Waiting for answer This question has not been answered yet. You can hire a professional tutor to get the answer.
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.