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? [Hint: calculate the Kp for the reaction Hg(l)=Hg(g)]
To calculate the vapor pressure of mercury at 25°C, we can use the relationship between ΔG°f (standard Gibbs free energy of formation) and the equilibrium constant (Kp) of the reaction.
The equation for the reaction you provided is: Hg(l) ⇌ Hg(g)
The equilibrium constant (Kp) relates the concentrations of the reactants and products in a gaseous reaction at equilibrium. In this case, Kp represents the ratio of the partial pressure of mercury gas (Hg(g)) to the partial pressure of mercury in the liquid state (Hg(l)).
The relationship between Kp and ΔG°f at a given temperature is given by the equation:
ΔG°f = -RT * ln(Kp)
Where:
- ΔG°f is the standard Gibbs free energy change for the reaction (in this case, for the formation of mercury gas).
- R is the ideal gas constant (8.314 J/(mol*K)).
- T is the temperature in Kelvin.
First, we need to convert the provided ΔG°f value from KJ/mol to J/mol:
ΔG°f = 31.85 KJ/mol = 31.85 × 10^3 J/mol
Next, we need to convert the temperature from degrees Celsius to Kelvin:
T = 25°C + 273.15 = 298.15 K
Now, we can rearrange the equation to solve for Kp:
ln(Kp) = -ΔG°f / (RT)
Substituting the values, we get:
ln(Kp) = -(31.85 × 10^3 J/mol) / (8.314 J/(mol*K) * 298.15 K)
Next, we use the natural logarithm function (ln) to find the value of ln(Kp):
ln(Kp) ≈ -4.11
To find the value of Kp, we take the exponential of both sides of the equation:
Kp ≈ e^(ln(Kp)) ≈ e^(-4.11)
Using a scientific calculator or software, we can evaluate e^(-4.11) to find Kp:
Kp ≈ 0.016
Therefore, the vapor pressure of mercury at 25°C is approximately 0.016 atm (or any other unit of pressure, depending on what units you used for the equilibrium constant Kp).