The activation energy of an uncatalyzed reaction is 91kJ/mol . The addition of a

Question

The activation energy of an uncatalyzed reaction is 91kJ/mol . The addition of a catalyst lowers the activation energy to 61 kJ/mol .Assuming that the collision factor remains the same, by what factor will the catalyst increase the rate of the reaction at 35degrees C?


The activation energy of an uncatalyzed reaction is 91kJ/mol . The addition of a catalyst lowers the activation energy to 61kJ/mol .Assuming that the collision factor remains the same, by what factor will the catalyst increase the rate of the reaction at 111 degrees C?

Explanation / Answer

We know

k = Ae-E/RT

R = k*product of reactants concentration

where R = rate of reaction

Ri/Rf = ki/kf

Where i & f denotes before and after catalyst respectively

a) ki = Ae-91*10^3/8.314*(273+35)

kf = Ae-61*10^3/8.314*(273+35)

Since collision factor is same so A will remain same

Ri/Rf = ki/kf = 8.166*10-6

=> Rf = 1.22*105 Ri

SO rate will increase by a factor of 1.22*105

b) At 111 C i.e. at 384 K

ki = Ae-91*10^3/8.314*(384)

kf = Ae-61*10^3/8.314*(384)

Again A will remain same since collision factor doesn't change

Ri/Rf = ki/kf = 8.3*10-5

=> Rf = 1.2*104 Ri

  SO rate will increase by a factor of 1.2*104

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