Rate constant for a reaction are `2.5xx10^(-4)` atm and `1` atm at temperature `327^(@)C` and `527^(@)C` respectively. Calculate activation energy in `Kj`.
The rate constants of a reaction are 1xx10^(-3)sec^(-1)" and "2xx10^(-3)sec^(-1)" at "27^(@)C" and "37^(@)C respectively. Calculate the activation energy of the reaction.
The rate constant for a reaction is 1.5 xx 10^(-7) at 50^(@) C and 4.5 xx 10^(7)s^(-1) at 100^(@) C . What is the value of activation energy?
The specific rate constant for a reaction increases by a factor of 4, if the temperature is changed from 27^(@)C" to "47^(@)C . Find the activation energy for the reaction.
The rate constant for a first order reaction increases from 4 xx 10^(-2) to 8 xx 10^(-2) when the temperature changes from 27^(@) C to 37^(@) C. Calculate energy of activation for the reaction.
The rate constant of a reaction increases by five times on increase in temperature from 27^@C to 52^@C . The value of activation energy in kJ "mol"^(-1) is ______ (Rounded-off to the nearest integer)
The rate constant of a reaction is 1.5 xx 10^(7)s^(-1) at 50^(@) C and 4.5 xx 10^(7)s^(-1) at 100^(@) C. Calculate the value of activation energy for the reaction (R=8.314 J K^(-1)mol^(-1))
The rate constant of a reaction is 1.2 xx10^(-3) s^(-1) at 30^@C and 2.1xx10^(-3)s^(-1) at 40^@C . Calculate the energy of activation of the reaction.
The rate constant of a reaction is 1.2xx10^(-3)sec^(-1)" at "30^(@)C" and "2.1xx10^(-3)sec^(-1)" at" 40^(@)C . Calculate the energy of activation of the reaction.
