I am having a hard time with this problem. I was wondering where I went wrong. The problem says The equilibrium constant, Kc, for the reaction of H2 and N2 is 1.7×10^2 at 225 ∘C.
3H2(g)+N2(g)←→2NH3(g) If the system at equilibrium contains 0.16M H2 and 0.026M N2, what is the concentration of NH3? I know the equation we use is Kc. And I set it up where Kc is times the reactants. I got .018 and the book says that it is wrong.
I did this earlier but I can't find what I posted. Perhaps I punched the wrong button. I didn't get 0.018 either.
Kc = (NH3)^2/(N2)(H2)^3
0.017 = x^2/(0.026)(0.16)^3
and solve for x.
0.14M
To solve this problem, we can use the equilibrium constant expression and the given concentrations of H2 and N2 to find the concentration of NH3.
First, let's write the balanced equation for the reaction:
3H2(g) + N2(g) ⇌ 2NH3(g)
The equilibrium constant expression, Kc, for this reaction is:
Kc = [NH3]^2 / ([H2]^3 * [N2])
Given:
[H2] = 0.16 M
[N2] = 0.026 M
Kc = 1.7 × 10^2
Substituting these values into the equilibrium constant expression:
1.7 × 10^2 = [NH3]^2 / (0.16^3 * 0.026)
Now, let's solve for [NH3].
[NH3]^2 = 1.7 × 10^2 * 0.16^3 * 0.026
[NH3]^2 = 1.7 × 0.016^3 * 0.026
[NH3]^2 = 1.7 × (1.6 × 10^-2)^3 * 2.6 × 10^-2
[NH3]^2 = 1.7 × 4.096 × 10^-6 × 2.6 × 10^-2
[NH3]^2 ≈ 1.7 × 1.06336 × 10^-7
[NH3]^2 ≈ 1.80312 × 10^-7
Taking the square root of both sides:
[NH3] ≈ √(1.80312 × 10^-7)
[NH3] ≈ 1.342 × 10^-4 M
Therefore, the concentration of NH3 in the equilibrium system is approximately 1.342 × 10^-4 M.
I'm not sure where you went wrong in your calculation, but it seems that your answer of 0.018 M differs significantly from the correct answer. Please check your calculations and make sure you used the correct values and equations when solving the problem.