[fender1289]

The question asks to predict the signs of ∆H and ∆S for the reaction 2O(g)->O2(g)

first, I'm not sure how to "predict" ∆H; I looked the ∆Hformations in the back of the book and calculated it using the heat of formation of the products minus that of the reactants. I get -249 kj/mol. How would I "predict" the sign of ∆H?

second, I'm running into a contradiction for ∆S: this reaction has decreasing positional probability; i.e. 2 molecules to 1 molecule, and thus ∆S should be negative. However, according to the equation ∑Sreaction=∑Sproducts - ∑Sreactants, ∆S for the reaction is actually positive: 205.41 - 161= 44 J/Kmol. Which sign is right?

Thank you for the help.

[renge ishyo]

Don't forget that when you calculate these values you must multiply the value in the table by the coefficient of each species in your balanced chemical equation!

The calculations would be
:delta: H = [(1)(0.00 kJ/mol)] - [(2)(249.2 kJ/mol)] = - 498.4 kJ/mol

and
 :delta: S = [(1)(205.2 kJ/mol*K)] - [(2)(161.1 J/mol*K)] = - 117 J/mol*K

These values are what you would expect. You would expect a large negative enthalpy of formation because bonds are formed in the product whereas no energy would be needed to break the bonds of the reactants because they are free oxygen atoms. You would expect the entropy to be negative because of the whole "2 vs. 1" reasoning that you gave above.