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Rate constant K of a reaction varies with temperature according to the equation log K = constant. \(\frac{E_a}{2.303 R} \frac{1}{T}\) |
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Answer» Where Ea is the energy of activation for the reaction. When a graph is plotted for log K versus \(\frac{1}{T}\), a straight line with a slope 6670 K is obtained. Calculate the energy of activation for this reaction. State units (R = 8.314 JK-1 mol-1) Slope of the line \(\frac{-E_a}{2.303 R}\) = -6670 K Ea = 2.303 × 8.314 (JK-1 mol-1) × 6670 K = 127711.4 J mol-1. |
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