temperature k Increasing temperature increases the rate constant k. The relationship is given by the Arrhenius equation k = Ae-Ea/RT where A is a constant R is gas constant and Ea is activation energy. Increasing the temperature increases the value of the rate constant k The Arrhenius equation can be rearranged ln k = constant – Ea/(RT) k is proportional to the rate of reaction so ln k can be replaced by ln(rate) From plotting a graph of ln(rate) or ln k against 1/T the activation energy can be calculated from measuring the gradient of the line ln (Rate) 1/T Gradient = – Ea/ R Ea = – gradient x R E. -3.1 -2.6 -2.1 -1.6 0.0029 0.003 0.0031 0.0032 0.0033 0.0034 ln (Rate) 1/T use a line of best fit use all graph paper choose points far apart on the graph to calculate the gradient Temperature T (K) 1/T time t (s) 1/t Ln (1/t) 297.3 0.003364 53 0.018868 -3.9703 310.6 0.00322 24 0.041667 -3.1781 317.2 0.003153 16 0.0625 -2.7726 323.9 0.003087 12 0.083333 -2.4849 335.6 0.00298 6 0.166667 -1.7918 x1 ,y1 x2 ,y2 gradient = y2 -y1 x2 -x1 In above example gradient =-5680 The gradient should always be -ve Ea = – gradient x R (8.31) = – -5680 x8.31 = 47200 J mol-1 The unit of Ea using this equation will be J mol-1 . Convert into kJ mol-1 by dividing 1000 Ea = +47.2 kJ mol-1 E

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5.1.1 How fast?

Effect of temperature on rate constants (j) a qualitative explanation of the effect of temperature change on the rate of a reaction and hence the rate constant (see 3.2.2 f–g) M0.3 (k) the Arrhenius equation: (i) the exponential relationship between the rate constant, k and temperature, T given by the Arrhenius equation, k = Ae–Ea/RT (ii) determination of Ea and A graphically using: ln k = –Ea/RT + ln A derived from the Arrhenius equation. M0.1, M0.4, M2.2, M2.3, M2.4, M2.5, M3.1, M3.2, M3.3, M3.4 Ea = activation energy, A = pre-exponential factor, R = gas constant (provided on the Data Sheet) Explanation of A is not required. Equations provided on the Data Sheet. HSW5 Link between k and T.