# 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.

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