# Introduction to Equilibrium

> AP Chemistry · AP Chemistry CED Unit 7
> Source: https://www.owlsprep.com/study/ap-chemistry-u7-introduction-to-equilibrium/

This foundational subtopic introduces dynamic equilibrium, the equilibrium constant, reaction quotient, and prediction of reaction direction for reversible reactions. It is required for all subsequent equilibrium topics in AP Chemistry, appearing in both MCQ and FRQ sections.

**Prerequisites:** Reversible vs irreversible reaction classification; Balanced chemical equation stoichiometry; Partial pressure and concentration units for gases and solutions

## Learning objectives

- Define dynamic equilibrium and distinguish it from static equilibrium
- Write correct Kc and Kp equilibrium constant expressions for any balanced reaction
- Calculate the reaction quotient Q from non-equilibrium concentrations/partial pressures
- Compare Q and K to predict the direction a reaction will shift to reach equilibrium

## Dynamic Equilibrium

**Dynamic Equilibrium** — A steady state reached by a reversible reaction in a closed, constant-temperature system where the rate of the forward reaction equals the rate of the reverse reaction. No net change in concentrations or observable properties occurs, but forward and reverse reactions continue continuously.

The key distinction between dynamic equilibrium and static equilibrium is that reactions do not stop at equilibrium. For example, when liquid bromine is sealed in a flask, the brown color of bromine vapor stops changing intensity once equilibrium is reached: the rate of evaporation of liquid bromine equals the rate of condensation of gaseous bromine, so vapor concentration remains constant even as both processes continue.

**Worked example:** A student places solid iodine into a closed, temperature-controlled flask. After 30 minutes, the intensity of purple color from gaseous iodine stops changing. (a) Explain why the color stops changing even though solid iodine is still evaporating, (b) Is the system at dynamic equilibrium? Justify your answer.

1. Constant color intensity means the concentration of gaseous $I_2(g)$ is no longer changing, but this does not mean all processes have stopped. The reversible process occurring is:
2. $$I_2(s) \rightleftharpoons I_2(g)$$
3. As solid iodine evaporates, the concentration of $I_2(g)$ increases, which increases the rate of the reverse process (deposition of $I_2(g)$ back to solid).
4. Eventually, the rate of evaporation equals the rate of deposition. Because the rates are equal, the concentration of $I_2(g)$ does not change, so the color intensity remains constant.
5. Since both forward and reverse processes continue to occur at equal rates (rather than stopping), the system is at dynamic equilibrium.

> **Exam tip:** On AP MCQ, any answer that claims 'at equilibrium, all reaction stops' or 'concentrations of reactants equal concentrations of products' is always wrong. Remember: equal rates, not equal concentrations, and reactions never stop at equilibrium.

## The Equilibrium Constant and Law of Mass Action

The law of mass action defines the form of the equilibrium constant expression for any balanced reversible reaction. For the general balanced reaction:

The equilibrium constant $K$ is defined as the ratio of product concentrations (raised to their stoichiometric coefficients) over reactant concentrations (raised to their stoichiometric coefficients):

$K_c$ uses molar concentrations (mol/L) for all species. For gas-phase reactions, we can also write $K_p$, which uses partial pressures of gases (typically in atm) instead of concentrations, with the same ratio of products to reactants. A critical rule for writing any equilibrium expression is: **pure solids, pure liquids, and solvents in dilute solutions do not appear in the expression**. Their concentrations are constant at constant temperature, so they are incorporated into the value of $K$ and do not need to be included explicitly. The relationship between $K_p$ and $K_c$ is:

Where $\Delta n = \text{moles of gaseous products} - \text{moles of gaseous reactants}$, $R = 0.0821 \ \text{L·atm/mol·K}$, and $T$ is absolute temperature in Kelvin.

**Worked example:** Write the $K_c$ and $K_p$ expressions for the reversible decomposition of solid calcium carbonate to solid calcium oxide and carbon dioxide gas: $CaCO_3(s) \rightleftharpoons CaO(s) + CO_2(g)$.

1. First, identify all species and exclude pure solids per the rule: both $CaCO_3$ (reactant) and $CaO$ (product) are pure solids, so they are omitted from both expressions.
2. For $K_c$, the only species with a variable concentration is gaseous $CO_2$, so the expression becomes:
3. $$K_c = \frac{[CO_2]}{1} = [CO_2]$$
4. For $K_p$, the only species with a variable partial pressure is gaseous $CO_2$, so the expression becomes:
5. $$K_p = P_{CO_2}$$
6. Verify that products are in the numerator, matching the direction of the balanced reaction, and no pure solids are included: the expressions are correct.

> **Exam tip:** Always check for pure solids and liquids before writing a $K$ or $Q$ expression on FRQ. If you include them, your expression will be marked wrong, which is an easy point to avoid losing.

## The Reaction Quotient and Predicting Reaction Direction

The reaction quotient ($Q$) is a value calculated from non-equilibrium concentrations (or partial pressures) of reactants and products at any point in a reaction before equilibrium is reached. $Q$ has the exact same form as the equilibrium constant $K$: the only difference is that $K$ uses equilibrium concentrations, while $Q$ uses current non-equilibrium concentrations. By comparing $Q$ to $K$, we can predict which direction the reaction will proceed to reach equilibrium:

- If $Q < K$: The product terms (numerator) are too small, so the reaction proceeds forward (toward products) to increase $Q$ until $Q=K$.
- If $Q > K$: The product terms (numerator) are too large, so the reaction proceeds reverse (toward reactants) to decrease $Q$ until $Q=K$.
- If $Q = K$: The system is already at equilibrium, so no net change occurs.

> **Direction Memory Hook**
>
> If $Q < K$, you need more of the top (products) to increase $Q$ to $K$. If $Q > K$, you need more of the bottom (reactants) to decrease $Q$ to $K$. This avoids misremembering the rule.

**Worked example:** For the reaction $N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g)$, $K_c = 0.50$ at 400 K. A student mixes the gases at 400 K with the following initial concentrations: $[N_2] = 0.10$ M, $[H_2] = 0.10$ M, $[NH_3] = 0.050$ M. Predict which direction the reaction will shift to reach equilibrium.

1. Write the expression for $Q_c$, which matches the form of $K_c$:
2. $$Q_c = \frac{[NH_3]^2}{[N_2][H_2]^3}$$
3. Substitute the initial non-equilibrium concentrations into the expression:
4. $$Q_c = \frac{(0.050)^2}{(0.10)(0.10)^3} = \frac{0.0025}{0.0001} = 25$$
5. Compare the calculated $Q_c$ to the given $K_c$: $Q_c = 25 > K_c = 0.50$.
6. Since $Q > K$, the reaction will proceed in the reverse direction (toward reactants) to reach equilibrium.

**Check your understanding**

Test your understanding with this AP-style multiple choice question:

1. For the reversible reaction $2HI(g) \rightleftharpoons H_2(g) + I_2(g)$, $K_c = 0.020$ at a certain temperature. A reaction mixture has the following concentrations: $[HI] = 0.25$ M, $[H_2] = 0.15$ M, $[I_2] = 0.10$ M. Which of the following statements is correct?

   - A) The system is at equilibrium, so no shift occurs.
   - B) The reaction will shift forward to produce more $H_2$ and $I_2$.
   - C) The reaction will shift in reverse to produce more $HI$.
   - D) The reaction will shift forward initially, then shift reverse to reach equilibrium.

   *Why:* First calculate $Q_c = \frac{(0.15)(0.10)}{(0.25)^2} = 0.24$. Since $Q_c = 0.24 > K_c = 0.020$, the reaction shifts reverse to produce more HI.

**Worked example:** Nitrogen dioxide gas dimerizes to form dinitrogen tetroxide gas in a closed container at 25°C. The balanced reaction is $2NO_2(g) \rightleftharpoons N_2O_4(g)$. (a) Write the $K_c$ and $K_p$ expressions for this reaction. (b) The initial partial pressures of the gases are: $P_{NO_2} = 0.50$ atm, $P_{N_2O_4} = 0.25$ atm. $K_p = 8.8$ at 25°C. Calculate $Q_p$ and predict the direction of shift to reach equilibrium. (c) A student claims that 'after the system reaches equilibrium, no more $NO_2$ is converted to $N_2O_4$'. Is the student correct? Justify your answer.

1. (a) All species are gaseous, so all are included in the expressions:
2. $$K_c = \frac{[N_2O_4]}{[NO_2]^2} \quad \quad K_p = \frac{P_{N_2O_4}}{(P_{NO_2})^2}$$
3. (b) Substitute initial partial pressures into the $Q_p$ expression:
4. $$Q_p = \frac{0.25}{(0.50)^2} = \frac{0.25}{0.25} = 1.0$$
5. Since $Q_p = 1.0 < K_p = 8.8$, the reaction proceeds forward (toward products, forming more $N_2O_4$) to reach equilibrium.
6. (c) The student is incorrect. At equilibrium, the system is dynamic: the rate of conversion of $NO_2$ to $N_2O_4$ (forward reaction) equals the rate of conversion of $N_2O_4$ back to $NO_2$ (reverse reaction). Conversion still occurs continuously, but there is no net change in concentrations, so the student's claim is false.

## Common pitfalls

- **Wrong:** Including pure solids or pure liquids in the equilibrium constant expression with their concentration values.
  - Why it fails: Students are used to including all species in stoichiometry and rate law problems, so they forget the rule that constant concentrations do not appear.
  - Correct: Before writing any $K$ or $Q$ expression, cross out all pure solids, pure liquids, and dilute solvents; only include aqueous or gaseous species with variable concentrations.
- **Wrong:** Claiming that at equilibrium, the concentration of reactants equals the concentration of products.
  - Why it fails: Students confuse equal rates of forward and reverse reaction with equal concentrations of reactants and products.
  - Correct: Always remember: at equilibrium, rates are equal, concentrations are constant (not necessarily equal). Only the ratio of concentrations (raised to stoichiometric coefficients) equals $K$.
- **Wrong:** Flipping the ratio, putting reactants in the numerator and products in the denominator for $K$ or $Q$.
  - Why it fails: Students mix up the direction of the reaction or mis-memorize the ratio order.
  - Correct: Always write the balanced reaction first, then put all species on the right (products) of the equilibrium arrow in the numerator, and all species on the left (reactants) in the denominator.
- **Wrong:** Forgetting to raise concentrations/partial pressures to their stoichiometric coefficients in $K$ or $Q$.
  - Why it fails: Students rush through calculations and skip checking exponents.
  - Correct: After writing the expression, check each term's exponent against the stoichiometric coefficient in the balanced reaction before plugging in any numbers.
- **Wrong:** Predicting a forward shift when $Q > K$.
  - Why it fails: Students reverse the direction rule because they misremember which value corresponds to which shift.
  - Correct: Always ask 'do I need to increase or decrease $Q$ to reach $K$?' If $Q < K$, increase $Q$ by making more products; if $Q > K$, decrease $Q$ by making more reactants.

## Cheatsheet

| Category | Formula / Rule | Notes |
| --- | --- | --- |
| Dynamic Equilibrium | $\text{rate}_{forward} = \text{rate}_{reverse}$ | Net concentration change is zero; reactions continue occurring |
| General Law of Mass Action | $K = \frac{[C]^c [D]^d}{[A]^a [B]^b}$ for $aA + bB \rightleftharpoons cC + dD$ | Omit pure solids, pure liquids, and dilute solvents |
| Concentration-based $K$ | $K_c$ | Uses molar concentrations (mol/L) for aqueous/gaseous species |
| Pressure-based $K$ | $K_p$ | Uses partial pressures (atm) for gaseous species |
| $K_p$-$K_c$ Conversion | $K_p = K_c(RT)^{\Delta n}$ | $\Delta n = \text{moles gaseous products} - \text{moles gaseous reactants}$; $R=0.0821$ L·atm/mol·K |
| Reaction Quotient | $Q = \frac{[C]^c [D]^d}{[A]^a [B]^b}$ | Same form as $K$, uses non-equilibrium concentrations |
| Direction: $Q < K$ | Shift forward (toward products) | Increase Q to match K |
| Direction: $Q = K$ | No net shift | System is at equilibrium |
| Direction: $Q > K$ | Shift reverse (toward reactants) | Decrease Q to match K |

## What's next

This introduction to equilibrium is the foundational prerequisite for all subsequent topics in Unit 7. Mastering the basics of writing $K$ expressions, calculating $Q$, and predicting direction from $Q$ vs $K$ is required to understand more advanced equilibrium topics, including acid-base and solubility equilibrium, which together account for nearly 15% of the total AP exam score. Next, you will learn to manipulate equilibrium constants, use ICE tables for equilibrium concentration calculations, and apply $Q$ vs $K$ comparisons to Le Châtelier's principle.

- [Direction of reversible reactions](https://www.owlsprep.com/study/ap-chemistry-u7-direction-of-reversible-reactions/)
- [Reaction Quotient Q](https://www.owlsprep.com/study/ap-chemistry-u7-reaction-quotient-q/)
- [Calculating the equilibrium constant K](https://www.owlsprep.com/study/ap-chemistry-u7-calculating-the-equilibrium-constant-k/)

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