A tabulation of more precise thermodynamic data than are presented in Appendix D lists the following values for H₂O(l) and H₂O(g) at 298.15 K, at a standard state pressure of 1 bar.

(a) Use these data to determine, in two different ways, Δ_rG° at 298.15 K for the vaporization: H₂O (l, 1 bar) ⇌ H₂O(g, 1 bar).

(b) Use the result of part (a) to obtain the value of K for this vaporization and, hence, the vapor pressure of water at 298.15 K.

(c) The vapor pressure in part (b) is in the unit bar. Convert the pressure to millimeters of mercury.

(d) Start with the value Δ_rG° = 8.590 kJ mol^-1, and calculate the vapor pressure of water at 298.15 K in a fashion similar to that in parts (b) and (c). In this way, demonstrate that the results obtained in a thermodynamic calculation do not depend on the convention we choose for the standard state pressure, as long as we use standard state thermodynamic data consistent with that choice.

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A$H, kJ mol-1 H₂O(1) -285.830 H₂O(g) -241.818 1.G kJ mol-1 Sº, J mol-1 K-1 -237.129 69.91 -228.572 188.825