Question

at equilibrium, a 1.0 liter container was found to contain 0.20 moles of a, 0.20 moles of b, 0.40 moles of c and 0.40 mole of d. if 0.10 moles of a and 0.10 moles of b are added to this system, what will be the new equilibrium concentration of a? a(g) b(g) c(g) d(g) (a) 0.37 mol/l (b) 0.47 mol/l (c) 0.87 mol/l (d) 0.23 mol/l (e) 0.15 mol/l

Answer 1

We can set up the reaction quotient Qc using the initial concentrations of A, B, C, and D:

Qc = [C]^c[D]^d/[A]^a[B]^b = (0.40)^2/(0.20)^2 = 2.00

Since Qc = Kc (at equilibrium), we can use this to find the equilibrium concentration of A after adding 0.10 moles of A and 0.10 moles of B:

Kc = [C]^c[D]^d/[A]^a[B]^b = 2.00

0.20 + 0.10 = 0.30 moles of A
0.20 + 0.10 = 0.30 moles of B
0.40 moles of C (unchanged)
0.40 moles of D (unchanged)
Total moles = 1.40 moles

Substitute the new concentrations into the equilibrium expression and solve for [A]:

Kc = [C]^c[D]^d/[A]^a[B]^b
2.00 = (0.40)^2/[A]^2(0.30)
2.00 = 0.44/[A]^2
[A]^2 = 0.44/2.00
[A] = sqrt(0.22)
[A] = 0.47 mol/L

Therefore, the new equilibrium concentration of A is 0.47 mol/L (option b).