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24 March, 00:11

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.

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  1. 24 March, 00:18
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    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
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