At the boiling point of water, liquid and vapor are inequilibrium at a pressure

Question

At the boiling point of water, liquid and vapor are inequilibrium at a pressure of 1 atm. Use H° andS° for the reaction, to calculate the boiling point of water. How doesthe answer compare to the actual boiling point? Does this supportthe statement that H and S are nearly temperatureindependent? Please EXPLAINyour answer.

Explanation / Answer

Recall the equation: G = Go + RT ln Q at equilibrium Go = 0 and Q = K, so we have G = RT lnK also recall that if we assume that H° and S°don't change much with temperature, G = H° -TS° so that we have G = H° - TS° = RT lnK we are after T here, so we isolate T T = H° / (R lnK + S°) also note that if we write the equilibrium constant expression forthe reaction, we get K = PH2O ---->pure liquids andsolids are omitted from the equation because their concentrationsdon't change over time then, we have T = H° / (R lnPH2O + S°) substituting the values , we get H° = -241.82 kJ -(-285.83 kJ) = 44.01kJ-------->all reactants have 1 mol lnPH2O = ln 1 = 0 S° = 188.83 J/K - 69.91 J/K = 118.92 J/K = 0.119 kJ/K-------->all reactants have 1 mol T = (44.01 kJ) / (0.119 kJ/K) = 370 K = 97oC, since only 1 significant figure is given ( 1atm), we could round off the answer to 1 x102oC The answers agree within the uncertainties specified.this supports thestatement that H and S are nearly temperatureindependent :-|

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