Chemical Equilibrium and Le Chatelier's Principle
The equilibrium condition
A reversible reaction reaches equilibrium when the forward and reverse rates are equal and concentrations stop changing, although both reactions continue. For aA + bB <=> cC + dD, Kc = [C]^c[D]^d / ([A]^a[B]^b), with pure solids and pure liquids omitted because their concentration is constant. K depends only on temperature.
Q and Le Chatelier
Q uses the same expression with current concentrations. If Q < K the reaction proceeds forward, if Q > K it goes reverse, and Q = K means equilibrium. Le Chatelier: the system opposes the change.
- Adding a reactant or removing a product shifts the equilibrium forward.
- Compressing a gas mixture shifts toward the side with fewer gas moles; only gases are counted.
- Heating shifts in the endothermic direction and changes K itself: K rises for endothermic reactions and falls for exothermic ones.
- A catalyst speeds both directions equally - no shift, no change in K, no change in yield.
- An inert gas added at constant volume has no effect; at constant pressure it dilutes the mixture and shifts toward more gas moles.
Worked example
H2(g) + I2(g) <=> 2HI(g). 1.00 mol of each reactant in a 1.00 L vessel; at equilibrium [HI] = 1.50 M. HI increased by 1.50 M, so by the 1:1:2 ratio each reactant fell by 0.75 M: [H2] = [I2] = 0.25 M. Kc = (1.50)^2/(0.25 x 0.25) = 2.25/0.0625 = 36. Because the number of gas moles is unchanged, pressure changes do not shift this equilibrium.
Traps
Putting solids or water into the K expression. Confusing a shift in position with a change in K - only temperature changes K. Using moles instead of molarity in Kc. Saying a catalyst increases yield.