Question

the activation energy for the isomerization of methyl isonitrile (see the figure (Figure 1)) is 160 kJ/mol.

1. Calculate this fraction for a temperature of 512 K

2. What is the ratio of the fraction at 512 K to that at 499 K

Answer 1

Part 1

activation energy Ea = 160 kJ/mol = 160000 J/mol

Temperature T = 512 K

From Arrhenius equation

k = A * exp (-Ea/RT)

A = pre-exponential factor

R = gas constant = 8.314 J/K-mole

k = rate constant

Fraction of molecules having energy 160 kJ/mol

k/A = exp(-Ea/RT)

= exp(-160000/(8.314 x 512)

Y1 = 4.744 x 10^-17

Part 2

Similarly at 499 K

Fraction of molecules having energy 160 kJ/mol

k/A = exp(-Ea/RT)

= exp(-160000/(8.314 x 499)

Y2 = 1.782 x 10^-17

Y1/Y2 = 4.744 x 10^-17 / 1.782 x 10^-17

= 2.662