(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?
1 Answer | Add Yours
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.
`x = e^((-160 x 1000 J/(mol))/(8.314 J/(molK) 500 K)) = 1.92 x 10^(-17) .`
`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.`
We’ve answered 317,286 questions. We can answer yours, too.Ask a question