Question

At 400K the energy of activation of a reaction is decreased by 0.4 Kcal in the presence of a catalyst hence rate will be? and how?
(1) Increased by 1.65 times
(2) Increased by 2.73 times
(3) Decreased by 1.18 times
(4) Decreased by 1.7 times

Answer 1

Answer on Question #56036 - Chemistry - Physical Chemistry

Solution:

According to Arrhenius equation:

$$k = A e ^ {- \frac {E _ {a}}{R T}}$$

Accept that $E_{a1}$ – energy of activation of uncatalytic reaction with rate constant $k_{1}$, $E_{a2} = E_{a1} - 0.4$ (Kcal) – energy of activation of catalytic reaction with rate constant $k_{2}$.

$$\text{Then} \quad \frac {k _ {2}}{k _ {1}} = A e ^ {\frac {- E _ {a 2}}{R T}} / A e ^ {\frac {- E _ {a 1}}{R T}} = e ^ {\frac {E _ {a 1} - E _ {a 2}}{R T}}$$ $$e = 1.9872 \text{cal} \cdot \text{K}^{-1} / \text{Mol}^{-1}, E_{a1} - E_{a2} = 0.4 \text{ Kcal} = 400 \text{ cal}$$ $$\frac {k _ {2}}{k _ {1}} = e ^ {\frac {400}{400 \cdot 1.987}} = 1.654$$

**Answer:** (1) rate increased by 1.65 times

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