Introduction to Equilibrium

AP Chemistry· AP Chemistry CED — Equilibrium· 14 min read

1. Dynamic Equilibrium★★☆☆☆⏱ 4 min

📘 Definition

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.

2. The Equilibrium Constant and Law of Mass Action★★★☆☆⏱ 5 min

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

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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):

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$K_{c}$ 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:

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Where $Δ n = moles of gaseous products - moles of gaseous reactants$ , $R = 0.0821 L \cdotp atm/mol \cdotp K$ , 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: $C a C O_{3} (s) ⇌ C a O (s) + C O_{2} (g)$ .

1
First, identify all species and exclude pure solids per the rule: both $C a C O_{3}$ (reactant) and $C a O$ (product) are pure solids, so they are omitted from both expressions.
2
For $K_{c}$ , the only species with a variable concentration is gaseous $C O_{2}$ , 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 $C O_{2}$ , 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.

3. The Reaction Quotient and Predicting Reaction Direction★★★☆☆⏱ 5 min

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.

📐 Worked Example

For the reaction $N_{2} (g) + 3 H_{2} (g) ⇌ 2 N H_{3} (g)$ , $K_{c} = 0.50$ 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, $[N H_{3}] = 0.050$ 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:
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$Q_c = \frac{(0.050)^2}{(0.10)(0.10)^3} = \frac{0.0025}{0.0001} = 25$
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.

✓ Quick check

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

For the reversible reaction $2 H I (g) ⇌ H_{2} (g) + I_{2} (g)$ , $K_{c} = 0.020$ at a certain temperature. A reaction mixture has the following concentrations: $[H I] = 0.25$ 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 $H I$ .
- D) The reaction will shift forward initially, then shift reverse to reach equilibrium.
Reveal answer
C —
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 $2 N O_{2} (g) ⇌ N_{2} O_{4} (g)$ . (a) Write the $K_{c}$ and $K_{p}$ expressions for this reaction. (b) The initial partial pressures of the gases are: $P_{N O_{2}} = 0.50$ atm, $P_{N_{2} O_{4}} = 0.25$ 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 $N O_{2}$ is converted to $N_{2} O_{4}$ '. 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:
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$Q_p = \frac{0.25}{(0.50)^2} = \frac{0.25}{0.25} = 1.0$
5
Since $Q_{p} = 1.0 < K_{p} = 8.8$ , the reaction proceeds forward (toward products, forming more $N_{2} O_{4}$ ) to reach equilibrium.
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(c) The student is incorrect. At equilibrium, the system is dynamic: the rate of conversion of $N O_{2}$ to $N_{2} O_{4}$ (forward reaction) equals the rate of conversion of $N_{2} O_{4}$ back to $N O_{2}$ (reverse reaction). Conversion still occurs continuously, but there is no net change in concentrations, so the student's claim is false.

4. Common Pitfalls

Wrong move:

Including pure solids or pure liquids in the equilibrium constant expression with their concentration values.

Why:

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 move:

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 move:

Claiming that at equilibrium, the concentration of reactants equals the concentration of products.

Why:

Students confuse equal rates of forward and reverse reaction with equal concentrations of reactants and products.

Correct move:

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 move:

Flipping the ratio, putting reactants in the numerator and products in the denominator for $K$ or $Q$ .

Why:

Students mix up the direction of the reaction or mis-memorize the ratio order.

Correct move:

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 move:

Forgetting to raise concentrations/partial pressures to their stoichiometric coefficients in $K$ or $Q$ .

Why:

Students rush through calculations and skip checking exponents.

Correct move:

After writing the expression, check each term's exponent against the stoichiometric coefficient in the balanced reaction before plugging in any numbers.

Wrong move:

Predicting a forward shift when $Q > K$ .

Why:

Students reverse the direction rule because they misremember which value corresponds to which shift.

Correct move:

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.

5. Quick Reference Cheatsheet

Category	Formula / Rule	Notes
Dynamic Equilibrium	$rate_{f or w a r d} = rate_{r e v er se}$	Net concentration change is zero; reactions continue occurring
General Law of Mass Action	$K = \frac{[ C ] ^{c} [ D ] ^{d}}{[ A ] ^{a} [ B ] ^{b}}$ for $a A + b B ⇌ c C + d D$	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	$K_{p} = K_{c} (R T)^{Δ n}$	$Δ n = moles gaseous products - moles gaseous reactants$ ; $R = 0.0821$ 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

When this came up on past exams

AI-estimated based on syllabus patterns — cross-check with official past papers for accuracy. Use only as revision-focus signals.

2023 · MCQ
Dynamic equilibrium conceptual question
2022 · FRQ
Write Kc, compare Q and 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.