Graph between log₁₀ k and (1/T) is linear of slope S. Hence E_a is:

2.303 RS

The correct answer is option 4. 2.303 RS.

We are given that the graph between \(\log_{10} k\) and \(\frac{1}{T}\) is linear with a slope \(S\).

From the Arrhenius equation:

\(k = A \cdot e^{-\frac{E_a}{RT}} \)

Taking the logarithm (base 10) of both sides:

\(\log_{10} k = \log_{10} A - \frac{E_a}{2.303RT} \)

Comparing this equation with the linear equation:

\(\log_{10} k = m \cdot \left(\frac{1}{T}\right) + c \]

We can see that the slope of the linear equation is \(\frac{-E_a}{2.303RT}\).

Therefore, the slope (\(S\)) is equal to \(\frac{-E_a}{2.303RT}\).

Rearranging the equation, we find:

\( E_a = -2.303RT \cdot S \)

Thus, the correct answer is: (4) \(2.303RS\)