Ah, the vapor pressure of gaseous H2O above BaCl2-H2O at 298K. Well, let's put on our chemistry hats and get cracking!
To calculate the vapor pressure, we can use the good ol' Gibbs free energy equation:
ΔG = ΔH - TΔS
Where ΔG is the change in Gibbs free energy, ΔH is the enthalpy change, T is the temperature in kelvin, and ΔS is the entropy change.
Now, we're given that the enthalpy change (ΔH) is -63.11 kJ/mol (negative because it's an exothermic reaction) and the entropy change (ΔS) is +148 J/(K-mol). However, we need to convert the units of ΔH to J/mol, so it becomes -63110 J/mol.
But before we continue, let's convert the temperature to kelvin. We have 298K, so no conversion needed there.
Now, plug in the values into our equation:
ΔG = (-63110 J/mol) - (298K)(148 J/(K-mol))
Now, let's do the math:
ΔG ≈ -63110 J/mol - 44104 J/mol
≈ -107,214 J/mol
Now, we can use the equation ΔG = -RTlnP to solve for the vapor pressure (P). R is the gas constant, which is 8.314 J/(mol·K), and T is the temperature in kelvin (298K).
-107214 J/mol = -(8.314 J/(mol·K))(298K)lnP
Now, solving for P:
lnP = -107214 J/mol / (8.314 J/(mol·K) * 298K)
lnP ≈ -40.63
Using the magic of e, we can now find P:
P ≈ e^(-40.63)
≈ 5.6 x 10^(-18)
And there you have it! The vapor pressure of gaseous H2O above BaCl2-H2O at 298K is approximately 5.6 x 10^(-18) (I hope that it's as low as my chances of becoming a stand-up comedian!).