text{K}K). The gradient of the line of best fit has the numerical value of −14,800-14,800−14,800. What is the activation energy, EaE_{\text{a}}Ea, in kJ mol−1\text{kJ mol}^{-1}kJ mol−1? Use R=8.314 J K−1 mol−1R = 8.314 \text{ J K}^{-1} \text{ mol}^{-1}R=8.314 J K−1 mol−1.
+1.78+1.78+1.78
+1.23×10−4+1.23 \times 10^{-4}+1.23×10−4
+1.23×105+1.23 \times 10^{5}+1.23×105
+123+123+123
