Question

The activation energy of an uncatalyzed reaction is 91 kJ/mol . The addition of a catalyst lowers the activation energy to 59 kJ/mol .

A.) Assuming that the collision factor remains the same, by what factor will the catalyst increase the rate of the reaction at 26 ∘C ? Express the ratio to two significant digits

B.) Assuming that the collision factor remains the same, by what factor will the catalyst increase the rate of the reaction at 120 ∘C ? Express the ratio to two significant digits.

Answer 1

Solution-

Here we use Arrhenius equation:

k = A exp (-Ea/RT)

[here A is collision factor, Ea is activation energy, R is molar gas constant and T is temperature in K]

Reaction rate = k[concentration]order of reaction

Thus Rate(catalyzed)/Rate(uncatalyzed)

= k(catalyzed)/k(uncatalyzed)

= exp [(Ea(uncatalyzed) - Ea(catalyzed))/RT]

Ea(uncatalyzed) = 91 kJ/mol = 91000 J/mol

Ea(catalyzed) = 59 kJ/mol = 59000 J/mol

1. T = 26 deg C = 299 K

Rate(catalyzed)/Rate(uncatalyzed)

= exp [(91000 - 59000)/(8.314 x 299)] = 3.90 x 10^5

2. T = 120 deg C = 393 K

Rate(catalyzed)/Rate(uncatalyzed)

= exp [(91000 - 59000)/(8.314 x 393)] = 1.79 x 10^4