AP Chemistry 7.4: Reaction Quotients and Le Châtelier's Principle

The reaction quotient Q is more than a simple direction predictor. It can also be used before setting up an ICE table, after a disturbance, or when determining how pressures and concentrations must change before equilibrium is restored.

• Khan Academy's current Unit 7 includes using Q to determine reaction direction and then calculating equilibrium partial pressures.

Again:

• Q < K → reaction proceeds right
• Q > K → reaction proceeds left
• Q = K → equilibrium
• This rule works because equilibrium is reached when:
• Q becomes K

Consider:

• H₂(g) + I₂(g) ⇌ 2HI(g)
• Suppose:
• Kc = 50
• Current concentrations:
• [H₂] = 0.50 M
• [I₂] = 0.50 M
• [HI] = 1.00 M
• Calculate Q:
• Q = [HI]² / ([H₂][I₂])
• Q = (1.00)² / [(0.50)(0.50)]
• Q = 4.0
• Since:
• 4.0 < 50
• the reaction proceeds to the right.
• Therefore:
• H₂ decreases
• I₂ decreases
• HI increases
• until Q becomes 50.

Q can respond immediately when concentrations or pressures change.

• Suppose:
• A ⇌ B
• with:
• Q = [B]/[A]
• If you suddenly add A, the denominator increases.
• Therefore Q decreases.
• Now:
• Q < K
• so the system shifts right, consuming some A and producing B.
• This mathematical explanation is often stronger than simply saying:
• “the equilibrium shifts to use up added reactant.”

For:

• A ⇌ B
• removing B decreases the numerator.
• Therefore:
• Q decreases.

If Q becomes less than K:

• reaction shifts right
• to produce more B.

Adding B increases the numerator.

• Q increases.
• If:
• Q > K
• the reaction shifts left.

For gases, a change in volume changes partial pressures and therefore may change Q.

• Suppose:
• N₂(g) + 3H₂(g) ⇌ 2NH₃(g)
• There are:
• 4 moles of gaseous species on the left
• and:
• 2 moles of gaseous species on the right.
• If the container volume suddenly decreases at constant temperature, all gas partial pressures initially increase.
• Because Q depends on different powers of those pressures, the change can cause Q to move away from K.
• The system then shifts until Q returns to K.
• In this reaction, compression favors the side with fewer gas particles:
• products

Le Châtelier's principle is useful, but Q provides a mathematical reason for the shift.

• Instead of only memorizing:
• “adding reactant shifts right,”
• you can think:
• adding reactant changes Q → compare new Q with K → predict direction.
• This becomes especially useful in complicated problems.

Calculate Q using the equilibrium-expression form, including coefficients as exponents.

Do not change K just because Q changed.

At constant temperature, a disturbance changes Q first. The reaction then shifts until:

Q = K again

A disturbance usually causes:

• Q ≠ K
• Then the reaction moves until:
• Q = K
• again.

Le Châtelier's principle helps predict how an equilibrium system responds when it is disturbed.

• If a system at equilibrium experiences a change in concentration, pressure, volume, or temperature, the system responds by shifting in the direction that partially counteracts that disturbance.
• Khan Academy's current lesson covers changes in concentration, volume, temperature, and pressure-related calculations.

Consider:

• A + B ⇌ C + D

If A is added, the system has extra reactant.

• The reaction generally shifts:
• right
• to consume some of the added A.
• Therefore:
• products increase
• while some reactants are consumed.

Removing A makes the system shift:

• left
• to produce more reactants.

Adding C generally shifts equilibrium:

• left
• to consume some product.

Removing C generally shifts:

• right
• to produce more product.

Suppose more reactant is added.

• Immediately after addition, its concentration jumps upward.
• The reaction then shifts right and consumes some of the added reactant.
• However, its final concentration does not necessarily return exactly to the original value.
• The system partially counteracts the disturbance.

Pressure/volume changes mainly matter when gases are present.

For gases:

• decreasing volume → pressure increases
• increasing volume → pressure decreases
• When volume decreases, equilibrium tends to shift toward the side with fewer moles of gas.
• When volume increases, equilibrium tends to shift toward the side with more moles of gas.

N₂(g) + 3H₂(g) ⇌ 2NH₃(g)

• Left:
• 1 + 3 = 4 mol gas
• Right:
• 2 mol gas
• Decrease volume:
• system favors fewer gas particles.
• Shift:
• right
• Increase volume:
• system favors more gas particles.
• Shift:
• left

Consider:

• H₂(g) + I₂(g) ⇌ 2HI(g)
• Left side:
• 2 total moles gas
• Right side:
• 2 total moles gas
• A volume change affects both sides in a way that does not favor one direction based simply on gas-particle count.
• So there is no equilibrium shift from that volume change under the idealized AP treatment.

Consider:

• CaCO₃(s) ⇌ CaO(s) + CO₂(g)

Only gaseous species count when considering compression/expansion effects.

Do not count the solid coefficients as gas particles.

This situation requires careful wording.

• If an inert gas is added to a rigid container at constant volume, the total pressure rises, but the partial pressures of the reacting gases do not change.
• Therefore Q does not change, so there is generally:
• no equilibrium shift
• This is a common AP trap.
• If conditions involve changing volume to maintain constant total pressure, the result can differ because the reacting gas partial pressures change.

Always pay attention to what stays constant.

Temperature is different from concentration, pressure, or volume because changing temperature can actually change the value of K.

• A useful method is to treat heat as part of the reaction.

For an exothermic reaction:

• Reactants ⇌ Products + heat

Increasing temperature is like adding product.

• Therefore equilibrium shifts:
• left
• Cooling removes heat, so equilibrium shifts:
• right

For an endothermic reaction:

• Reactants + heat ⇌ Products
• Heating adds a reactant-like term.
• Therefore equilibrium shifts:
• right
• Cooling shifts:
• left

This relationship is essential.

For an endothermic forward reaction:

• increasing temperature favors products.
• Therefore:
• K increases

For an exothermic forward reaction:

• increasing temperature favors reactants.
• Therefore:
• K decreases

Temperature is the major Le Châtelier variable here that changes K.

A catalyst does not shift equilibrium.

A catalyst lowers activation energy for both forward and reverse reactions.

• Therefore both directions become faster.
• The system reaches equilibrium more quickly, but:
• equilibrium concentrations do not change;

K does not change;

• the favored side does not change.
• This connects directly to Unit 5.

Consider:

• CaCO₃(s) ⇌ CaO(s) + CO₂(g)
• If enough pure CaCO₃ is already present, adding more CaCO₃ does not change the equilibrium expression because pure solids are omitted.
• Therefore it does not shift the equilibrium by changing Q.
• This is another common AP trap.

AP questions may show concentration vs. time.

• Suppose product is suddenly added.
• The product concentration shows an immediate vertical jump.
• Then the system shifts left, so:
• product concentration decreases somewhat
• and:
• reactant concentration increases gradually.
• Eventually new constant concentrations are reached.
• A sudden vertical concentration change usually represents something directly added, removed, diluted, or compressed—not the gradual chemical response itself.

Do not say the system “tries to undo the change completely.” It only shifts toward a new equilibrium.

Do not apply pressure rules to solids and liquids.

Do not automatically say adding an inert gas shifts equilibrium.

Do not say a catalyst changes K or equilibrium composition.

Most importantly:

only temperature changes K for a given reaction in this context.

For concentration:

• add something → shift away from it
• remove something → shift toward it

For gases:

• smaller volume → fewer gas moles favored
• larger volume → more gas moles favored
• For temperature:
• treat heat as reactant or product

These rules are best understood using Q, rather than memorized blindly.