Question

The activation energy of an uncatalyzed reaction is 95 kJ / mol. The addition of a catalyst lowers the activation energy to 55 kJ / mol. Assuming that the collision factor remains the same, by what factor will the catalyst increase the rate of the reaction at125∘C?

Answer 1

Answer:

The rate of the catalyzed reaction will increase by a 1.8 × 10⁵ factor.

Explanation:

The rate of a reaction (r) is proportional to the rate constant (k). We can find the rate constant using the Arrhenius equation.

$k=A.e^{-Ea/R.T}$

where,

A: collision factor

Ea: activation energy

R: ideal gas constant

T: absolute temperature (125°C + 273 = 398 K)

For the uncatalized reaction,

$kU=A.e^{-95\times 10^{3} kJ/mol /(8.314J/K.mol).398K}=3.4\times 10^{-13}A$

For the catalized reaction,

$kC=A.e^{-55\times 10^{3} kJ/mol /(8.314J/K.mol).398K}=6.0\times 10^{-8}A$

The ratio kC to kU is 6.0 × 10⁻⁸A/3.4 × 10⁻¹³A = 1.8 × 10⁵