Question

The rate constant of a reaction at $400$ K and $500$ K are $0.04$ min$^{-1}$ and $0.08$ min$^{-1}$ respectively. The activation energy of the reaction is about \ ($2.303R=19$ J K$^{-1}$ mol$^{-1}$, $\log 2 = 0.3010$)

• A. $8.5$ kJ
• B. $6.5$ kJ
• C. $10.5$ kJ
• D. $11.5$ kJ
• E. $15$ kJ

Hint:

If rate constant doubles with temperature rise, activation energy can be quickly estimated using Arrhenius relation.

Answer 1

The Correct Option is D

Concept: Chemistry - Arrhenius Equation. [ \log\frac{k_2}{k_1}=\frac{E_a}{2.303R}\left(\frac{1}{T_1}-\frac{1}{T_2}\right) ]
Step 1: Substitute values. [ k_1=0.04,\quad k_2=0.08 ] [ T_1=400K,\quad T_2=500K ] [ \log\frac{0.08}{0.04}=\log 2=0.3010 ]
Step 2: Use formula. [ 0.3010=\frac{E_a}{19}\left(\frac{1}{400}-\frac{1}{500}\right) ] [ 0.3010=\frac{E_a}{19}\left(\frac{100}{200000}\right) ] [ 0.3010=\frac{E_a}{19}(0.0005) ]
Step 3: Solve for $E_a$. [ E_a=\frac{0.3010 \times 19}{0.0005} ] [ E_a \approx 11438\text{ J mol}^{-1} ] [ E_a \approx 11.5\text{ kJ mol}^{-1} ]
Step 4: Final answer. [ \boxed{11.5\text{ kJ mol}^{-1}} ]
Hence, correct option is (D).