A series of experiments is carried out with compounds D and E. Using the data obtained, the rate equation for the reaction between the two compounds is deduced to be
rate=k[D][E] \text{rate} = k[\text{D}][\text{E}] rate=k[D][E]In one experiment at 45 ∘C45\ ^\circ\text{C}45 ∘C, the initial rate of reaction is 3.8×10−4 mol dm−3 s−13.8 \times 10^{-4}\ \text{mol}\ \text{dm}^{-3}\ \text{s}^{-1}3.8×10−4 mol dm−3 s−1 when the initial concentration of D is 0.40 mol dm−30.40\ \text{mol}\ \text{dm}^{-3}0.40 mol dm−3 and the initial concentration of E is 0.19 mol dm−30.19\ \text{mol}\ \text{dm}^{-3}0.19 mol dm−3.
Calculate a value for the rate constant kkk at this temperature and give its units.
An equation that relates the rate constant, kkk, to the activation energy, EaE_{\text{a}}Ea, and the temperature, TTT, is
lnk=−EaRT+lnA \ln k = \frac{-E_{\text{a}}}{RT} + \ln A lnk=RT−Ea+lnAUse this equation and your answer from part (a) to calculate a value, in kJ mol−1\text{kJ}\ \text{mol}^{-1}kJ mol−1, for the activation energy of this reaction at 45 ∘C45\ ^\circ\text{C}45 ∘C.
For this reaction lnA=24.2\ln A = 24.2lnA=24.2.
The gas constant R=8.31 J K−1 mol−1R = 8.31\ \text{J}\ \text{K}^{-1}\ \text{mol}^{-1}R=8.31 J K−1 mol−1.
(If you were unable to complete part (a), you should use the value of 1.2×10−31.2 \times 10^{-3}1.2×10−3 for the rate constant. This is not the correct value.)
