Consider the following equilibrium at a certain temperature: A(g) yields 2B(g), K=0.280. If initially a flasks contained equal concentrations of A (g) and B (g), 0.200 M each, what will their concentrations be when equilibrium is established?
2007-03-28 17:38:00 · 1 answers · asked by MARSHAL 2 in Science & Mathematics ➔ Chemistry
Suppose that x mol/L of A reacts to form B. Then the final concentration of A is 0.200 - x and the concentration of B is 0.200 + 2x.
K = 0.280 = [B]^2 / [A] = (0.200 + 2x)^2 / (0.200 - x)
So (0.200 + 2x)^2 = 0.280 (0.200 - x)
<=> 0.0400 + 0.800x + 4x^2 = 0.0560 - 0.280x
<=> 4x^2 + 1.080x - 0.0160 = 0
<=> x^2 + 0.270 x - 0.00400 = 0
<=> x = (-0.270 ± √(0.270^2 - 4.1.(-0.00400)))/2
<=> x = -0.2841 or 0.0141
The first figure gives a negative final concentration for B so can be discarded. Hence the concentrations are
[A] = 0.186 M and [B] = 0.228 M.
2007-03-28 17:55:03 · answer #1 · answered by Scarlet Manuka 7 · 0⤊ 0⤋