How is rate constant of a reaction related to its activation energy?

Explanation

Arrhenius (1889) developed a mathematical relationship between temperature and the rate constant on the basis of the observations from the large number of experiments. The temperature dependence of the rate constant, k is expressed algebraically as follows.
`k prop e^(-Ea//RT)`
`k=Ae^(-Ea//RT)`

This equation is known as Arrhenius equation. Here, A is a constant of proportionality and is known as frequency factor:-This-factor is related to the number of binary molecular collisions per second per litre. E is the activation energy, R is gas constant and T is the absolute temperature. Taking logarithm, egn. (1) may be written as ,
In k=In A+ In `e^(-Ea//RT)`
In k=In A-`(E_a)/(RT)In e`
In `k=In A-(E_a)/(RT)` [`because`In e=1] ...(ii)
Converting to common logarithm, we get
`logk=logA-(E_a)/(2.303R)1/T` ... (iii)
or logk=`(-E_a)/(2.303R)1/T+logA`
This equation is of the form y = mx + ci.e., the equation of a straight line. Thus, if a plot of log k versús `1//T` is a straight line, the valdity of the equation is confirmed. From eqn: (lin) we havė,
Slope of the line = `(E_a)/(2.303R)`
Thus, by knowing the value of the slope and gas constant R, the activation energy `E_` can be calculated.
Alternatively, `E_a`, can be determined by using another form of the Arrhenius equation in which A is eliminated. This is because frequency factor A cannot be determined accurately for the reactions. From eqn. (iii)
In k=In A-`(E_a)/(RT)`
In k = In `k_1` and `k_2` are the values of rate constants at temperatures `T_1` and `T_2`, then,
In `k_1=In A-(E_a)/(RT_1)` ...(iv)
and `In k_2=In A-(E_a)/(RT_2)`
In Subtracting eqn. (iv) from egn. (v), we get
In `k_2`-In `k_1`=`-(E_a)/(RT_2)-(-E_a)/(RT_1)`
In `k_2-In k_1=(E_a)/(RT_1)-(E_a)/(RT_2)`
In `(k_2)/(k_1)=(E_a)/(2.303R)[(T_2-T_1)/(T_1T_2)]`
Converting to common iogarithm,
log `(k_2)/(k_1)=(E_a)/(2.303R)[(T_2-T_1)/(T_1T_2)]`