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For an uncatalysed elementary reaction, a plot of ln k versus (1/T)*10³ is a straight line through the points (3.2, -5.2) and (3.8, -8.8). When a catalyst is added, the reaction at 300 K proceeds at the same rate as the uncatalysed reaction at 400 K (same pre-exponential factor). By approximately how much (in kcal) has the catalyst lowered the activation energy?
- about 3 kcal
- about 12 kcal
- about 6 kcal
- about 9 kcal
Correct answer: about 3 kcal
Solution
Slope = (-8.8+5.2)/(3.8-3.2) = -6 per (1/T)*10³, so Ea = 6000*R ~ 11.9 kcal/mol. Equal rates give Ea(cat) = (300/400)Ea, so the drop = 0.25*Ea ~ 3 kcal.
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