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 calculate the approximate concentration of ammonia at equilibrium, we need to use the equilibrium constant expression (K) and the stoichiometry of the balanced equation. Here's how you can do it step-by-step:
Step 1: Write the balanced equation for the reaction:
4NH3(g) + 3O2(g) -> 2N2(g) + 6H2O(g)
Step 2: Write the equilibrium constant expression (K):
K = [N2]^2[H2O]^6 / [NH3]^4[O2]^3
Step 3: Assign a variable to the equilibrium concentration of ammonia. Let's call it "x" in this case.
Step 4: Use the given information that the initial concentration of ammonia is 12M. Therefore, the equilibrium concentration of ammonia ([NH3]) is 12M - x.
Step 5: Substitute the equilibrium concentrations into the equilibrium constant expression:
10^80 = ([N2]^2[H2O]^6) / ([NH3]^4[O2]^3)
10^80 = ([N2]^2[H2O]^6) / ((12M - x)^4(12M)^3)
Step 6: Since all the reactants and products have initial concentrations equal to 12M, we can assume that x is small compared to 12M. This allows us to make an approximation: (12M - x) ≈ (12M).
Step 7: Substitute the approximation into the equilibrium constant expression:
10^80 = ([N2]^2[H2O]^6) / ((12M)^4(12M)^3)
10^80 = ([N2]^2[H2O]^6) / (12^7M^7)
Step 8: Rearrange the equation to solve for x:
x^4 = ([N2]^2[H2O]^6) / (10^80 * (12^7M^7))
Step 9: Since the question is asking for the approximate concentration of ammonia, we will assume x is very small compared to 12M. Therefore, we can approximate the equilibrium concentration of ammonia as 12M.
So, the approximate concentration of ammonia at equilibrium is approximately 12M.
To determine the approximate concentration of ammonia (NH3) at equilibrium, we need to use the equilibrium constant expression (K) and the initial concentrations of the reactants and products.
The balanced chemical equation for the reaction is:
4NH3(g) + 3O2(g) -> 2N2 + 6H20
The equilibrium constant expression (Kc) for this reaction is:
Kc = [N2]^2 * [H2O]^6 / [NH3]^4 * [O2]^3
Given that the equilibrium constant (K) is equal to 10^80 and the initial concentrations of all reactants and products are 12M, we can set up the equation as follows:
10^80 = ([N2]eq^2 * [H2O]eq^6) / ([NH3]eq^4 * [O2]eq^3)
Since the initial concentration of all species is 12M and assuming there is no change in volume during the reaction, the equilibrium concentrations ([N2]eq, [H2O]eq, [NH3]eq, and [O2]eq) will also be equal. Let's denote this common concentration as 'x.'
Now, we can rewrite the equation using the value of 'x':
10^80 = (x^2 * x^6) / (x^4 * x^3)
Simplifying the equation:
10^80 = (x^8) / (x^7)
10^80 = x
Then, we take the 80th root of both sides to determine the approximate concentration of ammonia at equilibrium:
x ≈ 10^(80/8)
x ≈ 10^10
Therefore, the approximate concentration of ammonia at equilibrium is approximately 10^10 M.