According to the Arrhenius rate equation, rate constant \(k\) is equal to \(A.e^{-E_a/RT}\). Which of the following options represents the graph of \(ln k \) vs \(\frac{1}{T}\)?

The correct answer is option 1.

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

\(k = Ae^{-E_a/RT}-------(1)\)

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

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

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

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

\(⇒ ln k = ln A - \left(\frac{E_a}{RT}\right) --------(2)\)

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

\(ln k_1 = ln A - \left(\frac{E_a}{RT_1}\right) --------(3)\)

\(\text{ & }ln k_2 = ln A - \left(\frac{E_a}{RT_2}\right) --------(4)\)

Subtracting equation (3) from equation (4), we get

\(lnk_2 - lnk_1 = -\frac{E_a}{RT_2} -\left(-\frac{E_a}{RT_1}\right)\)

\(⇒ lnk_2 - lnk_1 = \frac{E_a}{RT_1} - \frac{E_a}{RT_2}\)

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

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

\(⇒  ln\frac{k_2}{k_1}  = \frac{E_a}{R}\left[\frac{T_2 - T_1}{T_1T_2}\right] -----(5)\)

Thus, knowing the values of rate constants k₁ and k₂ at two different temperatures T₁ and T₂, the value of E_a can be calculated. Alternatively, knowing the rate constant at any one temperature, its value at another temperature can be calculated provided the value of E_a is known for that reaction.

To test the validity of Arrhenius equation, let us consider equation (ii). It may be written as

\(ln k = -\frac{E_a}{RT} + ln A --------(6)\)

The equation (vii) is in the form , i.e., the equation of a straight line. Thus, if a plot of \(\text{ln k vs }\frac{1}{T}\) is a straight line, the validity of the equation is confirmed.