Question

A certain reaction has an activation energy of 31.18 kJ/mol. At what Kelvin temperature will the reaction proceed 6.00 times faster than it did at 311 K?

Answer 1

Given:

Ea (activation energy ) = 31.18 kJ/mol

Let K2 / K1 = 6

K1 is the rate constant at Temperature T 1 = 311 K

We have to find T2

We use Arrhenius equation

Ln (k2/k1) = -Ea / R [ 1/T2 – 1/T1]

Here Ea is activation energy in J/mol

R is gas constant. value is 8.314 J per K per mol

Lets plug given value in above equation to get T2

Ln (6) = - 31180 J/mol / 8.314 J per K per mol [ 1/T2 – 1/311]

1.792 = - 3750.301 ( 311 – T2 / 311 T2 )

( 311 – T2 / 311 T2 ) = - 0.000478

311 – T2 = - 0.000478 x 311 T2

311 – T2 = - 0.1486 T2

T2 = 311 + 0.1486

= 311.15 K

So the Temperature at which reaction rate increases by 6 is 311.15 K