Question 17.10: The value of ΔG°f at 25 °C for gaseous mercury is 31.85 kJ/m......
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_{p} for the reaction
Hg(l)\xrightleftharpoons{} 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(g) (31.85 kJ/mol). We can calculate K_{p} from the equation ΔG° = -RT ln K_{p}, as in Worked Example 17.9.
| IDENTIFY | |
| Known | Unknown |
| Hg (g), ΔG°_{f} = 31.85 kJ/mol | P_{Hg} |
| T = 25 °C | |
K_{p} = antiln (-12.86)=e^{-12.86}=2.6\times 10^{-6}
Since K_{p} is defined in units of atmospheres, the vapor pressure of mercury at 25 °C is 2.6\times 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.