# (a) The activation energy for the isomerization of methyl isonitrile (Figure 14.7) is 160 kJ/mol. Calculate the fraction of methyl isonitrile molecules that has an energy of 160.0 kJ or greater...

1. (a) The activation energy for the isomerization of methyl isonitrile (Figure 14.7) is 160 kJ/mol. Calculate the fraction of methyl isonitrile molecules that has an energy of 160.0 kJ or greater at 500 K. (b) Calculate this fraction for a temperature of 520 K. What is the ratio of the fraction at 520 K to that at 500 K?

mvcdc | Student, Graduate | (Level 2) Associate Educator

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The Arrhenius equation gives the relationship of the rate constant of the reaction with the activation energy of a chemical reaction.

It is given by the following formula:

`k = Ae^((-E_a)/(RT))`

where k is the rate constant, A is a constant (called a pre-factor), T is the temperature in the absolute scale (K), Ea is the activation energy of the reaction and R is the universal gas constant (J/Kmol version). The term beside A gives the fraction of molecules having an energy greater than or equal to the activation energy:

`x = e^((-E_a)/(RT))`

where x is the said fraction of molecules.

Now, we know that Ea = 160 kJ/mol and R = 8.314 J/mol K.

To get the fraction of molecule with energy greater than or equal to Ea, we simply substitute to the equation and solve for x.

At 500K:

`x = e^((-160 x 1000 J/(mol))/(8.314 J/(molK) 500 K)) = 1.92 x 10^(-17) .`

At  520K:

`x = e^((-160 x 1000 J/(mol))/(8.314 J/(molK) 520K) = 8.46 x 10^(-17))`

Hence, the ratio of fraction at 520K to fraction at 500K is:

`8.46/1.92 = 4.41.`

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