Calculate the vapor pressure of Hg at 14 oC (in atm).
Hg(l) ↔ Hg(g) . . . ΔHo = 61.32 kJ and ΔSo = 98.83 J/K
I would try this.
dGo = dHo - TdSo
Then dGo = -RTlnK and solve for Kp
Then Kp = pHg(g)
To calculate the vapor pressure of mercury (Hg) at 14 °C, we can use the Clausius-Clapeyron equation, which relates the vapor pressure of a substance to its enthalpy and entropy of vaporization. The Clausius-Clapeyron equation is given by:
ln(P2/P1) = -(ΔHvap/R) * (1/T2 - 1/T1)
Where:
P1 is the initial vapor pressure
P2 is the final vapor pressure
ΔHvap is the enthalpy of vaporization
R is the ideal gas constant (0.0821 L*atm/(mol*K))
T1 is the initial temperature (in Kelvin)
T2 is the final temperature (in Kelvin)
In this case, we want to find the vapor pressure of Hg at 14 °C. We need to convert this temperature to Kelvin by adding 273.15, so T2 = 14 + 273.15 = 287.15 K.
Given:
ΔHvap = 61.32 kJ = 61.32 × 10^3 J
ΔS = 98.83 J/K
T1 = boiling point of mercury = 356.7 °C = 356.7 + 273.15 = 629.85 K
Now substitute the values into the Clausius-Clapeyron equation and solve for P2:
ln(P2/P1) = -(61.32 × 10^3 J)/(0.0821 L*atm/(mol*K)) * (1/287.15 K - 1/629.85 K)
Simplifying the equation further:
ln(P2/P1) = -748.29 * (0.001815 - 0.001587)
ln(P2/P1) = -748.29 * 0.000228
ln(P2/P1) = -0.17072
Now we can solve for P2/P1 by taking the exponential of both sides:
P2/P1 = e^(-0.17072)
Using a calculator, we find:
P2/P1 ≈ 0.842
To solve for P2, we can assume P1 = 1 atm (since the vapor pressure of Hg is negligible at typical room temperatures). Therefore:
P2 ≈ P1 * (0.842) = 1 atm * 0.842
P2 ≈ 0.842 atm
Therefore, the vapor pressure of Hg at 14 °C is approximately 0.842 atm.