Activation energy \((E_a)\) are rate constant (\(k_1\) and \(k_2\)) of a chemical reaction at two different temperatures (\(T_1\) and \(T_2\)) are related by:

\(ln \frac{k_2}{k_1} = \frac{E_a}{R}\left[\frac{1}{T_1} - \frac{1}{T_2}\right]\)

The correct answer is option 3. \(ln \frac{k_2}{k_1} = \frac{E_a}{R}\left[\frac{1}{T_1} - \frac{1}{T_2}\right]\).

Most of the chemical reactions are accelerated by increase in temperature. It has been found that for a chemical reaction with rise in temperature by 10°, the rate constant is nearly doubled.        

The effect of temperature on the rate of a reaction and hence on the rate constant k, was proposed by Arrhenius (1889). The equation is called Arrhenius equation, is usually written in the form

\(k Ae^{-E_a/RT}\) ---------(i)

where,

the pre-exponential factor, \(A\) is a constant and is called frequency factor (because it gives the frequency of binary collisions and the reacting molecules per second per litre),

\(E_a\) is the energy of activation,

\(R\) is gas constant

\(T\) is the absolute temperature

The two quantities ‘\(E_a\)’ and ‘\(A\)’ are together called Arrhenius parameters. The energy of activation \((E_a)\) is an important quantity as it is characteristic of the reaction. Using the above equation its value can be calculated.

Taking logarithm on both sides of equation (i), we get

\(ln k = ln\left(Ae^{-E_a/RT}\right)\)

\(⇒ ln k = ln A + ln (e^{-E_a/RT})\)

\(⇒ ln k = ln A + \left(\frac{-E_a}{RT}\right)\)

\(∴ ln k = ln A - \frac{E_a}{RT}\) --------(ii)

If the value of the rate constant at temperature T₁ and T₂ are k₁ and k₂ respectively, then we have

\(⇒ ln k_1 = ln A - \frac{E_a}{RT_1}\) ---------(iii)

and \(⇒ ln k_2 = ln A - \frac{E_a}{RT_2}\) -----------(iv)

Subtracting equation (iii) from equation (iv), we get

\(ln k_2 - ln k_1 = -\frac{E_a}{RT_2} - \left(-\frac{E_a}{RT_1}\right)\)

\(⇒ ln k_2 - ln k_1 = \frac{E_a}{RT_1} - \left(-\frac{E_a}{RT_2}\right)\)

\(⇒ ln \frac{k_2}{k_1} = \frac{E_a}{RT_1} - \left(-\frac{E_a}{RT_2}\right)\)

\(⇒ ln \frac{k_2}{k_1} = \frac{E_a}{R} \left[\frac{1}{T_1} - \frac{1}{T_2}\right]\)