Consider the reaction 4NH₃(g)+ 3O₂(g) →2N₂(g)+ 6H₂O(g),K = 10⁸⁰ at a certain temperature. Initially,all reactants and products have concentrations equal to 12 M.At equilibrium,the approximate concentration of oxygen is

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At equilibrium, the concentration of each species can be calculated using the equilibrium constant expression: K = [N₂]2[H₂O]6/[NH₃]4[O₂]3 Substituting the initial concentration of 12 M for each species, we can simplify the expression to: 10⁸⁰ = (2x)2(6x)6/ (12-4x)4(12-3x)3 where x is the change in concentration of oxygen at equilibrium. Solving for x using a numerical method such as the quadratic formula or a graphical method, we get x ≈ 3 M. Therefore, the approximate concentration of oxygen at equilibrium is 3 M.

Consider the Reaction 4NH3(g)+ 3O2(g) →2N2(g)+ 6H2O(g),K = 1080 at a Certain

Consider the Reaction 4NH₃(g)+ 3O₂(g) →2N₂(g)+ 6H₂O(g),K = 10⁸⁰ at a Certain