What is the equilibrium constant at 25 degrees Celsius.?

NetherCraft 0

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

Also Check This  H&R Block emerald card direct deposit time?

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.


Leave a Reply

Your email address will not be published. Required fields are marked *