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(a) The activation energy for the isomerization of methyl isonitrile (Figure 14.7)...

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moos140 | eNotes Newbie

Posted May 11, 2013 at 5:40 PM via web

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  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? 

     

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mvcdc | Student, Graduate | (Level 1) Associate Educator

Posted May 12, 2013 at 5:59 AM (Answer #1)

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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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