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QUESTION

At 25 degrees Celsius, by what factor is the rate increased by a catalyst that reduces the activation energy by 1.00 kJ/mol?

The rate is increased by a factor of 1.50.

For this problem, we use the Arrhenius equation:

k = Ae^(-E_a/(RT))

In logarithmic form,

ln k = lnA – E_a/(RT)

Let k_2 and E_2 be the rate constant and activation energy for the catalyzed reaction, and k_1 and E_1 be the values for the uncatalyzed reaction. Then

1. lnk_2 = lnA – E_2/(RT)

2. lnk_1 = lnA – E_1/(RT)

Subtracting 2 from 1, we get

ln(k_2/k_1) = -E_2/(RT) + E_1/(RT) = (E_1 – E_2)/(RT)

E_1 – E_2 = "1 kJ·mol⁻¹" = "1000 J·mol⁻¹"

ln(k_2/k_1) = (1000 cancel("J·mol⁻¹"))/(8.314 cancel("J·K⁻¹mol⁻¹") × 298.15 cancel("K")) = 0.4034

k_2/k_1 = e^0.4034 = 1.50

The rate will increase by a factor of 1.50.