A certain reaction has an activation energy of 50.75 kJ/mol. At what Kelvin temperature will the reaction proceed 3.00 times faster than it did at 313 K?

Question

can you please show me the formula with the answer

Answer 1

The formula is the Arrhenius equation:

ln (k2/k1) = -Ea/R (1/T2 - 1/T1)

k1 and k2 are the rate constants at temperature 1(T1) and temperature 2(T2), respectively.

R = 8.314 J/Kmole = 0.008314 kJ/Kmol to keep units consistent with units of Ea

Ea = activation energy = 50.75 kJ/mol

T1 = 313

T2 = ?

k2/k1 = 3 in order for rate at T2 to be 3x the rate at T1, the rate constant will be 3x what it was at T1

ln (3) = -50.75/0.008314 (1/T2 - 1/313)

1.0986 = -6104 (1/T2 - 0.003195)

1.0986 = -6104/T2 + 19.50

-18.40 = -6104 / T2

T2 = 332K