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#### Question

(b) Rate constant ‘k’ of a reaction varies with temperature ‘T’ according to the equation:

`logk=logA-E_a/2.303R(1/T)`

Where E_{a} is the activation energy. When a graph is plotted for `logk Vs. 1/T` a straight line with a slope of −4250 K is obtained. Calculate ‘E_{a}’ for the reaction.(R = 8.314 JK^{−1} mol^{−1})

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