The apparent equilibrium constant for the reaction
A + B <==> 2C
is Kc= 4.11 at 298.2 K. Given that the initial concentrations of A, B, and C are .10M, .10M, and zero, respectively, find the equilibrium concentrations of A, B, and C.
I had this long, drawn out equation written up that I solved using the quadratic formula, but it didn't yield the correct answer so I'm assuming I went about it the wrong way. How would I solve this?
You solve it with the long drawn out equation, from this relationsip.
Kc= [2x]^2 /(.1-x)^2
To solve for the equilibrium concentrations of A, B, and C, you can use the concept of ICE (Initial, Change, Equilibrium) and the expression for the equilibrium constant.
1. Start by writing down the balanced chemical equation: A + B ⇌ 2C.
2. Assume the initial concentrations of A and B to be 0.10 M, and the initial concentration of C to be zero.
3. Use the ICE table to determine the changes in concentration at equilibrium. Let's assume that the change in concentration of A and B is "x" M, and the change in concentration of C is "2x" M.
Initial Change Equilibrium
A 0.10 -x 0.10 - x
B 0.10 -x 0.10 - x
C 0 +2x 2x
4. Write the expression for the equilibrium constant (Kc) in terms of the equilibrium concentrations:
Kc = ([C]^2) / ([A] * [B])
Given that Kc = 4.11, substitute the equilibrium concentrations into the expression:
4.11 = (2x)^2 / ((0.10 - x) * (0.10 - x))
5. Now, solve the equation for "x" by rearranging it algebraically:
4.11 * (0.10 - x) * (0.10 - x) = (2x)^2
Expand the equation and simplify:
4.11 * (0.01 - 0.20x + x^2) = 4x^2
0.0411 - 0.0822x + 4.11x^2 = 4x^2
0.0822x = 0.0411
x = 0.0411 / 0.0822
x ≈ 0.50 M
6. Substitute the value of "x" back into the ICE table to calculate the equilibrium concentrations:
[A] = 0.10 - x ≈ 0.10 - 0.50 ≈ 0.05 M
[B] = 0.10 - x ≈ 0.10 - 0.50 ≈ 0.05 M
[C] = 2x ≈ 2(0.50) ≈ 1.0 M
Therefore, the equilibrium concentrations of A, B, and C are approximately 0.05 M, 0.05 M, and 1.0 M, respectively.