For a reaction, consider the plot of ln k versus 1/T given in the figure where k is rate constant and T is temperature. If the rate constant of this reaction at 500 K is 10−4 s−1, then the rate constant at 600 K is:
According to Arrhenius equation,
Taking ln on both sides,
This equation is in the form of y=mx+c
The plot of ln k vs 1T gives a strraight line with negative slope.
From the graph we know,
Slope =m=−EaR=−6909 K
Let k1 and k2 are the rate constant of the reaction at temperature T1 and T2