The activation energy of a hypothetical reaction \(A → Product\) is 12.49 kcal/mol. If the temperature is raised from 295 to 305, the rate of the reaction increases by

100%

The correct answer is option 2. 100%.

The given reaction is \(A → Product\)

From Arrhenius's equation,

\(k_{295} = Ae^{\frac{-E_a}{295R}}\)

and \(k_{305} = Ae^{\frac{-E_a}{305R}}\)

∴ \(\frac{k_{305}}{k_{295}} = e^{-\frac{E_a}{R}\left(\frac{1}{305} − \frac{1}{295}\right)}\)

\(\frac{k_{305}}{k_{295}} ≈ 2\)

∴ \(\frac{2k_{295} − k_{295}}{k_{295}} × 100 = 100\)%