These are five different changes at equilibrium:
1) Double the concentrations of both products and then double the container volume
"No shift"
2) Double the container volume add more A
"Rightward shift"
3) Double the concentration of B and halve the concentration of C
"No shift"
4) Double the concentrations of both products
"Leftward shift"
5) Double the concentrations of both products and then quadruple the container volume
"No shift"
Explanation:
0) Equilibrium reaction
A(s) ⇌ B(g)+C(g)
In an equlibrium reaction the equilibrium constant is calculated from the species in gas or aqueous state. The concentration of the solid substances remains basically constant, so their concentrations are included in the equilibrium constant.
Hence, the equilibrium equation for this equation is given by the product of the concentrations of the products B and C, each raised to the power 1, because that is the stoichiometric coefficient of each one in the chemical equation.
Kc = [B] [C]
Following Le Chatelier principle, when a disturbance is produced in a chemical reaction at equilibrium such disturbance will be counteracted by a change that minimizes its effect trying to restore the equilibrium.
That will let us analyze the given changes.
1) Double the concentrations of both products and then double the container volume
Since the equilibrium is proportional to the concentration of both products, see what the given changes cause.
The concentration of each species is proportional to the number of moles and inversely related to the volume. If you first double the concentration (without changing the volume) means that your are doubling the amount of moles, if then you doubles the volume you are restoring the original concentrations, and there is not a net change in the concentrations.
Hence, since the concentrations remain the same the equilibrium is not affected: no shift.
2) Double the container volume add more A.
You need to assume that adding more A, which is a solid compound, does not change the volume for the reaction. A normal assumption since the gas substances occupies a large volume compared with the solid substances.
As the concentration is inversely related to the volume, doubling the container volume will cut in half the concentrations of the gas products, B and C.
Since, the equilibrium is directly proportional to those concentrations, reducing the concentrations of both products will shift the equilibrium to the right, to produce more products, seeking to increase their concentrations and restore the equilibrium.
Conclusion: rightward shift.
3) Double the concentration of B and halve the concentration of C:
Call [B₁] the original concentration of B at equilibrium. When you double the concentration you get [B₂] = 2 [B].
Call [C₁] the original concentration of C at equilibrium. When you halve its concentration you get [C₂] = [C₁] / 2
Then, when you make the new product you get [B₂] [C₂] = 2 [B₁] [C₁] / 2 = [B₁] [C₁]
So, the product (the equilibrium) has not been changed and there is no shift.
4) Double the concentrations of both products
Now, both product concentrations have been increased, which is the most simple case to analyze, since you know that increasing the concentrations of one side will require a shift to the other side.
This is, to restore the equilibrium, more B and C must react to produce more A. Thus, the reverse reaction will be favored, i.e. the the reaction shall shift to the left.
5) Double the concentrations of both products and then quadruple the container volume
Doubling the concentration of both products means that the product of both concentrations wil be quadrupled (2[B] × 2[C] = 4 {B] [C] )
Since concentrations and volume are inversely related, the effect of quadrupling the volume will balance the effect of doubling both concentrations, and the effect is cancelled, no producing a net unbalance at the equilibrium, so no shift is produced.
