A graph is plotted between \(ln k\) and \(\frac{1}{T}\), what is the value of slope?

slope \(= -\frac{E_a}{R}\)

The correct answer is option 2. slope \(= -\frac{E_a}{R}\).

The Arrhenius equation describes the temperature dependence of reaction rates. It states that the rate constant (\(k\)) of a reaction is exponentially related to the inverse of the temperature (\(T\)):

\[ k = Ae^{-\frac{E_a}{RT}} \]

Where:

\( A \) is the pre-exponential factor, which represents the frequency of collisions between reactant molecules.

\( E_a \) is the activation energy, which represents the minimum energy required for a reaction to occur.

\( R \) is the gas constant.

\( T \) is the absolute temperature in Kelvin.

Taking the natural logarithm of both sides of the equation, we get:

\[ \ln(k) = \ln(A) - \frac{E_a}{RT} \]

This equation has the form of a straight line equation, \(y = mx + c\), where:

\( y = \ln(k) \),

\( m = -\frac{E_a}{R} \),

\( x = \frac{1}{T} \), and

\( c = \ln(A) \).

So, when we plot \(\ln(k)\) against \(\frac{1}{T}\), we should get a straight line with slope \(m = -\frac{E_a}{R}\). This slope represents the ratio of activation energy to the gas constant (\(R\)).

Hence, the correct option is: 2. slope \(= -\frac{E_a}{R}\)

This relationship is fundamental in understanding how temperature affects reaction rates and is often used in kinetics studies to determine activation energies.