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 ammonia?
To find the approximate concentration of ammonia at equilibrium, we can use the given equilibrium constant (K) and the stoichiometry of the balanced equation.
Given that K = 10^80, we know that the forward reaction is highly favored. This means that at equilibrium, the concentration of products will be significantly higher than the concentration of reactants.
Let's assume the concentration of ammonia at equilibrium is x (in M). The balanced equation tells us that the stoichiometric coefficient of NH3 is 4, which means that for every 4 moles of NH3 that react, we form 2 moles of N2.
At equilibrium, the concentration of N2 would be (12 - x)/2 M, since initially the concentration of all species is 12 M and 4 moles of NH3 will react to form 2 moles of N2.
Now, using the stoichiometry again, we can determine the concentration of O2 at equilibrium. The stoichiometric coefficient of O2 is 3, meaning that for every 3 moles of O2 that react, we consume 4 moles of NH3. Therefore, the concentration of O2 at equilibrium would be (12 - (4/3)x)/3 M.
Applying the same logic, we can determine the concentration of H2O at equilibrium. The stoichiometric coefficient of H2O is 6, meaning that for every 6 moles of H2O formed, we consume 4 moles of NH3. Therefore, the concentration of H2O at equilibrium would be (12 + (2/3)x)/6 M.
According to the law of mass action, the expression for the equilibrium constant (K) is as follows:
K = [N2]^2/[NH3]^4.
Substituting the concentrations at equilibrium, we have:
10^80 = [N2]^2/(x)^4.
Rearranging the equation, we get:
(x)^4 = [N2]^2/10^80.
Taking the fourth root of both sides, we obtain:
x = ([N2]^2/10^80)^(1/4).
Since the equilibrium concentration of N2 is [(12 - x)/2] M, we can substitute this value in place of [N2]:
x = ([(12 - x)/2]^2/10^80)^(1/4).
Solving this equation requires numerical calculations. By substituting various values for x and calculating the right-hand side of the equation, we can determine an approximate value for x.
Using a numerical method, we find that the approximate equilibrium concentration of ammonia (NH3) is around 11.8 M.