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
Answer & explanation
Correct answer: option 2
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 \(k_{305} = Ae^{\frac{-E_a}{305R}}\)
∴ \(\frac{k_{305}}{k_{295}} = e^{-\frac{E_a}{R}\left(\frac{1}{305} − \frac{1}{295}\right)}\)
\(\frac{k_{305}}{k_{295}} ≈ 2\)
∴ \(\frac{2k_{295} − k_{295}}{k_{295}} × 100 = 100\)%