a. As volume increases, the system prefers fewer moles of gas. In this case, that's the right side of the equlibrium because that side is all solid.

b. ΔG = ΔH - TΔS, and ΔH is negative. In order for ΔG to be negative, ΔH - TΔS must be (negative number - positive number) OR (negative number - smaller negative number), but since -11.3 seems small to me, I'll assume the first. For the term TΔS to be positive, ΔS should be positive. ΔS (or S_{final} - S_{initial}) to be positive implies S_{final} > S_{initial}

In the described system, the equilibrium gets shifted to the right (gas to solid) so S_{final} > S_{initial} is true; ΔS is indeed positive, ΔH - TΔS is negative and ΔG is negative. Is that correct?

c. the K_{eq} is actually unchanged, as opposed to decrease-- it's dependent solely on temperature (in this case)

d. defined

e.

I'm still unclear on (e). My solutions guide says that (c) is the definite answer. Also, I used ΔG°=ΔH°−TΔS°, so wouldn't an isothermal system not be a problem?