[nernst]

1. Calculate the  :delta:S for decomposition of ozone from oxygen:
2O₃(g)  3O₂(g)

S^o = 205 J/molK for O₂ and 239 J/molK for O₃ at 25^oC.

I calculated that  :delta:S =  :delta:S = 137 J/molK because I calculated it this way:

3(205J/molK) - 2(239J/molK)


2. In the reaction, 2NO(g) + O₂(g)  2NO₂(g)  :delta:G^o = -77.4 kJ and  :delta:S^o= -146.5 at 251 K and 1 atm.

This reaction is product(?) favored under standard condtions at 251 K.

The standard enthalpy change  :delta:H^o for the reaction of 1.55 moles of NO(g) at this temperature would be ___?

I calculated that  :delta:H = -96.68 kJ because I calculated that:

1.55 mol NO * ((-77.4kJ)/(2mol NO)) = -59.98 =  :delta:G^o

And since  :delta:G =  :delta:H -  :delta:S(T):

-59.98kJ =  :delta:H - (-.1465kJ)(251K)
Hence, :delta:H = -96.68 kJ

[plankk]

First is correct.

About second task: remember that enthropy is inversely proportional to the number of moles.

[nernst]

First is correct.

About second task: remember that enthropy is inversely proportional to the number of moles.

does that mean i just divide 1/1.55 mol?

[MrTeo]

-59.98kJ =  :delta:H - (-.1465kJ)(251K)
Hence, :delta:H = -96.68 kJ

First of all a of -146.5 kJ/K⋅mol sounds very strange to me...
Personally I think that it's -146.5 J/K⋅mol... anyway I would suggest you to find the enthalpy variation for 2 mol and then calculate the for 1.55 mol, but if you want you can also apply what plankk said, I just think doing this way is more straightforward...  But you can't use the absolute temperature of 251 K as they ask you (notice the 0 superscript) to calculate enthalpy at 298.15 K and 1 atm (normal conditions).

Moreover I don't really understand what did you mean with that question mark after product... if you hadn't clear what it meant consider simply that a reaction with an equilibrium moved to the right at about -22º C must be esothermic ()...

Here is the answer if you don't understand my explanation or if you simply want to check your answers: