The rate constant (k) of a rection is measured different temperatures (T) , and the data are plotted in the given figure . The activation energy of the reaction in `kJ"mol"^(-1)` is : (R is gas constant )

The rate constant of a reaction at temperature 200 K is 10 times less than the rate constant at 400 K. What is the activation energy (E_(a)) of the reaction ? (R = gas constant)

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

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)]

The rate constant (K') of one reaction is double of the rate constant (K") of another reaction. Then the relationship between the corresponding activation energies of the two reactions (E_(a)^(') "and" E_(a)^('')) will be

The rate constant K_(1) of a reaction is found to be double that of rate constant K_(2) of another reaction. The relationship between corresponding activation energies of the two reaction at same temperature (E_(1) "and" E_(2)) can be represented as:

i) Write the mathematics expressions relating the variation of the rate constant of a reaction with temperatures. ii) How can you graphically find the activation energy of the reaction from the above expression? iii) The slope of the line in the graph of log k(k=rate constant) versus 1//T is -5841 . Calculate the activation energy of the reaction.