Reaction quotient Q — AP Chemistry Study Guide

For: AP Chemistry candidates sitting AP Chemistry.

Covers: Definition of reaction quotient $Q$ , calculation of $Q_{c}$ and $Q_{p}$ for homogeneous and heterogeneous equilibria, comparison of $Q$ to equilibrium constant $K$ , prediction of reaction direction, and calculation of $Q$ after system perturbation.

You should already know: Definition of the equilibrium constant K, how to balance chemical reaction equations, molarity and partial pressure for gases.

A note on the practice questions: All worked questions in the "Practice Questions" section below are original problems written by us in the AP Chemistry style for educational use. They are not reproductions of past College Board / Cambridge / IB papers and may differ in wording, numerical values, or context. Use them to practise the technique; cross-check with official mark schemes for grading conventions.

1. What Is Reaction quotient Q?

The reaction quotient ( $Q$ ) is a dimensionless quantity that describes the relative ratio of products to reactants (adjusted for stoichiometry) in a reaction system at any given point in time, regardless of whether the system has reached equilibrium. Unlike the equilibrium constant $K$ , which can only be calculated from equilibrium concentrations or partial pressures, $Q$ can be evaluated for any stage of a reaction before equilibrium is established.

Notation conventions match those for $K$ : $Q_{c}$ is used when calculating with molar concentrations of solutes and gases, while $Q_{p}$ is used exclusively for gas-phase systems with partial pressures of gases. The value of $Q$ always follows the stoichiometry of the balanced reaction equation, just like $K$ .

In the AP Chemistry Course and Exam Description (CED), content related to $Q$ falls within Unit 7 Equilibrium, which accounts for 7-9% of the overall AP exam score. $Q$ is tested in both multiple-choice (MCQ) and free-response (FRQ) sections, and it is regularly combined with Le Chatelier’s principle, ICE table calculations, and Gibbs free energy problems.

2. Calculating Q for Homogeneous and Heterogeneous Equilibria

Calculating $Q$ follows the exact same structural rules as calculating $K$ , with the only difference being that $Q$ uses non-equilibrium concentration/partial pressure values, while $K$ only uses equilibrium values. For a general balanced reaction: $a A + b B ⇌ c C + d D$ The concentration-based reaction quotient is: $Q_{c} = \frac{[ C ] ^{c} [ D ] ^{d}}{[ A ] ^{a} [ B ] ^{b}}$ For gas-phase reactions, the partial pressure-based reaction quotient is: $Q_{p} = \frac{( P _{C} ) ^{c} ( P _{D} ) ^{d}}{( P _{A} ) ^{a} ( P _{B} ) ^{b}}$ Just like with $K$ , pure solids, pure liquids, and solvent in dilute solutions are never included in the $Q$ expression, because their thermodynamic activity is always 1, so they do not affect the ratio. This rule holds for both homogeneous equilibria (all species in the same phase) and heterogeneous equilibria (species in multiple phases). Intuitively, $Q$ is just a snapshot of the current product-to-reactant ratio, adjusted for how many moles of each species appear in the reaction.

Worked Example

Problem: For the heterogeneous reaction $2 A g N O_{3} (s) ⇌ 2 A g (s) + 2 N O_{2} (g) + O_{2} (g)$ , a student measures partial pressures at an early point in the reaction: $P_{N O_{2}} = 0.40 a t m$ , $P_{O_{2}} = 0.20 a t m$ . Calculate $Q_{p}$ for this system.

First, eliminate pure solids from the expression: both $A g N O_{3}$ and $A g$ are pure solids, so they do not appear in $Q_{p}$ .
Write the $Q_{p}$ expression from the balanced reaction: $Q_{p} = (P_{N O_{2}})^{2} (P_{O_{2}})$
Substitute the given partial pressure values: $Q_{p} = (0.40)^{2} (0.20)$
Calculate: $(0.16) (0.20) = 0.032$
Final $Q_{p} = 0.032$

Exam tip: Always double-check that you omitted pure solids and pure liquids from your Q expression. AP exam MCQ distractors almost always include wrong options that incorrectly add solid terms to the ratio.

3. Comparing Q and K to Predict Reaction Direction

The core purpose of calculating $Q$ is to predict which direction a reaction will proceed to reach equilibrium, by comparing $Q$ to the equilibrium constant $K$ (which is fixed at a given temperature). The logic of the comparison is straightforward: $K$ is the equilibrium ratio of products to reactants, so the current ratio $Q$ tells us how far we are from equilibrium:

If $Q < K$ : The numerator (product terms) is too small relative to the denominator (reactant terms), so the reaction proceeds forward (shifts right) to make more products, increasing $Q$ until it equals $K$ .
If $Q = K$ : The system is already at equilibrium, so no net change occurs.
If $Q > K$ : The numerator (product terms) is too large relative to the denominator (reactant terms), so the reaction proceeds in reverse (shifts left) to consume products, decreasing $Q$ until it equals $K$ .

This comparison is the most concrete way to predict reaction direction, and AP exam graders require this reasoning for full credit on FRQ questions about equilibrium shifts.

Worked Example

Problem: For the reaction $2 H I (g) ⇌ H_{2} (g) + I_{2} (g)$ , $K_{c} = 0.016$ at 300°C. A reaction mixture has $[H I] = 0.60 M$ , $[H_{2}] = 0.040 M$ , $[I_{2}] = 0.040 M$ at 300°C. Predict the direction the reaction will proceed to reach equilibrium.

Write the $Q_{c}$ expression for the balanced reaction: $Q_{c} = \frac{[ H _{2} ] [ I _{2} ]}{[ H I ] ^{2}}$
Substitute the non-equilibrium concentrations: $Q_{c} = \frac{( 0.040 ) ( 0.040 )}{( 0.60 ) ^{2}} = \frac{0.0016}{0.36} \approx 0.0044$
Compare $Q_{c}$ to $K_{c}$ : $0.0044 < 0.016$ , so $Q < K$
Conclusion: The concentration of products is too low, so the reaction will proceed in the forward direction to make more $H_{2}$ and $I_{2}$ to reach equilibrium.

Exam tip: If you ever mix up the shift rule, reason it out from the ratio: $Q = [p r o d u c t s] / [r e a c t an t s]$ . If $Q < K$ , you need more products to get to K, so you go forward.

4. Calculating Q After a System Perturbation

A common AP exam question asks you to predict how an equilibrium system will shift after a disturbance (e.g., adding a reactant, changing volume, adding a product). To solve this, you calculate $Q$ immediately after the perturbation (before any shift has occurred), then compare to the original $K$ (K only changes if temperature changes). This method is more reliable than memorizing Le Chatelier’s general rules and is required for full credit on justification questions.

Key rules for this calculation:

Only change the concentration/partial pressure of the species directly affected by the perturbation (e.g., if you add HI, only change [HI] for the Q calculation).
All other species keep their original equilibrium concentrations, because the shift hasn’t happened yet.
K remains the same unless the problem states the temperature of the system changed.

Worked Example

Problem: The reaction $N_{2} (g) + 3 H_{2} (g) ⇌ 2 N H_{3} (g)$ is at equilibrium at 500°C with $K_{p} = 0.16$ . Equilibrium partial pressures are $P_{N_{2}} = 0.40 a t m$ , $P_{H_{2}} = 0.40 a t m$ , $P_{N H_{3}} = 0.020 a t m$ . Enough $N_{2}$ is added to increase $P_{N_{2}}$ to 0.80 atm immediately after addition. Calculate the new Q and predict the direction of shift.

Write the $Q_{p}$ expression: $Q_{p} = \frac{( P _{N H_{3}} ) ^{2}}{P _{N_{2}} ( P _{H_{2}} ) ^{3}}$
Update the affected partial pressure: only $P_{N_{2}}$ changes; $P_{H_{2}}$ and $P_{N H_{3}}$ stay at their original equilibrium values.
Substitute values: $Q_{p} = \frac{( 0.020 ) ^{2}}{( 0.80 ) ( 0.40 ) ^{3}} = \frac{0.0004}{( 0.80 ) ( 0.064 )} = \frac{0.0004}{0.0512} \approx 0.0078$
Compare Q to K: $Q_{p} = 0.0078 < K_{p} = 0.16$ , so the reaction will shift forward (right) to consume the added $N_{2}$ and reestablish equilibrium.

Exam tip: When volume is changed for a gas-phase reaction, scale all partial pressures by the same factor (pressure is inversely proportional to volume) before calculating Q.

5. Common Pitfalls (and how to avoid them)

Wrong move: Including the concentration of a pure solid or pure liquid in the Q expression, e.g., writing $Q = [C a O] [C O_{2}] / [C a C O_{3}]$ for calcium carbonate dissociation. Why: Students memorize the exclusion rule for K but forget it applies equally to Q. Correct move: Every time you write a Q expression, cross out any pure solids, pure liquids, or solvent (for dilute solutions) before plugging in values.
Wrong move: Reversing the direction of shift when comparing Q and K, e.g., stating that Q > K means shift right. Why: Students mix up which side of the ratio is which, or memorize the rule backwards. Correct move: Always reason it out: Q = [products]/[reactants]. If Q > K, [products] are too high, so you need to make more reactants (shift left) to get to K.
Wrong move: Changing the value of K when comparing Q after a concentration or volume perturbation. Why: Students confuse temperature changes (which change K) with other perturbations (which do not change K). Correct move: Only update K if the problem states the temperature of the system changed; for all other changes, K stays the same for the Q vs K comparison.
Wrong move: Calculating Q with new equilibrium concentrations instead of non-equilibrium concentrations immediately after a perturbation. Why: Students forget that Q measures the system before the shift occurs. Correct move: When asked for Q after a perturbation, only adjust the concentration/partial pressure of the species that was changed by the disturbance; all other species keep their original equilibrium values for the Q calculation.
Wrong move: Raising stoichiometric coefficients to the wrong power in the Q expression, e.g., writing $Q = [N O_{2}] / [N_{2} O_{4}]$ instead of $[N O_{2}]^{2} / [N_{2} O_{4}]$ . Why: Students rush and forget that each term is raised to the power of its stoichiometric coefficient, just like in K. Correct move: After writing the Q expression, check every term’s exponent against the balanced reaction before plugging in numbers.

6. Practice Questions (AP Chemistry Style)

Question 1 (Multiple Choice)

For the reaction $2 S O_{3} (g) ⇌ 2 S O_{2} (g) + O_{2} (g)$ , $K_{p} = 1.8 \times 1 0^{- 5}$ at 500 K. A reaction mixture contains $P_{S O_{3}} = 0.10 a t m$ , $P_{S O_{2}} = 0.050 a t m$ , $P_{O_{2}} = 0.0020 a t m$ . Which of the following correctly predicts the reaction direction and justifies it? A) The reaction will proceed forward, because $Q_{p} = 5.0 \times 1 0^{- 5} > K_{p}$ B) The reaction will proceed forward, because $Q_{p} = 1.0 \times 1 0^{- 5} < K_{p}$ C) The reaction will proceed reverse, because $Q_{p} = 5.0 \times 1 0^{- 5} > K_{p}$ D) The reaction will proceed reverse, because $Q_{p} = 1.0 \times 1 0^{- 5} < K_{p}$

Worked Solution: First, write the Qp expression from the balanced reaction: $Q_{p} = \frac{( P _{S O_{2}} ) ^{2} ( P _{O_{2}} )}{( P _{S O_{3}} ) ^{2}}$ . Substitute the given partial pressures: $Q_{p} = \frac{( 0.050 ) ^{2} ( 0.0020 )}{( 0.10 ) ^{2}} = \frac{5.0 \times 1 0 ^{- 6}}{0.01} = 5.0 \times 1 0^{- 5}$ . Compare Qp to Kp: $5.0 \times 1 0^{- 5} > 1.8 \times 1 0^{- 5}$ , so Q > K. This means there are too many products, so the reaction proceeds in the reverse direction to reach equilibrium. The correct answer is C.

Question 2 (Free Response)

The dissociation of solid calcium carbonate follows the reaction: $C a C O_{3} (s) ⇌ C a O (s) + C O_{2} (g)$ . At 800°C, $K_{p} = 0.236 a t m$ for this reaction. (a) Write the expression for $Q_{p}$ for this reaction. (b) A sealed 1.0 L reaction vessel at 800°C initially contains 10.0 g of $C a C O_{3} (s)$ , 5.0 g of $C a O (s)$ , and $C O_{2} (g)$ at a partial pressure of 0.150 atm. Calculate Qp and predict the direction the reaction will proceed to reach equilibrium. Justify your answer. (c) The temperature of the vessel is increased to 900°C, and $K_{p}$ increases to 0.625. If the partial pressure of CO2 is still 0.150 atm immediately after the temperature change, what direction will the reaction proceed now?

Worked Solution: (a) Pure solids ( $C a C O_{3}$ and $C a O$ ) have activity of 1, so they are omitted from the Qp expression. The expression is: $Q_{p} = P_{C O_{2}}$ (b) Substitute the given partial pressure: $Q_{p} = 0.150$ . Compare to Kp: $Q_{p} = 0.150 < K_{p} = 0.236$ . Since Q < K, the product $C O_{2}$ is at too low a partial pressure, so the reaction proceeds forward to produce more $C O_{2} (g)$ until equilibrium is reached. (c) The new Kp is 0.625, and Qp is still 0.150. $Q_{p} = 0.150 < K_{p} = 0.625$ , so the reaction still proceeds forward to produce more $C O_{2}$ .

Question 3 (Application / Real-World Style)

Hemoglobin (Hb) binds oxygen in the lungs to form oxyhemoglobin $(H b O_{2})$ according to the reaction: $H b (a q) + O_{2} (a q) ⇌ H b O_{2} (a q)$ , $K_{c} = 1.8 \times 1 0^{6} M^{- 1}$ at body temperature. At high altitude, the concentration of dissolved O2 in lung fluid drops to $3.2 \times 1 0^{- 5} M$ . If [Hb] = $1.2 \times 1 0^{- 6} M$ and $[H b O_{2}] = 6.5 \times 1 0^{- 4} M$ at this high altitude, what is Q for the system, and will the system shift to produce more $H b O_{2}$ or decompose $H b O_{2}$ ? Explain what this means for oxygen transport at high altitude.

Worked Solution: Write the Qc expression: $Q_{c} = \frac{[ H b O _{2} ]}{[ H b ] [ O _{2} ]}$ . Substitute values: $Q_{c} = \frac{6.5 \times 1 0 ^{- 4}}{( 1.2 \times 1 0 ^{- 6} ) ( 3.2 \times 1 0 ^{- 5} )} \approx 1.7 \times 1 0^{7} M^{- 1}$ Compare Qc to Kc: $1.7 \times 1 0^{7} > 1.8 \times 1 0^{6}$ , so Q > K. The system shifts left to decompose $H b O_{2}$ . In context, this means lower oxygen concentrations at high altitude reduce the amount of oxyhemoglobin that can form, decreasing the total amount of oxygen that can be transported to body tissues.

7. Quick Reference Cheatsheet

Category	Formula	Notes
General concentration Q ( $Q_{c}$ )	$Q_{c} = \frac{[ C ] ^{c} [ D ] ^{d}}{[ A ] ^{a} [ B ] ^{b}}$ for $a A + b B ⇌ c C + d D$	Uses molar concentrations, applies to aqueous and gaseous systems
General partial pressure Q ( $Q_{p}$ )	$Q_{p} = \frac{( P _{C} ) ^{c} ( P _{D} ) ^{d}}{( P _{A} ) ^{a} ( P _{B} ) ^{b}}$ for $a A + b B ⇌ c C + d D$	Applies only to gas-phase systems
Rule for pure substances	Pure solids/pure liquids are not included	Activity of pure substances = 1, so they cancel out
Q < K prediction	Reaction proceeds forward (right)	Too few products, Q increases to reach K
Q = K prediction	System is at equilibrium	No net change in concentrations/partial pressures
Q > K prediction	Reaction proceeds reverse (left)	Too many products, Q decreases to reach K
Q after perturbation	Use values immediately after disturbance	K remains unchanged unless temperature changes
Q for reversed reaction	$Q_{r e v er se} = 1/ Q_{f or w a r d}$	Same scaling rule as K
Q for reaction multiplied by n	$Q_{n e w} = (Q_{or i g ina l})^{n}$	Same scaling rule as K

8. What's Next

Reaction quotient Q is the foundational tool for all subsequent topics in equilibrium, and it connects equilibrium to thermodynamics across the AP Chemistry course. Next, you will apply Q calculation and Q vs K comparison to solve ICE table problems that calculate equilibrium concentrations from initial non-equilibrium conditions, and you will use Q to justify shifts from perturbations in Le Chatelier’s principle FRQ questions. Without mastering this chapter, you cannot correctly predict equilibrium shifts or solve any dynamic equilibrium problem that starts from non-equilibrium conditions. Beyond Unit 7, Q is a core term in the Gibbs free energy equation for non-standard conditions, which is used to predict reaction spontaneity at any concentration or partial pressure. Follow-on topics:

Reaction quotient Q — AP Chemistry Study Guide

1. What Is Reaction quotient Q?

2. Calculating Q for Homogeneous and Heterogeneous Equilibria

Worked Example

3. Comparing Q and K to Predict Reaction Direction

Worked Example

4. Calculating Q After a System Perturbation

Worked Example

5. Common Pitfalls (and how to avoid them)

6. Practice Questions (AP Chemistry Style)

Question 1 (Multiple Choice)

Question 2 (Free Response)

Question 3 (Application / Real-World Style)

7. Quick Reference Cheatsheet

8. What's Next

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