Question

The activation energy of a particular reaction is 245 kJ/mol. How many degrees above room temperature (25°C) would the reaction need to be heated in order to see a 10 fold increase in the rate constant?

Answer 1

Arrhenus 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 10 fold increase in the rate constant

Hence, Initial rate constant = k1

Final rate constant k2 = 10 k1

Initial temperature T1 = 25oC = 25 + 273 K = 298 K

Final temperature T2 = ?

Ea = activation energy = 245 kJ/mol = 245000 J/K/mol

Substitute all these velues in eq (1),

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

In (10k1/k1) = [245000/8.314] [ (1/298) - (1/T2)]

[ (1/298) - (1/T2)] = In(10) x [8.314/245000]

1/T2 = (1/298) -  In(10) x [8.314/245000]

T2 = 305.1 K

T2 = 32.1 ^oC

Therefore, 7.1 ^oC above room temperature (25°C) the reaction need to be heated in order to see a 10 fold increase in the rate constant.