Consider the freezing of liquid water at -10oC. For this process what are the signs for Δ H, Δ S and Δ G?

A. ΔH = +, ΔS = -, ΔG = 0

B. ΔH = +, ΔS = -, ΔG = –

C. ΔH = -, ΔS = +, ΔG = 0

D. ΔH = -, ΔS = +, ΔG = –

E. ΔH = -, ΔS = -, ΔG = –

I guess this is an exothermic reaction, because the heat is released from the system so ΔH is negative.

Since the water is going from a liquid to a solid phase the entropy should get smaller, so ΔS is negative.

And ΔG is negative as well because the freezing of water is spontaneous at -10 degrees celsius

Did I got this right?

So the correct answer should be E?