Question

A certain reaction has an activation energy of 53.03 kJ/mol. At what Kelvin temperature will the reaction proceed 7.00 times faster than it did at 365 K?

Answer 1

This problem can be solved by using Arrhenius equation.

Arrhenius equation k = A e^-Ea/RT

where k = rate of reaction

A = collision frequency

Ea = activation energy

R= universal gas constant = 8.314 J/K/mol

T = temperature

Arrhenius equation can be written as

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

Given that Ea = 53.03 kJ/mol = 53030 J/mol

Initial rate of reaction k1 = k1

Final rate of reaction k2 = 7k1

Initial temperature T1 = 365 K

Final temperature T2 = ?

Substitute all these values in eq (1),

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

In (7k1/k1) = [53030/8.314] [ (1/365) - (1/T2)]

[(1/365) - (1/T2)] = [In ( 7)] x [8.314 / 53030 ]

(1/375) - [In ( 7)] x [8.314 / 53030 ] = 1/T2

On solving ,

T2 = 423 K

Therefore,

at 423 Kelvin temperature, the reaction proceed 7.00 times faster than it did at 365 K.