# What is the equilibrium constant at 25 degrees Celsius.?

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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

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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.