a. Calculate the fraction of methyl isonitrile molecules that have an energy of 160.0 kJ or...
a. Calculate the fraction of methyl isonitrile molecules that have an energy of 160.0 kJ or greater at 504 K .
answer:
| f1 = |
2.61×10?17 |
b. Calculate this fraction for a temperature of 519 K .
c. What is the ratio of the fraction at 519 K to that at 504 K ?
Homework Answers
b) According to Boltzmann distribution
The fraction of moles (f) = e-Ea/RT
Where Ea is the activation energy, R is the universal gas constant and T is the absolute temperature.
i.e. ln(f2/f1) = Ea/R * (1/T1 - 1/T2)
Here, f1 = 2.61*1017 molecules = 2.61*1017/6.023*1023 mol = 4.33*10-7 mol
i.e. ln(f2/4.33*10-7 mol) = 160*103 J/(8.314 J mol-1 K-1) * (1/504 - 1/519) K
i.e. ln(f2/4.33*10-7 mol) = 1.1036
i.e. f2/4.33*10-7 mol = e1.1036 = 3.015
i.e. f2 = 4.33*10-7 * 3.015 * 6.023*1023 molecules = 7.86*1017 molecules
(c) The ratio of the fraction at 519 K to 504 K
= f2/f1 = 3.015
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