The following reaction has an equilibrium constant (Keq) of 160.

Question

2 NO2(g) <--> 2 NO(g) + O2(g)
What will be the reaction quotient (Q) and in which direction will the reaction proceed if the partial pressure of NO2 is 5.0*10^-4 atm, NO is 0.080 atm, and O2 is 0.020 atm?

Work:

Keq= [NO]^2 [O2]/[NO2]

Keq=[0.080]^2 [.020]/ 5.0*10^-4

Keq=1.28*10^-4/5.0*10^-4

Keq=.256 which is less than 1 meaning that the reactants are favored moving the reaction towards the left.

I am not sure if .256 is also the reaction quotient or if it's a totally different process to find it.

Answer 1

Nevermind I figured it out