The equilibrium constant for the gas-phase isomerization of borneol (c10h17oh) to isoborneol at 503 k is 0.106. a mixture consisting of 7.50 g of borneol and 15.0 g of isoborneol in a container of volume 6.3 dm3 is heated to 503 k and allowed to come to equilibrium. calculate the mole fractions of the two substances at equilibrium.

The solution is as follows:

K = [Partial pressure of isoborneol]/[Partial pressure of borneol] = 0.106

The molar mass of isoborneol/borneol is 154.25 g/mol

Mol isoborneol = 15 g/154.25 = 0.0972 mol

Mol borneol = 7.5 g/154.25 = 0.0486 mol

Use the ICE approach

borneol → isoborneol

I 0.0972 0.0486

C - x + x

E 0.0972 - x 0.0486 + x

Total moles = 0.1458

Using Raoult's Law,

Partial Pressure = Mole fraction*Total Pressure

[Partial pressure of isoborneol] = [ (0.0972-x) / 0.1458]*P

[Partial pressure of borneol] = [ (0.0486+x/0.1458) ]*P

0.106 = [ (0.0972-x) / 0.1458]*P / [ (0.0486+x/0.1458) ]*P

Solving for x,

x = 0.0832

Thus,

Mol fraction of borneol = (0.0486+0.0832) / 0.1458 = 0.904

Mol fraction of isoborneol = (0.0972-0.0832) / 0.1458 = 0.096