Consider the reaction: 3A(g) + B(s) ↔ 5C(s) + 2D(g). In a 5.00 L container, the reactants and products are at equilibrium. There are 0.41 mol A, 0.58 mol B, 1.11 mol C, and 0.46 mol D. What is the equilibrium constant?
0.41 mol A, 0.58 mol B, 1.11 mol C, and 0.46 mol D.
moles/L = M = 0.41/5.00 L = ? for A.
For B M = 0.58/5.00 = ?
For C M = 1.11/5.00 = ?
For D M = 0.46/5.00 = ?
Substitute the M for A, B, C, D into the Keq expression and solve. Post your work if you get stuck.
Keq = (C)^5 * (D)^2/(A)^3 * B
thank you very much this was a big help
To solve this problem, we need to understand the equilibrium constant expression for this reaction, which is given by:
Kc = ([C]^5[D]^2) / ([A]^3[B])
Where [A], [B], [C], and [D] represent the molar concentrations of A, B, C, and D, respectively.
In the given problem, we have the following molar concentrations:
[A] = 0.41 mol / 5.00 L = 0.082 M
[B] = 0.58 mol / 5.00 L = 0.116 M
[C] = 1.11 mol / 5.00 L = 0.222 M
[D] = 0.46 mol / 5.00 L = 0.092 M
Now we can substitute these values into the equilibrium constant expression to calculate the equilibrium constant (Kc):
Kc = (0.222^5 * 0.092^2) / (0.082^3 * 0.116)
Kc = 0.0721 / 0.0913
Kc ≈ 0.788
Therefore, the equilibrium constant for this reaction is approximately 0.788.