Question

A certain reaction has an activation energy of 64.28 kJ/mol. At what Kelvin temperature will the reaction proceed 4.50 times faster than it did at 329 K?

Answer 1

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)

From the given data,

Initial rate of rection = k1

Final rate of reaction k2 = 4.5 k1

Initial temperature T1 = 329 K

Final temperature T2 = ?

activation energy Ea = 64.28 kJ/mol = 64280 J/mol

Substitute all these values in eq (1),

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

In (4.5 k1/k1 ) = [ 64280 /8.314] [(1/329) - (1/T2)]

[(In 4.5) (8.314 / 64280)] = [(1/329) - (1/T2)]

1/T2 = [(1/329) ] - [(In 4.5) (8.314 / 64280)]

T2 = 351.5 K

Therefore,

At 351.5 K the reaction proceed 4.50 times faster than it did at 329 K.