Question 16.10: CALCULATING A VAPOR PRESSURE FROM ΔG° The value of ΔG°f at 2...
CALCULATING A VAPOR PRESSURE FROM ΔG°
The value of ΔG°_{f} at 25 °C for gaseous mercury is 31.85 kJ/mol. What is the vapor pressure of mercury at 25 °C?
STRATEGY
The vapor pressure (in atm) equals K_{sp} for the reaction
Hg(l) \rightleftharpoons Hg(g) K_{p} = P_{Hg}
Hg(l) is omitted from the equilibrium constant expression because it is a pure liquid.
Because the standard state for elemental mercury is the pure liquid, ΔG°_{f} = 0 for Hg(l) and ΔG° for the vaporization reaction simply equals ΔG°_{f} for Hg(31.85 kJ/mol). We can calculate K_{p} from the equation ΔG° = -RT ln K_{p}, as in Worked Example 16.9.
ln K_{p} = \frac{- ΔG°}{RT} = \frac{-(31.85 × 10^{3} J/mol)}{[8.314 J/(K · mol)](298 K)} = -12.86.
K_{p} = antiln (-12.86) = e^{-12.86} = 2.6 × 10^{-6}
Since K_{p} is defined in units of atmospheres, the vapor pressure of mercury at 25 °C is 2.6 × 10^{-6} atm(0.0020 mm Hg). Because the vapor pressure is appreciable and mercury is toxic in the lungs, mercury should not be handled without adequate ventilation.
BALLPARK CHECK
ΔG°is positive, so the vaporization reaction should not proceed very far before reaching equilibrium. Thus, K_{p} should be less than 1, in agreement with the solution.
