Consider the reaction below:

4NH3(g) + 3O2(g) -> 2N2 + 6H20, K=10^80 @ certain temp

initially, all the reactants and products have concentrations equal to 12M. At equilibrium, what is the approximate concentration of oxygen?

k=10^80 is a big number, so this is a forward reaction (very close to completion). So figure the limiting reactant: Looks like NH3 goes first, so when the 12M*Volume is used up, one will have used 3/4*12*volume of O2, which is 9 liters, leaving 3 liters, so 3M is the final oxygen concentration

Oh ok I understand now. Thanks!

To determine the approximate concentration of oxygen at equilibrium, we need to use the given equilibrium constant (K) and the initial concentrations of the reactants and products.

Let's denote the initial concentration of O2 as [O2]₀, and the change in concentration at equilibrium as "x" (since we don't know the exact concentration yet). The reaction stoichiometry tells us that the change in concentration for O2 is -3x (since the coefficient of O2 in the balanced equation is 3).

At equilibrium, the concentrations of the reactants and products can be expressed in terms of the initial concentrations and the changes:

[O2] = [O2]₀ - 3x

The equilibrium constant (K) for the given reaction is 10^80. The equilibrium expression for the reaction is:

K = ([N2]² * [H2O]⁶) / ([NH3]⁴ * [O2]³)

Since the initial concentration of all the species is 12M, we can substitute these values into the equilibrium expression:

10^80 = ([N2]² * [H2O]⁶) / (12⁴ * [O2]³)

Now, we can solve this equation to find the concentration of oxygen (x) at equilibrium:

10^80 = (1 * 1) / (12⁴ * (12 - 3x)³)

Simplifying further:

10^80 = 1 / (12⁴ * (12 - 3x)³)

Multiplying both sides by (12⁴ * (12 - 3x)³):

12⁴ * (12 - 3x)³ = 1 / 10^80

Taking the cube root of both sides:

12 - 3x = (1 / 10^80)^(1/3)

Simplifying the right side:

12 - 3x = 10^(-80/3)

Now, by rearranging the equation, we can solve for x:

3x = 12 - 10^(-80/3)

Dividing by 3:

x = (12 - 10^(-80/3)) / 3

Now, we can substitute this value of x back into the equation for [O2]:

[O2] = [O2]₀ - 3x

Substituting the given initial concentration ([O2]₀ = 12M), and solving:

[O2] = 12 - 3 * [(12 - 10^(-80/3)) / 3]

This will give us the approximate concentration of oxygen at equilibrium.