The equilibrium constant for the gas-phase isomerization of borneol (c10h17oh) to isoborneol at 503

3 answers
Question:

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.

Answers

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

It dont even matter because at the end of thre day we gone life like it ius s

2b2t

explanation:

2b2t

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