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Graph between log k and (1)/(T) (k is ra...

Graph between log k and `(1)/(T)` (k is rate constant in `s^(-1)` and T is the temperature in K) is a straight line. As shown in figure if OX = 5 and slope of the line `=- (1)/(2.303)`then `E_(a)` is :

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`log_(10) k =log_(10) A-(E_(a))/(2.303 RT)" Slope "=-(E_(a))/(2.303 R) =(-1)/(2.303)" "therefore E_(a) =R =2` cal
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