Question

The rate constant of a reaction at 500 K is 0.02 and it increased to 0.07 at 700 K. The \(E_a\) value of the reaction is about [Given: \(R = 8.3 \, J\,K^{-1}\,mol^{-1}\) and \(\log(7/2) = 0.544\)].

• A. $180.1~kJ~mol^{-1}$
• B. $36.2~kJ~mol^{-1}$
• C. $54.6~kJ~mol^{-1}$
• D. $42.8~kJ~mol^{-1}$
• E. $18.2~kJ~mol^{-1}$

Hint:

Always ensure units for $R$ and $E_a$ (Joules vs KiloJoules) are consistent.

Answer 1

Answer: (E)

Step 1: Concept
Use the Arrhenius Equation: $\log(k_2/k_1) = \frac{E_a}{2.303R} \left( \frac{T_2 - T_1}{T_1 T_2} \right)$.

Step 2: Meaning
$k_1=0.02, k_2=0.07, T_1=500, T_2=700$. $\log(0.07/0.02) = \log(7/2) = 0.544$.

Step 3: Analysis
$0.544 = \frac{E_a}{2.303 \times 8.3} \left( \frac{200}{350000} \right)$
$E_a \approx \frac{0.544 \times 19.1139 \times 350000}{200} \approx 18200~J/mol = 18.2~kJ/mol$.

Step 4: Conclusion
Hence, the activation energy is approximately $18.2~kJ~mol^{-1}$.
Final Answer: (E)