Consider the evaporation of water at 298 K and 1 atm pressure:

H20 (l) <---> H20 (g)

Given that ΔGOrxn= 8.6kJ/mol, what is the equillibrium vapor pressure of water at 298 K?

dGo = -RTlnK

8600 J = -8.314*298*K
Solve for K, then
K = pH2O
This gives pH2O in atmospheres; that x 760 converts to mm Hg. I get about 24 mm or so but that's approximate.

To find the equilibrium vapor pressure of water at 298 K, we can use the Gibbs free energy change (ΔG) and the equation:

ΔG = ΔG° + RTln(Q)

Where:
- ΔG is the Gibbs free energy change
- ΔG° is the standard Gibbs free energy change under standard conditions (298 K and 1 atm pressure)
- R is the gas constant (8.314 J/(mol·K))
- T is the temperature in Kelvin
- Q is the reaction quotient

In this case, the reaction is the evaporation of water from liquid to gas. We are given the ΔGOrxn (standard Gibbs free energy change) as 8.6 kJ/mol. We need to convert it to Joules:

ΔG° = 8.6 kJ/mol * 1000 J/1 kJ = 8600 J/mol

Now, we can use the equation above to solve for ln(Q):

ΔG = ΔG° + RTln(Q)

Rearranging the equation and substituting the known values, we have:

ln(Q) = (ΔG - ΔG°) / RT

Substituting the values:

ln(Q) = (0 - 8600 J/mol) / (8.314 J/(mol·K) * 298 K)

Calculating the right-hand side of the equation:

ln(Q) = -8600 J/mol / (2467.372 J/mol·K) ≈ -3.49

Now, to find Q (the reaction quotient), we take the exponential of both sides:

Q = e^ln(Q)

Q = e^(-3.49) ≈ 0.0302

The reaction quotient (Q) can also be expressed as the ratio of the partial pressures of the gaseous products to the partial pressure of the gaseous reactants. In this case, there is only one gaseous product (water vapor) and one gaseous reactant (water in the liquid phase):

Q = [H2O(g)] / [H2O(l)]

Since we are looking for the equilibrium vapor pressure of water (H2O(g)), we can approximate the equilibrium vapor pressure to be the value of Q.

Hence, the equilibrium vapor pressure of water at 298K is approximately 0.0302 atm.