The activation energy of a reaction is "58.3kJ/mole" .The ratio of the rate constants at 305K and "300K" is about (\(R=8.3Jk^{-1}mol^{-1}\)) (Antilog 0.1667=1.468).

The activation energy of a reaction is zero. Its rate constant at 280 K is 16 xx 10^(-6) s^(-1) the rate constant at 300k is

The activation energy of a reaction is 94.14 "kJ mol"^(-1) and the value of rate constant at 313 K is 1.8xx10^(-5) s^(-1) . Calculate the frequency factor A

The activation energy of a reaction is 94.14 kJ mol^(-1) and the value of rate constant at 313 K is 1.8 xx 10^(-5)s^(-1) . Calculate the frequency factor A.

For a reaction, the activation energy is zero. What is the value of rate constant at 300 K if k=1.6 xx 10^(6)s^(-1) at 280 K (R= 8.31 JK^(-1)mol^(-1)) ?

Calculate the rate constant of a reaction at 293 K when energy of activation is 103 kJ "mol"^(-1) and the rate constant at 273 K is 7.87xx10^(-7)s^(-1) (R=8.314xx10^(-3)kJ"mol"^(-1)KJ^(-1))

The energy of activation of a reaction is 140 kJ "mol"^(-1) . If its rate constant at 400 K is 2.0xx10^(-6) s^(-1) , what is the value at 500 K