Question

The activation energy for a hypothetical reaction, $A⟶$ Product, is $12.49kcal/mol$. If the temperature is raised from $295K$ to $305K$, the rate of reaction increases by:

• A. $60$%
• B. $100$%
• C. $50$%
• D. $20$%

Answer 1

The correct option is B $100$%

$k_{295}=A{e}^{\frac{-E_a}{295R}}{k}_{305}=A{e}^{\frac{-E_a}{305R}}\frac{k_{305}}{k_{295}}=e^{-\frac{E_a}{R}(\frac{1}{305}-\frac{1}{295})}\sim 2$

$\therefore \frac{k_{305}-k_{295}}{k_{295}}\times 100$

$\therefore \frac{2k_{295}-k_{295}}{k_{295}}\times 100=100\%$