Calculate the free energy (ΔG˚) of reaction (kJ) for the equation as written at 298 K (Dependence of temperature on enthalpy can be disregard). If the answer is negative, enter the sign and then the magnitude. Express your answer to three significant figures.

CaO(s) + H2O(l) → Ca(OH)2(s)

I know this part (Hess's Law stuff):
FYI the numbers were given enthalpy values.

DeltaG=DeltaH-TDeltaS
DeltaH=-986.1-[-635.1-285.83]=-65.17
DeltaS=83.4-[39.7+69.91]=-26.17
Then you plug it in the equation and solve for DeltaG, but when they say to disregard the temperature, does that mean I don't plug in the 298 to T or do I?

If the delta G" is delta Go, I think you are supposed to look up the value of delta Go in a set of tables. DGo for the rxn = [DGo products] -[DGo reactants]

Hmmm...well the problem gives us a table of values for DeltaH and DeltaS for the compounds, but purposely left the DeltaG for the compounds blank, so I'm assuming they want you to use what values they give you to solve for it.

When it says to disregard the dependence of temperature on enthalpy, it means that you should assume that the enthalpy change (ΔH) is constant with respect to temperature. In other words, the ΔH value you have been given is already calculated at the temperature of 298 K.

Therefore, to calculate ΔG at 298 K, you can plug in the given values for ΔH and ΔS into the equation ΔG = ΔH - TΔS, where T is the temperature in Kelvin.

Using the values you provided:
ΔH = -65.17 kJ
ΔS = -26.17 kJ/K
T = 298 K

Plug these values into the equation to calculate ΔG:
ΔG = -65.17 kJ - (298 K)(-26.17 kJ/K)
ΔG = -65.17 kJ + 7771.66 kJ
ΔG = 7706.49 kJ

The free energy change ΔG for the given reaction at 298 K is 7706.49 kJ.