Consider the reaction system:
Br₂ (g) + Cl₂ (g) ---> 2 BrCl (g)
Initially, a 2.00 L reaction vessel contains 0.15 mol of each gas. Kc = 7.0 at 400 K.
Since we have 0.15 mol of each gas and a 2.00 L reaction vessel, the initial concentration of each gas is:
molarity = moles of solute/volume of container in L
molarity = 0.15 mol/2.00 L
molarity = 0.075 M
For this reaction the expresion of the equilibrium constant is:
Br₂ (g) + Cl₂ (g) ---> 2 BrCl (g) Kc = [BrCl]²/[Br₂]*[Cl₂]
We are adding a product, so the equilibrium will shift to the left. The system will try to compensate the adittion of the product. To explain with evidence we have to find the equilibrium concentrations of our compounds. To do so we have to work with the ICE table.
We are adding at the initial point 0.05 mol of BrCl, so we have:
number of moles of BrCl = 0.15 mol + 0.05 mol = 0.20 moles
Molarity of BrCl = 0.20 moles/2.00L
Molarity of BrCl = 0.10 M
Br₂ (g) + Cl₂ (g) ---> 2 BrCl (g)
I 0.075 M 0.075 M 0.10 M
C +x +x - 2 x
E 0.075 M + x 0.075 M + x 0.10 M - 2 x
Replacing these values in the Kc expression:
Kc = [BrCl]²/[Br₂]*[Cl₂]
7.0 = (0.10 - 2x)² /[(0.075 + x)*(0.075 + x)]
Now we have to work with that equation and solve it for x.
7.0 = (0.10 - 2x)² /(0.075 + x)²
7.0 * (0.075 + x)² = (0.10 - 2x)²
7.0 * (0.075² + 2 * 0.075 * x + x²) = 0.10² - 2 * 0.10 * 2x + (2x)²
7.0 * (0.005625 + 0.15 x + x² = 0.010 - 0.40 x + 4 x²
0.039375 + 1.05 x + 7 x² = 0.010 - 0.40 x + 4 x²
3 x² + 1.45 x + 0.029375 = 0
We got this quadratic equation and we have to find its roots.
x1 = -0.5028 x2 = 0.01947
We will select the positive root. Now we can find the equilibrium concentrations.
[Br₂]eq = 0.075 M + x = 0.075 M + 0.0195 M
[Br₂]eq = 0.0945 M
[Cl₂]eq = 0.075 M + x = 0.075 M + 0.0195 M
[Cl₂]eq = 0.0945 M
[BrCl]eq = 0.10 M - 2 x = 0.10 M - 2 * 0.0195 M
[BrCl]eq = 0.061 M
As we see the concentration of the reactants increased and the concentration of the product decreased.