The activation energy for the isomerization of methyl isonitrile is 160 kj/mol. part acalculate the fraction of methyl isonitrile molecules that have an energy of 160.0 kj or greater at 496 k .f1=part bcalculate this fraction for a temperature of 528 k .f2=part cwhat is the ratio of the fraction at 528 k to that at 496 k ? f2/f1=

(a)

(b)

(c)

Explanation:

Using the expression,

Where,

is the activation energy

A is the pre-exponential factor

R is Gas constant having value = 8.314 J / K mol  

The factor,  is the fraction of molecules which have value equal to or grater than activation energy.

Thus,

(a) Given, = 160.0 kJ/mol = 160000 J/mol (As 1 kJ = 1000 J)

T = 496 K

Thus, applying values as:

(b) Given, = 160.0 kJ/mol = 160000 J/mol (As 1 kJ = 1000 J)

T = 528 K

Thus, applying values as:

(c) Ratio :