Question

If a first order reaction has a rate constant of 4.15 X 10^-2 s^-1 at a temperature of 24.5^oC, what would the value of k be if the reaction temperature was changed to 52^oC given that the activation energy is 67.8 kJ/mol?

Answer 1

Given, K₁ = 4.15 x 10^-2 s^-1 , T₁ = 24.5⁰C = 297.5 K , T₂ = 52⁰C = 325 K, E_a = 67.8 KJ/mol = 67800 J/mol , K₂ = ?

We have the relation as---

log K₂/ K₁ = E_a / 2.303 R [ 1/T₁ - 1/T₂]

=> log K₂ / 4.15 x 10^-2 s^-1 = 67800 J/mol / 2.303 x 8.314 JK^-1mol^-1 [ 1/ 297.5 K - 1 / 325 K ]

=> log K₂ / 4.15 x 10^-2 s^-1 = 67800 /19.147142 [ 325-297.5 / 297.5 x 325]

=> log K₂ / 4.15 x 10^-2 s^-1 = 3540.998442x [ 27.5/96687.5]

=> log K₂ / 4.15 x 10^-2 s^-1 = 97377.45716/96687.5

=> log K₂ / 4.15 x 10^-2 s^-1 = 1.00713595

=> K₂ / 4.15 x 10^-2 s^-1 = antilog ( 1.00713595]

=> K₂ / 4.15 x 10^-2 s^-1= 10.16566866

=> K₂ = 42.18752492 x 10^-2 s^-1