The rate constant for the first order decomposition of `(H)_(2) (O)_(2)` is given by the following equation: `log k=14.34-1.25 xx 10^4(K) / (T)` Calculate `E_{a}` for this reaction and at what tem perature will its half-period be 256 minutes?

The rate constant of a reaction at 298 K is given by 3.2xx10^(-3)s^(-1) .Find the order of reaction?

The rate constant for the decomposition of (N)_(2) (O)_(5) at various temperatures is given below: Draw a graph between log kand 1 / (T) and calculate the values of (A) and E_{a} . Predict the rate constant at 30^oC and 50^oC . (##VPU_HSS_CHE_XII_C04_E02_022##)

The rate constant of a first order reaction is 231 xx 10^(-5)- s^(-1) How long will 4 g of this reactant reduce to 2 g?

For a certain reaction log K (S^(-1))= 14- (1.25 xx 10^(4)K)/(T) Hence it can be said that The E_(a) for the reaction is

For a certain reaction log K (S^(-1))= 14- (1.25 xx 10^(4)K)/(T) Hence it can be said that The pre-exponential factor is

For a certain reaction log K (S^(-1))= 14- (1.25 xx 10^(4)K)/(T) Hence it can be said that K will be equal to 10^(14) if

Calculate the half-life of a first order reaction from their rate constants given below. 2min^-1

The decomposition of hydrocarbon follows the equation k=(4.5xx10^11s^-1) e^-28000K/t Calculate E_(a)^(*)

Calculate the half-life of a first order reaction from their rate constants given below. 4 years^-1 .