The slope of the graph drawn between ln k and 1/T as per Arrhenius equaiton gives the value (R= gas constant, `E_(a)=` Activation energy) .

In the graph drawn between log K and 1/T, intercept equalst o

The slope of the graph between log P and log V at constant temperature for a given mass of gas is

The slope of the graph between log P and log V at constant temperature for a given mass of gas is

What is the slope of the straight line for the graph drawn between Ink and y, where k is the rate constant of a reaction at temperature T?

The slope of straight line graph between ln k us (1)/(T) is equal to 2.4 xx 10^(4)K . Calculate the activation energy of the reaction at 400 K. (K represents rate constant)