7.3 Reaction Quotient and Equilibrium Constant

The reaction quotient $Q$ tells you the ratio of products to reactants at any moment, using the same math form as the equilibrium constant $K$. Compare $Q$ to $K$ to predict which way a reaction will shift: if $Q<K$, it moves toward products; if $Q>K$, it moves toward reactants; and if $Q=K$, it is at equilibrium. For AP Chemistry, leave pure solids and liquids out of $Q$ and $K$ expressions.

Why This Matters for the AP Chemistry Exam

Q and K are central tools for Unit 7, which carries a meaningful share of the exam. Writing a correct equilibrium expression and comparing Q to K shows up in both multiple-choice reasoning and free-response questions where you explain the direction a reaction proceeds. You will also use this skill to represent chemical systems with graphs, particle diagrams, and balanced equations, and to support claims about reaction behavior with data.

A few exam realities worth knowing now:

• You will not have to convert between Kc and Kp, but you do need to notice which one a question uses.
• Equilibrium problems where a dissolved species is in balance with that same species in the gas phase are not assessed.

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Key Takeaways

• Q measures the product-to-reactant ratio at any instant; K is that same ratio specifically at equilibrium.
• Use Qc with molar concentrations and Qp with partial pressures for gas-phase reactions.
• For a A + b B ⇌ c C + d D, products go on top and reactants on bottom, each raised to its coefficient.
• Pure solids and pure liquids never appear in Q or K expressions.
• Q < K shifts toward products, Q > K shifts toward reactants, Q = K means no net change.

What Is the Reaction Quotient?

Reactions are not always sitting at equilibrium. They can be on their way there, already there, or pushed past it. The reaction quotient Q lets you describe exactly where a reaction stands at a chosen moment by measuring the relative amounts of reactants and products at that specific point in time.

Q comes in two forms:

• Qc: uses molar concentrations (for reactions in solution)
• Qp: uses partial pressures (for gas-phase reactions)

For the general reaction a A + b B ⇌ c C + d D, the law of mass action gives:

Concentration-based (Qc and Kc):

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

Pressure-based (Qp and Kp) for gas-phase reactions:

$K_p = \frac{(P_C)^c(P_D)^d}{(P_A)^a(P_B)^b}$

where $P_C$, $P_D$, $P_A$, and $P_B$ are the partial pressures of the gases.

Pure Solids and Pure Liquids Are Left Out

Pure solids and pure liquids do not appear in Q or K expressions. Their concentrations stay effectively constant no matter how much of the substance is present, so they do not affect the ratio. A large chunk and a small chunk of the same solid have the same "concentration."

For the reaction CaCO₃(s) ⇌ CaO(s) + CO₂(g):

• The expression is simply $K_c = [\text{CO}_2]$ or $K_p = P_{\text{CO}_2}$
• The solids CaCO₃ and CaO do not show up at all.

Q vs. K: Same Math, Different Moment

Q and K use identical expressions, but they describe different moments. Q uses concentrations or partial pressures at a chosen time t. K uses equilibrium concentrations only. So Q is the product-to-reactant ratio at any point, while K is that ratio specifically at equilibrium.

That difference is what makes Q useful. By comparing Q to K, you can predict whether a reaction will move forward (make more products), move in reverse (make more reactants), or stay put because it is already at equilibrium.

This idea returns throughout Unit 7. When you reach Le Chatelier's principle later in the unit, most shifts (temperature being the main exception) come down to a change in Q that pushes the system back toward K.

Equilibrium vs. Non-Equilibrium Concentrations

Concentrations that are not at equilibrium will readjust. The reaction makes more product or more reactant until the system returns to equilibrium. Think of a seesaw: if one side has too much weight, the system shifts to rebalance. That same picture explains how Q moves back toward K by shifting right (toward products) or left (toward reactants).

Comparing Q and K to Predict Direction

The main job of Q and K together is telling you which direction a reaction will move. Three cases cover everything:

1. Q > K: there are relatively more products than the equilibrium ratio allows, so the reaction shifts left (toward reactants).
2. Q = K: the ratios match, so the system is at equilibrium with no net change.
3. Q < K: there are relatively more reactants than the equilibrium ratio allows, so the reaction shifts right (toward products).

Why This Works Mathematically

For A + B ⇌ C + D, $Q_c = \frac{[C]_t[D]_t}{[A]_t[B]_t}$.

When Q > K, the numerator $[C]_t[D]_t$ is too high compared to its equilibrium value, so it needs to decrease while $[A]_t[B]_t$ increases. The reaction uses excess C and D to form A and B until the system reaches equilibrium.

When Q < K, the numerator is too low and the denominator is too high. Excess reactants react to form products, raising Q back up to K. This matches the seesaw picture from before.

How to Use This on the AP Chemistry Exam

Problem Solving

Worked example with concentrations:

Consider the reaction 2 NOBr ⇌ 2 NO + Br₂.

If Kc = 0.0142 and the initial concentrations are 1.0 M NOBr, 0.2 M NO, and 0.8 M Br₂, which way will the reaction proceed?

Calculate Q and compare it to Kc:

$Q = \frac{[\text{NO}]^2[\text{Br}_2]}{[\text{NOBr}]^2} = \frac{(0.2)^2(0.8)}{(1.0)^2} = 0.032$

Since 0.032 > 0.0142, Q > K, so the reaction proceeds to the left to make more reactants and reach equilibrium. The system started with relatively too much product, so it shifts toward reactants.

Working With Gases (Qp and Kp)

For gas-phase reactions, use partial pressures. Take ammonia synthesis:

N₂(g) + 3 H₂(g) ⇌ 2 NH₃(g)

$K_p = \frac{(P_{\text{NH}_3})^2}{(P_{\text{N}_2})(P_{\text{H}_2})^3}$

where $P_{\text{NH}_3}$, $P_{\text{N}_2}$, and $P_{\text{H}_2}$ are the partial pressures of each gas. The same comparison rules apply:

• Qp < Kp: shift toward products
• Qp = Kp: at equilibrium
• Qp > Kp: shift toward reactants

Common Trap

You will not be asked to convert between Kc and Kp, but you do need to read which one the question uses and write the matching expression. Mixing up concentrations and partial pressures, or forgetting to leave out solids and liquids, are the usual point-losers here.

Common Misconceptions

• Q and K are the same thing. They share the same formula but not the same meaning. Q is the ratio at any moment; K is the ratio only at equilibrium.
• Solids and liquids belong in the expression. Pure solids and pure liquids are always left out of Q and K. Only gases and aqueous species appear.
• A large Q means more products in an absolute sense. Q is a ratio, not a raw amount. What matters for direction is how Q compares to K, not its size alone.
• Q > K means the reaction is "done." It just means there is relatively too much product, so the reaction shifts toward reactants until Q equals K.
• Coefficients are optional in the expression. Each concentration or pressure must be raised to its stoichiometric coefficient, or the value of Q will be wrong.
• Initial and equilibrium concentrations are interchangeable. Use the concentrations at the moment you care about for Q, and equilibrium concentrations only for K.

Related AP Chemistry Guides

• 7.1 Introduction to Equilibrium
• Unit 7 Overview: Equilibrium
• 7.2 Direction of Reversible Reactions
• 7.4 Calculating the Equilibrium Constant
• 7.11 Introduction to Solubility Equilibria
• 7.5 Magnitude of the Equilibrium Constant