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?
i set up an ICE table then got:
10^80 = [(12+2x)^2 (12+6x)^6]/[(12-4x)^4 (12-3x)^3]
But then it looks waaay to complicated. Am I doing it wrong?
Please help. Thanks
see other post.
To determine the approximate concentration of oxygen at equilibrium, we need to use the concept of the equilibrium constant (K) and the given initial concentrations. The equilibrium constant expression for the given reaction is:
K = [N2]^2 * [H2O]^6 / [NH3]^4 * [O2]^3
The given value of K is 10^80, indicating a very large equilibrium constant. This implies that the reaction strongly favors the products over the reactants at equilibrium.
Initially, all reactants and products have concentrations of 12 M. Let's assume the concentration of oxygen at equilibrium is "x" M.
Substituting the initial and equilibrium concentrations into the equilibrium constant expression, we have:
10^80 = ([N2]^2 * [H2O]^6) / ([NH3]^4 * x^3)
Since all the concentrations are initially equal to 12 M, we can further simplify the equation:
10^80 = (12^2 * 12^6) / (12^4 * x^3)
10^80 = 12^8 / (x^3)
x^3 = 12^8 / 10^80
x^3 = (12/10)^8
x^3 = 1.792
x ≈ 1.12
Therefore, the approximate concentration of oxygen at equilibrium is 1.12 M.