Question

18. Please try the problem again, your previous tries are listed below. Try 1: * -8 Use the data found in the Thermodynamic Values link under Reference Materials to calculate AHº and AS™ at 25.0 °C for CH3OH (1) → CH3OH (g) Although the AH° and AS values do change slightly with temperature, assume that the change is not large enough to significantly affect the outcome of the following calculation. Use the AH° and AS™ values to determine the temperature (in °C) at which this phase change occurs (find the boiling point). X Y our answer, -8, is incorrect by more than 10%. Submit Cancel to Assignment Menu DSA Main Menu

Answer 1

I don't exactly know which data is available for you at the REFERENCE MATERIALS section. If they are enthalpies of formation and entropies, we can solve this like this:

The standard enthalpy of formation of gaseous methanol is −201.3 kJ/mol, while the value for the liquid is −238.4 kJ/mol. This means that the ∆H° can be calculated as the difference between the enthalpies of formation of product and reactant:

$\Delta H^{o}=\Delta _{f}H^{o}_{MeOH(g)}-\Delta _{f}H^{o}_{MeOH(l)}=-201.3kJ/mol-(-238.4kJ/mol)=37.1kJ/mol$

The standard entropies are: 239.9 J/(mol K) for the gas and 127.2 J/(mol K) for the liquid. The ∆S° can be calculated as we did with the enthalpy:

$\Delta S^{o}=S^{o}_{MeOH(g)}-S^{o}_{MeOH(l)}=239.9\frac{J}{mol\cdot K}-127.2\frac{J}{mol\cdot K}=112.7\frac{J}{mol\cdot K}$

With these values we can calculate the boiling point taking into account that it is defined at the temperature at which the liquid and gaseous phases are in equilibrium, which is equivalent to say that it is the temperature at which ∆G° = 0. ∆G° is given by:

$\Delta G^{o}=\Delta H^{o}-T\Delta S^{o}$

If we make ∆G° = 0, we get:

$0=\Delta H^{o}-T\Delta S^{o}$

$T=\frac{\Delta H^{o}}{\Delta S^{o}}=\frac{37100\frac{J}{mol}}{112.7\frac{J}{mol\cdot K}}=329K=56^{o}C$