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Using the thermodynamic properties of methanol, calculate its normal boiling point

So the standard formations for methanol in liquid are:

H= -239.2 kJ/mol

G= -166.6 kJ/mol

S= 127 J/Kmol

In gas:

H= -201 kJ/mol

G= -162.3 kJ/mol

S= 239.8 J/Kmol

This is all that is given, it was a test question

These are the standard enthalpy, entropy, free energy of formation of methanol in liquid and gas at 25 degrees celsius or 298K

1 Answer

  • hfshaw
    Lv 7
    9 years ago
    Favorite Answer

    Given the data in the question, we can calculate the standard enthalpy and entropy of vaporization, ΔH°_vap and ΔS°_vap:

    ΔH°_vap = -201 kJ/mol - (-239.2kJ/mo)l = 38.2 kJ/mol

    ΔS°_vap = (239.8 - 127)J/(K*mol) = 112.8 J/(K*mol)

    To a first approximation we can assume that ΔH°_vap and ΔS°_vap are independent of temperature. (In reality, they *do* depend on temperature.)

    At the standard boiling point of a substance, the liquid and gas phases are in equilibrium, so the free energy change for the reaction:

    liquid -> gas

    is zero, so at the boiling temperature, and making the above assumption that the entropy and enthalpy of vaporization are independent of temperature:

    0 = ΔG = 0 = ΔH°_vap - T_boiling*ΔS°_vap

    T_boiling = ΔH°_vap/ΔS°_vap

    In this case,

    T_boiling = (38,200 J/mol)/(112.8 J/(K*mol)) = 338.7 K = 65.5°C.

    (In reality, the boiling point of methanol at 1 atm pressure is 64.7°C, so even though we made an approximation, we are not far off.)

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