Rate constant, k of a reaction varies with temperature as: `log k = Constant- E_a/(2.303RT)` where`E_a` is the activation energy. When a Graph is plotted for log k vs 1/T ,a straight line with a slope of -6670K is obtained. Calculate the energy of activation for this reaction. (`R = 8.314 JK^-1 mol^-1`)

Rate constant 'k' of a reaction varies with temperature 'T' according to the equation: log k= log A- (E_a)/(2.303R) (1/T) where E_a is the activation energy. When a graph is plotted for log k vs. 1/T , a straight line with a slope of - 4250 K is obtained. Calculate E_a for the reaction (R = 8.314 JK^-1 mol^-1)

If a graph is plotted between Iog k and 1/T the slope of the straight line so obtained is given by

The rate constant for the first order reactior becomes three times when the temperatur is raised from 20^@ C to 50^@ C. Calculate the energy of activation for the reaction.

For a first order reaction, the plot of log K versus 1/T gives straight line. The slope of the line has been found to be a - 8.95 xx 10^-3 K. The activation energy for the reaction is

The rate constant of a reaction increases four times when the temperature changes from 300 K to 320 K, Calculate the activation energy for the reaction. (R = 8.314 Jmol ^(-1) K ^(-1))

The rate constant for first order reaction becomes six times when the temperature is raised from 50 K to 400 K. Calculate the activation energy for the reaction. (R= 8.314 J/K/mol)