Slope of the graph given is \(-9.2 J/mol\). The activation energy of the reaction is :

\(175.8\, \ J/mol\)

The correct answer is option 2 - \(175.8\, \ J/mol\)

The given graph is:

The complete statement is: For a plot of log k vs 1/T, the slope is equal to −Eₐ / (2.303 R).

slope = −Eₐ / (2.303R)
Given slope = −9.2

−9.2 = −Eₐ / (2.303R)

Eₐ = 9.2 × 2.303 × 8.314

Eₐ ≈ 175.8 J mol⁻¹

Thus, Option (2) is correct.