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?
0M
sorry actually 12M
To determine the approximate concentration of ammonia (NH3) at equilibrium, we can use the equilibrium constant (K) and the stoichiometric coefficients of the balanced equation.
The given balanced equation is:
4NH3(g) + 3O2(g) -> 2N2(g) + 6H2O(g)
The equilibrium constant expression for this reaction is:
K = [N2]^2 [H2O]^6 / [NH3]^4 [O2]^3
Given that K has a value of 10^80 at a certain temperature, and that the initial concentrations of all the reactants and products are equal to 12M, we can use these values to solve for the equilibrium concentration of ammonia.
Let's assume the equilibrium concentration of ammonia is x M.
Substituting the initial concentrations into the equilibrium constant expression, we get:
10^80 = (12 - x)^4 (12)^3 / x^4 (12)^3
Simplifying the equation, we have:
10^80 = (12 - x)^4 / x^4
To approximate the concentration of ammonia at equilibrium, we can now solve this equation approximately using logarithms. Taking the logarithm of both sides:
log(10^80) = log[(12 - x)^4 / x^4]
80 log(10) = log[(12 - x)^4] - log[x^4]
80 = 4 log(12 - x) - 4 log(x)
20 = log(12 - x) - log(x)
log[(12 - x) / x] = 20
Taking the antilog of both sides:
(12 - x) / x = 10^20
Multiplying both sides by x:
12 - x = 10^20 x
Rearranging the equation:
10^20 x + x = 12
x(10^20 + 1) = 12
x ≈ 12 / (10^20 + 1)
So, the approximate concentration of ammonia at equilibrium is approximately 12 / (10^20 + 1) M.