Question

The activation energy for the isomerization of methyl isonitrile is 160 kJ/mol.

a.Calculate the fraction of methyl isonitrile molecules that have an energy of 160.0kJ or greater at 500K ?

b. Calculate this fraction for a temperature of 520K ?

c. What is the ratio of the fraction at 520K to that at 500K? show all work.

Answer 1

K = Ae^-(Ea/RT)
In Arrhenius equation,the factor e^(-Ea/RT) gives the fraction of molecules having energy equal to or greater than the activation enrgy Ea.
assume if x is equal to
x = e^(-Ea/RT)
ln x = -Ea/RT
2.303*log x = -Ea/RT
log x = -Ea/(2.303RT)

a) At 500 K

Ea = 160 KJ or 160000 J
T = 497 K

log x = (-160000)/(2.303*8.314*497)

         = -16.814
x = 10^-16.814 = 1.53*10^-17

At 520 K
log x = (-160000)/(2.303*8.314*522)
log x = -16.008
x = 9.817*10^-17

The ratio of fraction at 520K to fraction at 500K is

Ratio = (9.817*10^-17)/(1.53*10^-17)

         = 6.416