The rate constant for the first order decompoistion of a certain reaction is described by the equation
`log k (s^(-1)) = 14.34 - (1.25 xx 10^(4)K)/(T)`
(a) What is the energy of activation for the reaction?
(b) At what temperature will its half-life periof be `256 min`?

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

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) At what temperature, rate constant is equal to pre-exponential factor?

Rate constant for the decomposition of ethylene oxide into CH_(4) and CO may be described be the equation. logk(s)^(-1)=14.34-(1.25xx10^(4))/(T) (a) What is the energy of activation of this reaction ? (b) What is the value of k at 670K ?

The rate constant for the first order decomposition of ethylene oxide into CH_(4) and CO, may be describe by log k("in" s^(-1))=14.34-(1.25xx10^(4)K)/T The activation energy of the 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) 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) A two-step mechanism has been suggested for the reaction of nitric oxide and bromine: NO(g) + Br_(2)(g) overset(k_(1))rarr NOBr_(2)(g) NOBr_(2)(g)+NO(g) overset(k_(2))rarr 2NOBr(g) The observed rate law is, rate = k[NO]^(2)[Br_(2)] . Hence, the rate-determining step is