GENERAL CHEMISTRY · CHEMICAL EQUILIBRIA

Write the system.
Compare Q with K.

Learn all five official topics through condition ledgers, worked examples, closed-note retrieval, and twenty original five-choice questions.

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5guided lessons

20practice questions

15study objectives

5choices per item

A repeatable equilibrium routine

System. Expression. Compare. Verify.

01Write the system
Balance the reversible equation, label every phase, and inventory the actual post-mixing conditions.
02Build the expression
Include variable gases and aqueous solutes, omit constant-activity pure phases, and use coefficients as exponents.
03Compare or solve
Use Q versus K for direction, an ion threshold for precipitation, or one reaction extent for an ICE table.
04Verify the boundary
Check physical roots, approximations, temperature, gas constraints, phase presence, and rate-versus-composition claims.

Five prerequisite-aware lessons

Let the quotient diagnose the system.

Each lesson pairs two semantic ledgers with a fully worked example and closed-note retrieval before multiple-choice practice.

LESSON 1 · 23 MIN

Study + retrieve

Equilibrium from molecular motion to Q and K

Recognize dynamic equilibrium, construct phase-aware expressions, and use Q versus K without confusing composition with reaction rate.

ESSENTIAL QUESTIONWhat changes microscopically, what remains constant macroscopically, and which expression tests the system now?

Molecular equilibrium ledger

Observable and molecular evidence
Evidence	Supported conclusion	Unsupported leap
Forward rate = reverse rate > 0	Dynamic equilibrium	The reactions have stopped
Concentrations stay constant	No net macroscopic change	Reactant and product concentrations are equal
Large K	Products are favored at equilibrium	The reaction is fast

Build and interpret Q or K
Decision	Rule	Boundary
Include a species?	Aqueous solutes and gases contribute	Omit pure solids and liquids already present
Set an exponent?	Use the balanced coefficient	Do not use a subscript from the formula
Predict direction?	Q < K forward; Q > K reverse	Q = K means no net direction

EQUILIBRIUM LEDGER · CAPTIONS INCLUDED

Recognize dynamic equilibrium

A closed reversible system is at equilibrium when forward and reverse reactions continue at equal nonzero rates. Concentrations are constant over time, but reactant and product concentrations need not be equal.

Equal rates, not equal concentrations
Constant does not mean stopped

Build the expression from the equation

Place variable product terms over variable reactant terms and raise each to its balanced coefficient. Omit pure solids and pure liquids because their activities remain constant while those phases are present.

Coefficients become exponents
Include gases and aqueous solutes

Separate Q, K, composition, and speed

Q uses the current mixture; K is Q at equilibrium for one reaction and temperature. Q below K drives a net forward response, while K magnitude describes favored composition rather than the time required to equilibrate.

Q < K → forward
Large K does not mean fast

Worked example

For CO(g) + H₂O(g) ⇌ CO₂(g) + H₂(g), Kc = 0.50. A mixture has [CO] = 0.20 M, [H₂O] = 0.40 M, [CO₂] = 0.10 M, and [H₂] = 0.20 M. Predict the net direction.

1
Write Qc = [CO₂][H₂]/([CO][H₂O]).
2
Substitute to obtain Qc = (0.10)(0.20)/((0.20)(0.40)) = 0.25.
3
Compare Qc = 0.25 with Kc = 0.50; Q is smaller, so the system reacts forward until Q returns to K.

ConclusionThe quotient diagnoses direction. It does not say that the forward reaction is fast or that every reactant will disappear.

Close the notes first

Retrieve the model.

01What two rates are equal at dynamic equilibrium?

The forward and reverse reaction rates.

Equal opposing rates keep macroscopic concentrations constant while molecular reactions continue.

02Why is CaCO₃(s) omitted from a K expression?

Its pure-solid activity is constant while the phase is present.

Omission does not mean the solid is absent or chemically uninvolved.

03What direction follows when Q is greater than K?

Net reverse reaction.

The mixture has too much quotient numerator relative to its equilibrium value.

LESSON 2 · 22 MIN

Study + retrieve

Acid-base equilibria as coupled reactions

Place Ka, Kb, and Kw inside the general K framework, combine proton-transfer equilibria, and explain common-ion response without replacing the acid-base regime map.

ESSENTIAL QUESTIONWhich proton-transfer equation is being measured, and how do its component constants combine?

Acid–base equilibrium ledger

Place acid–base constants in the general K framework
Reaction change	Constant operation	Check
Reverse an equation	K becomes 1/K	Numerator and denominator switch
Multiply every coefficient by n	K becomes Kⁿ	Fractional n is permitted
Add equations	Multiply their constants	Cancel intermediates first

Proton transfer and common-ion response
System	Relationship	Interpretation
HA + B⁻ ⇌ A⁻ + HB	K = K_a(HA)/K_a(HB)	Favors the weaker acid–base pair
Conjugate pair	K_aK_b = K_w	Same temperature and pair
Add a common ion	Q changes; K does not	Buffer capacity remains finite

EQUILIBRIUM LEDGER · CAPTIONS INCLUDED

Treat acid-base constants as K

Ka and Kb follow the same product-over-reactant rules as any equilibrium constant. Pure liquid water is omitted from the ordinary aqueous expression, while hydronium, hydroxide, weak species, and conjugates remain variable terms.

Write the reaction first
KaKb = Kw for one conjugate pair

Couple reactions multiplicatively

Reverse a reaction to invert K, scale every coefficient by n to raise K to n, and add reactions to multiply their constants. For HA + B⁻ ⇌ A⁻ + HB, the overall constant is Ka(HA)/Ka(HB).

Add equations → multiply K
Favor the weaker acid-base pair

Use Q to explain common ions

Adding a conjugate common ion changes Q and drives net reaction toward the un-ionized weak partner at constant temperature. A buffer responds stoichiometrically first and then re-equilibrates; it reduces pH change but has finite capacity.

Common ion changes Q, not K
Buffer is resistance, not immunity

Worked example

For HF + CN⁻ ⇌ F⁻ + HCN, Ka(HF) = 6.8 × 10⁻⁴ and Ka(HCN) = 6.2 × 10⁻¹⁰. Find the overall K and favored side.

1
Recognize HF as the reactant acid and HCN as the product acid.
2
Calculate K = Ka(HF)/Ka(HCN) = (6.8 × 10⁻⁴)/(6.2 × 10⁻¹⁰) = 1.1 × 10⁶.
3
Because K is much greater than 1, the products F⁻ and HCN are strongly favored under the compatible stated conditions.

ConclusionThe constant quantifies the same weaker-pair direction used in Brønsted-Lowry reasoning; it does not redefine pH or titration regimes.

Close the notes first

Retrieve the model.

01Where does pure liquid water appear in the ordinary Kb expression for NH₃?

It is incorporated into Kb and omitted as a variable term.

Its activity is effectively constant under the stated dilute-aqueous model.

02What happens to K when a reaction is reversed?

K becomes 1/K.

The product and reactant terms exchange numerator and denominator.

03Does adding a common ion change Ka at constant temperature?

No.

The disturbance changes Q and composition; the same-temperature equilibrium constant remains fixed.

LESSON 3 · 24 MIN

Study + retrieve

Precipitation from Ksp and ion thresholds

Connect dissolution stoichiometry to Ksp, evaluate post-mixing ion products, and distinguish common-ion or selective-precipitation evidence from a bare Ksp ranking.

ESSENTIAL QUESTIONWhat are the ion concentrations in the actual mixture, and has their stoichiometric product crossed Ksp?

Precipitation threshold ledger

Translate dissolution stoichiometry into Ksp
Dissolution	Ion concentrations from s	K_sp
MX(s) ⇌ M + X	s and s	s²
MX₂(s) ⇌ M + 2X	s and 2s	4s³
M₂X₃(s) ⇌ 2M + 3X	2s and 3s	108s⁵

Evaluate the actual post-mixing ion product
Comparison	State	Prediction
Q < K_sp	Unsaturated	No precipitation is favored
Q = K_sp	Saturated	Dissolution equilibrium
Q > K_sp	Supersaturated	Precipitation is favored
Selective threshold	Solve required added-ion concentration	Do not rank by K_sp alone

EQUILIBRIUM LEDGER · CAPTIONS INCLUDED

Translate dissolution into Ksp

Omit the pure solid, include each dissolved ion, and use dissolution coefficients as exponents. If solubility is s, each ion concentration is its coefficient times s, so Ksp is not generally equal to s.

Formula controls ion coefficients
Compare solubility only after solving for s

Test the mixture, not the stock bottles

After dilution, mixing, and any prior essentially complete reaction, calculate the ion product. Q below Ksp is unsaturated, Q equal to Ksp is saturated, and Q above Ksp predicts precipitation until equilibrium is restored.

Use total mixed volume
Q > Ksp → precipitation

Find controlled thresholds

A common ion generally lowers molar solubility. Selective precipitation compares the added counter-ion concentration required for each solid to reach Ksp; the smallest Ksp alone cannot decide when stoichiometries or starting ion concentrations differ.

Threshold before ranking
More pure solid does not change a saturated solution

Worked example

CaF₂(s) ⇌ Ca²⁺ + 2F⁻ has Ksp = 4.0 × 10⁻¹² in pure water. Find its molar solubility s.

1
Write Ksp = [Ca²⁺][F⁻]².
2
Use [Ca²⁺] = s and [F⁻] = 2s, giving Ksp = s(2s)² = 4s³.
3
Solve s = (Ksp/4)^(1/3) = (1.0 × 10⁻¹²)^(1/3) = 1.0 × 10⁻⁴ M.

ConclusionThe dissolution coefficients create the cubic relationship; setting s equal to Ksp would ignore both concentration factors and the exponent.

Close the notes first

Retrieve the model.

01What is Q relative to Ksp at the instant precipitation is thermodynamically favored?

Q is greater than Ksp.

The dissolved ion product must decrease toward Ksp by forming solid.

02What must happen before multiplying stock ion concentrations from two solutions?

Account for the total mixed volume and any prior reaction.

The quotient uses concentrations in the actual mixture.

03Why can two salts not always be ranked by Ksp alone?

Their dissolution stoichiometries and ion concentrations may differ.

Those factors change both molar-solubility algebra and precipitation thresholds.

LESSON 4 · 25 MIN

Study + retrieve

Equilibrium calculations by one reaction extent

Evaluate Q or K, transform constants with their reactions, convert between Kc and Kp, and solve a physical ICE-table model.

ESSENTIAL QUESTIONAre the supplied values initial or equilibrium values, and what one extent links every concentration change?

Equilibrium calculation ledger

Choose the correct constant operation
Task	Relationship	Required input
Evaluate K or Q	Products over reactants	Equilibrium values for K; current values for Q
Convert K_c to K_p	K_p = K_c(RT)^Δn	Ideal gases and stated temperature
Combine equations	Invert, exponentiate, multiply	Target equation exactly

One-extent ICE-table verification
Step	Action	Fail-closed check
Initial	Record every stated concentration	Account for prior mixing or reaction
Change	Use one x times signed coefficients	Preserve stoichiometric ratios
Equilibrium	Substitute into K and solve	Reject nonphysical roots
Approximation	Calculate percent change	Verify the chosen threshold afterward

EQUILIBRIUM LEDGER · CAPTIONS INCLUDED

Name the snapshot

Substituting equilibrium values evaluates K; substituting any other stated instant evaluates Q. Convert moles to molarity or gas amounts to partial pressures before using the matching expression.

Initial values usually give Q
Kc and Kp are not interchangeable labels

Transform equation and constant together

Every reversal, coefficient scaling, or equation addition changes K by its reciprocal, a power, or a product. For ideal gases, Kp = Kc(RT)^Δn and Δn counts only gaseous coefficients.

Match the target equation exactly
Temperature must be kelvins

Solve and verify the ICE table

Use Q or the initial inventory to choose direction, apply one signed reaction extent scaled by every coefficient, and solve. Reject negative concentrations, substitute back into K, and test any small-change approximation afterward.

One extent, coefficient-scaled changes
Physical root + substitution check

Worked example

A(g) ⇌ B(g) + C(g) begins with [A] = 0.100 M and no products. If Kc = 0.0250, find [B] at equilibrium.

1
Let x dissociate: [A] = 0.100 − x and [B] = [C] = x.
2
Substitute Kc = x²/(0.100 − x) = 0.0250 and solve the quadratic.
3
Select the physical root x = 0.0390 M, reject the negative root, and verify x²/(0.100 − x) = 0.0250.

ConclusionThe balanced equation makes both products change by x. The physical root leaves every concentration nonnegative and reproduces K.

Close the notes first

Retrieve the model.

01What does substituting initial nonequilibrium concentrations into the equilibrium expression calculate?

Q, not K.

K requires equilibrium composition at the stated temperature.

02If every coefficient is halved, what happens to K?

K is raised to the one-half power.

Scaling coefficients scales every exponent in the quotient.

03Which species count in Δn for Kp = Kc(RT)^Δn?

Only gaseous reactants and products.

Condensed and aqueous terms do not contribute to the gas-mole coefficient difference.

LESSON 5 · 22 MIN

Study + retrieve

Le Chatelier responses without shortcut errors

Use Q and constraints to analyze concentration, phase, gas-volume, inert-gas, temperature, and catalyst changes while keeping K and rate claims separate.

ESSENTIAL QUESTIONWhich variable changed immediately, did Q or K change, and what constraint controls the response?

Le Chatelier condition ledger

Stress, quotient, and net response
Stress	Immediate effect	Boundary
Add variable reactant	Q falls; net forward response	K stays fixed at constant temperature
Add variable product	Q rises; net reverse response	The addition is only partly offset
Add pure solid already present	Q does not change	Its amount changes, not equilibrium gas pressure
Compress gases	Favor fewer gaseous coefficients	No shift if totals are equal

Temperature, inert gas, and catalyst boundaries
Change	Equilibrium result	What happens to K?
Raise temperature	Favor the endothermic direction	Changes
Inert gas at constant volume	No reactive partial-pressure change	Unchanged
Inert gas at constant pressure	Expansion may favor more gas	Unchanged
Add catalyst	Same composition reached sooner	Unchanged

EQUILIBRIUM LEDGER · CAPTIONS INCLUDED

Track the immediate quotient change

Adding or removing a variable reactant or product changes Q immediately while K stays fixed at constant temperature. Adding more of a pure phase already present does not change Q; a shift offsets only part of a disturbance.

Disturbance first, net reaction second
Pure-phase amount is not a Q term

Name the gas constraint

Compression favors fewer gas moles and expansion favors more; equal gaseous coefficient sums give no volume-driven shift. An inert gas at constant volume leaves reactive partial pressures unchanged, while constant-pressure addition can expand the mixture.

Count gas coefficients only
Total pressure alone is insufficient

Separate temperature from catalysis

Heating favors the endothermic direction and changes K; cooling favors the exothermic direction. A catalyst accelerates both directions and reaches the same equilibrium sooner without changing K or equilibrium composition.

Temperature changes K
Catalyst changes time, not position

Worked example

For N₂(g) + 3H₂(g) ⇌ 2NH₃(g), ΔH = −92 kJ. Compare raising temperature with adding a catalyst.

1
The forward reaction is exothermic, so heating favors the reverse, endothermic direction and establishes a different K.
2
A catalyst lowers barriers for both directions and therefore shortens the time needed to equilibrate.
3
The catalyst does not change K or the final equilibrium composition at the same temperature.

ConclusionTemperature is an equilibrium-position stress; a catalyst is a kinetic change. Treating both as product-favoring confuses yield with speed.

Close the notes first

Retrieve the model.

01What happens to K after adding a reactant at constant temperature?

K stays unchanged.

The addition changes Q, and the system responds until Q again equals the same K.

02Does compression shift H₂(g) + I₂(g) ⇌ 2HI(g)?

No, not in the ideal model.

Both sides have two total moles of gas, so the volume factors cancel from Qp.

03What does a catalyst change about equilibrium?

Only the time required to approach it, not K or composition.

Both forward and reverse pathways are accelerated.

All twenty Chemical Equilibria problems

Diagnose before you calculate.

Question order and all five answer options shuffle each time. Reports automatically include the exact question, content version, skill, and seed.

20 PRACTICE QUESTIONS

Retrieve before you review.

Question order and all five answer options are shuffled when you begin. The correct answer stays attached to the same underlying choice.

Interpret results carefully

Raw accuracy directs review—not score prediction.

These original questions are draft and have not been calibrated to the official score scale. Use each explanation to repair the exact system, expression, quotient, threshold, reaction-extent, or condition decision.

DAT TRAIN does not claim topic quotas because the official manual does not publish them. The completed Acids and Bases route remains the home for pH, weak-equilibrium, buffer, and titration-regime methods.