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 a first order decomposition reaction is given by "log K=\(10-\frac{1000}{T}\) .Then,what will be activation energy in kcal/mol?

The rate constant 'K' for pseudo first order reaction is

The rate constant for a first order reaction becomes six times when the temperature is raised from 350 K to 400 K. Calculate the activation energy for the reaction [R=8.314JK^(-1)mol^(-1)]

For a first order reactin, rate constant is given is log k = 14 - (1.2 xx 10^(4))/(T) , then what will be value of temperature if its half life period is 6.93 xx 10^(-3) min ?

The rate constant for a first order reaction becomes six times when the temperature is raised from 350 to 400 K. Calculate the energy of activation for the reactions.

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 first order rate constant for the decomposition of CaCO_3 , at 700 K is 6.36 xx 10^(-3)s^(-1) and activation energy is 209 "kJ mol"^(-1) . Its rate constant (in s^(-1) ) at 600 K is "x" xx 10^(-6) . The value of x is ______ .(Nearest integer) [Given R=8.31 J K^(-1) "mol"^(-1) , log 6.36 xx 10^(-3)= -2.19, 10^(-4.79)= 1.62 xx10^(-5) ]