Use data from Appendix IIB in the textbook to calculate the equilibrium constants at 25 degrees Celsius for each of the following reactions. I have provided the data below.
N2(g) + O2(g) >< 2NO(g)
delta G for N2 – 0
delta H for N2 – 0
delta S for N2 – 191.6
delta G for O2 – 0
delta H for O2 – 0
S for O2 – 205.2
delta G for 2NO – 87.6
delta H for 2NO – 91.3
S for 2NO – 210.8
Express in Kelvin using 2 sig figs. I don’t even know which pieces of info to use. PLEASE HELP.
4 Answers

Use the equation G = RTln(k). Substitute the G value you have provided along with 8.314 J/mol*k for R and 298 K for T, and you end up with this:
G = 87.6 (2) = 175.2 (1000) = 175200 J /mol
175200 / 8.314 / 298 = 70.7
70.7 = ln (K)
K = 1.95 * 10^31

In a chemical equilibrium, the consumption and formation of substances has reached a balanced condition. The quantities of reactants and products have achieved a constant ratio, but they are almost never equal. There may be much more product or much more reactant.

its at equilibrium so we know delta g is 0 so you actually don t need to find delta g rxn
Source(s): My AP Chemistry teacher 
this above solution is wrong do not use it. must find deltaG of rxn before using equation.