What is the ∆G for the following reaction under nonstandard conditions where the following data is given? 2NH4Cl(s) + CaO(s) CaCl2(s)+ H2O + 2NH3(g)

Given:
CaO: ∆Hf = -635.09 kJ, ∆Sf =39.75 J/K
CaCl2: ∆Hf = -795.8 kJ, ∆Sf =104.6 J/K.
NH3: ∆Hf = -46.11 kJ, ∆Sf = 192.45,
H2O: ∆Hf = -285.830 kJ, ∆Sf = 69.91 ,
NH4Cl: ∆Hf = -314.43 kJ, ∆Sf = 94.6
A.
2.27 J
B.
-4.54 J
C.
4.54 J
D.
9.08 J
E.
-9.08 J

Answerthis solution will not form a precipitateexplanationsolubility equilbrium:         pbf₂(s)  ⇄ pb²⁺(aq) + 2 f⁻(aq)the solubility product expression is    (we exclude solids from the equilibrium expression).now suppose we have the trial ion product q, which we can define as    then, we can interpret  q as the product of the concentration of the ions that we have and ksp as the product of the ion concentrations that are needed to form a saturated solution.if q > ksp, then that implies that there will be a surplus of ions that exceed the amount that is required to form a saturated solution. a precipitate will then form (the precipitate continually  forms until the surplus ions are removed and q is decreased to ksp's value).if q = ksp, the saturated solution that forms is barely saturated and the precipitate will be minimal.if q < ksp, then the ion concentration will not be enough to form a saturated solution (so no precipitate).using our information, we calculate the trial ion product:     since q = 2.7  × 10⁻⁸ < ksp =  4  ×  10⁻⁸, a precipitate will not form.

Hello! data: p (pressure) = 1.25 atmv (volume) = ? (in l)t (temperature) = 298 kn (number of mols) = 0.900 molesr (gas constant) = 0.082 (atm*l/mol*k)apply the data to the clapeyron equation (ideal gas equation), see: i hope this . =)