The rate constant of a certain reaction is given by:
`log k= 5.4-(212)/(T)+2.17 log T`
Calculate `E_(a) at 127^(@)C`.

The rate constant for a first order decomposition reaction is given by log K=10 - 1000/T .Then,what will be activation energy in kcal/mol?

The rate constant for the decompoistion of a certain reaction is described by the equation: log k(s^(-1)) = 14 - (1.25 xx 10^(4) K)/(T) What is the effect on the rate of reaction at 127^(@)C , if in the presence of catalyst, energy of activation is lowered by 10 kJ mol^(-1) ?

The rate constant for the decompoistion of a certain reaction is described by the equation: log k(s^(-1)) = 14 - (1.25 xx 10^(4) K)/(T) What is the effect on the rate of reaction at 127^(@)C , if in the presence of catalyst, energy of activation is lowered by 10 kJ mol^(-1) ?

The rate constant for the decompoistion of a certain reaction is described by the equation: log k(s^(-1)) = 14 - (1.25 xx 10^(4) K)/(T) At what temperature, rate constant is equal to pre-exponential factor?

The rate constant for the decompoistion of a certain reaction is described by the equation: log k(s^(-1)) = 14 - (1.25 xx 10^(4) K)/(T) At what temperature, rate constant is equal to pre-exponential factor?

The rate constant for the decompoistion of a certain reaction is described by the equation: log k(s^(-1)) = 14 - (1.25 xx 10^(4) K)/(T) Pre-exponential factor for this reaction is

The rate constant for the decompoistion of a certain reaction is described by the equation: log k(s^(-1)) = 14 - (1.25 xx 10^(4) K)/(T) Pre-exponential factor for this reaction is

The rate constant for the decompoistion of a certain reaction is described by the equation: log k(s^(-1)) = 14 - (1.25 xx 10^(4) K)/(T) Energy of activation (in kcal ) is