J.R. S. answered 03/19/21
Ph.D. University Professor with 10+ years Tutoring Experience
a) The ∆Hºrxn = ∑∆Hºf products - ∑∆Hºf products. For the current reaction, we need to know the ∆Hºf for H2O(l) since it is a product. Looking it up, I find it to be -285.8kJ/mol. Solving for ∆Hºrxn we have...
(-485.8 + -285.5) - (-283.3 + 0) = -483.3 kJ = ∆H
Doing the same for ∆Sº after looking up values for H20(l) and O2(g) we have...
(178.7 + 69.9) - (148.5 + 205) = -104.9 J/K = ∆S
b) If ∆Gº is negative, it is spontaneous
∆Gº = ∆Hº - T∆Sº = 483.3 kJ = 298K(-0.1049 kJ/K) = -452.0 kJ YES it is spontaneous
Sorry, I've run out of time for now. Hopefully someone else will finish. If not, perhaps I can return at a later time and finish up. I think it unfair for them not to provide all the necessary data, however.