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A chemist starts a first-order reaction at 8:00 AM at 27 deg C. At 1:00 PM only 10% of the reaction is complete. He then raises the temperature to 127 deg C. At 4:00 PM, 50% of the reaction is complete (from the 1 PM restart). He must reach 90% completion by 5:00 PM and uses a catalyst at 4:00 PM and 127 deg C to achieve this. What is the activation energy of the original (uncatalysed) pathway? (Use: ln(10/9) = 0.1, ln(9/5) = 0.6, ln10 = 2.3, ln5 = 1.6, ln8 = 2; R = 2 cal/mol/K)

1. 10.14 kcal/mol
2. 5.52 kcal/mol
3. 2.64 kcal/mol
4. 7.92 kcal/mol

Correct answer: 5.52 kcal/mol

Solution

At T1=300 K (27 deg C), reaction runs for 5 h; 10% complete means 90% reactant remains. k1 = ln(N0/N)/t = ln(1/0.9)/5 = ln(10/9)/5 = 0.1/5 = 0.02 h⁻¹. At T2=400 K (127 deg C), reaction runs from 1PM to 4PM = 3 h; total completion reaches 50%, so 50% of original reactant remains (dropped from 90% to 50% remaining). k2 = ln(0.9/0.5)/3 = ln(9/5)/3 = 0.6/3 = 0.2 h⁻¹. Arrhenius: ln(k2/k1) = (Ea/R)(1/T1 - 1/T2). ln(0.2/0.02) = ln(10) = 2.3. (1/300 - 1/400) = 1/1200 K⁻¹. Ea = 2 * 1200 * 2.3 = 5520 cal/mol = 5.52 kcal/mol.

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