For a reaction with an activation energy of 52.0 kilojoules per mole and at a temperature of 35°c find the pre-exponential factor if the rate constant is 0.158 a) 0.161 в) 1.03 x 10% c) 6.42 x 1076 10 d) 2.42 x 10% e) 6.71 x 106

Explanation:

Using the expression,

Wherem  

is the activation energy

A is the pre-exponential factor

R is Gas constant having value = 8.314 J / K mol  

Thus, given that, = 52.0 kJ/mol = 52000 J/mol (As 1 kJ = 1000 J)

The conversion of T( °C) to T(K) is shown below:

T(K) = T( °C) + 273.15  

So,  

T = (35 + 273.15) K = 308.15 K  

Thus, applying values as:

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