Slope of the graph given is -9.2 J/mol . The activation energy of the…

The correct answer is Option (2) → 175.8 J/mol

The Arrhenius equation k = A e^−E_a/RT in logarithmic form is:

    ln k = ln A − E_a/RT   ⇒   log k = log A − E_a/(2.303 RT)

The graph in the question plots log k against 1/T, so its slope is −E_a/(2.303 R):

    −E_a/(2.303 R) = −9.2
    E_a = 9.2 × 2.303 × R
    E_a = 9.2 × 2.303 × 8.3 J K⁻¹ mol⁻¹ ≈ 175.8 J mol⁻¹

(With R = 8.314 J K⁻¹ mol⁻¹ the value is 176.2 J mol⁻¹; the nearest option is 175.8 J/mol.)