Question

When the temperature of a reaction \( A + B \rightarrow C \) is increased from \( 300\,K \) to \( 310\,K \) the rate constant increases by \( 12\% \). What is the Activation energy of the reaction?

• A. \( 163.9 \, \text{kJ/mol} \)
• B. \( 85.69 \, \text{kJ/mol} \)
• C. \( 192.17 \, \text{kJ/mol} \)
• D. \( 8.76 \, \text{kJ/mol} \)

Hint:

For small temperature changes, directly use Arrhenius form \( \ln(k_2/k_1) \) instead of approximationAlways convert energy to kJ/mol at the end.

Answer 1

The Correct Option is D

Step 1: Use Arrhenius equation.
\[ \ln\left(\frac{k_2}{k_1}\right)=\frac{E_a}{R}\left(\frac{1}{T_1}-\frac{1}{T_2}\right) \]

Step 2: Substitute given values.
Increase in rate constant is \(12\%\):
\[ \frac{k_2}{k_1}=1.12 \]
\[ T_1=300K,\quad T_2=310K,\quad R=8.314\,J/mol\,K \]
\[ \ln(1.12)=\frac{E_a}{8.314}\left(\frac{1}{300}-\frac{1}{310}\right) \]

Step 3: Calculate temperature term.
\[ \frac{1}{300}-\frac{1}{310} =\frac{310-300}{300\times310} =\frac{10}{93000} \]
\[ =1.075\times10^{-4} \]

Step 4: Calculate logarithmic term.
\[ \ln(1.12)\approx 0.1133 \]

Step 5: Substitute into equation.
\[ 0.1133=\frac{E_a}{8.314}\times 1.075\times10^{-4} \]

Step 6: Solve for \(E_a\).
\[ E_a=\frac{0.1133\times 8.314}{1.075\times10^{-4}} \]
\[ E_a\approx 8760\,J/mol \]
\[ E_a=8.76\,kJ/mol \]

Step 7: Final conclusion.
\[ \boxed{8.76\,kJ/mol} \]