Topic 7.10 Notes – Reaction Quotient and Le Châtelier’s Principle

1. What Q and K Represent

For a general reversible reaction:

$$\ce{aA + bB <=> cC + dD}$$

the equilibrium constant is

$$K = \frac{[C]^c [D]^d}{[A]^a [B]^b}$$

For gases, you might see $K_p$, which uses partial pressures instead of concentrations.

The reaction quotient, Q, has the exact same mathematical form.

The difference is timing:

• K uses equilibrium concentrations or pressures.
• Q uses concentrations or pressures at any moment.

That’s it. Same equation. Different data.

What Q tells you

• If $Q = K$ → the system is at equilibrium.
• If $Q < K$ → the reaction will shift forward (toward products).
• If $Q > K$ → the reaction will shift reverse (toward reactants).

Why? Because equilibrium is defined as the state where that ratio equals K. If it doesn’t match, the system changes concentrations until it does.

On tests, they often give you initial concentrations after a disturbance and ask for direction only. You calculate Q and compare. No ICE table needed if you’re only predicting direction.

2. Comparing Q and K to Predict Direction

Here’s the logic laid out clearly:

A quick example:

$$\ce{2SO2(g) + O2(g) <=> 2SO3(g)}$$

Suppose $K = 4.0$, and you calculate $Q = 0.8$.

Since $0.8 < 4.0$, the reaction shifts right. More $\ce{SO3}$ forms.

The reaction always moves in whatever direction makes Q move toward K. That single sentence explains Le Châtelier’s Principle mathematically.

3. How Concentration and Pressure Disturb Equilibrium

A disturbance changes Q, not K (as long as temperature stays constant). Once Q changes, the system shifts until Q = K again.

A. Changes in Concentration

Because Q is a ratio of products over reactants:

• Add reactant → denominator increases → Q decreases → shift right
• Add product → numerator increases → Q increases → shift left
• Remove reactant → denominator decreases → Q increases → shift left
• Remove product → numerator decreases → Q decreases → shift right

K does not change.

A common trap on multiple choice: adding a solid or pure liquid does nothing if it’s not in the expression. Only species in the equilibrium expression affect Q.

B. Changes in Pressure (Gases Only)

For gases, use $Q_p$.

If volume decreases, total pressure increases, and all partial pressures increase proportionally.

Whether Q increases or decreases depends on the total moles of gas on each side.

Rule to remember:

• Increase pressure → shift toward fewer moles of gas
• Decrease pressure → shift toward more moles of gas

Here’s why. Suppose pressure doubles. Each partial pressure doubles, but they’re raised to their coefficients. The side with the larger total exponent (more gas moles) changes Q more. The diagram below shows two common cases you’ll see on exams.

Pressure changes and moles of gas at equilibrium

On the left, increasing pressure shifts the system toward the side with fewer gas particles. On the right, both sides have the same total moles of gas, so there is no shift.

Important details:

• Only gases count.
• Solids and liquids are excluded.
• Equal moles of gas on both sides → no shift with pressure change.

Again, K stays constant.

4. Temperature Is the Exception

Temperature is different because it changes K itself.

K is temperature dependent. When temperature changes, the numerical value of K changes.

Think of heat as part of the reaction.

Exothermic (heat released)

Heat acts like a product.

• Increase temperature → K decreases → shift left
• Decrease temperature → K increases → shift right

Endothermic (heat absorbed)

Heat acts like a reactant.

• Increase temperature → K increases → shift right
• Decrease temperature → K decreases → shift left

What’s happening underneath: temperature changes the value of K, and the system shifts until Q equals the new K.

If you remember nothing else:

• Concentration or pressure change → Q changes
• Temperature change → K changes
• The system always adjusts until $Q = K$