GENERAL CHEMISTRY · THERMODYNAMICS AND THERMOCHEMISTRY

Name the system.
Close the ledger.

Learn all five official topics through signed energy 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 energy routine

Boundary. Sign. Model. Verify.

01Name the boundary
Identify the system, surroundings, reaction direction, phase, pressure constraint, and initial and final states.
02Assign the sign
Mark heat, work, enthalpy, and entropy terms for the defined system before combining magnitudes.
03Select the model
Choose first-law, Hess, Gibbs, standard-state, sensible-heat, calorimeter, or phase-change relationships.
04Verify the boundary
Check kelvins, J versus kJ, coefficients, phases, constant pressure or volume, and direction-versus-rate claims.

Five prerequisite-aware lessons

Let the boundary control the sign.

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

LESSON 1 · 24 MIN

Study + retrieve

Thermodynamic laws from a named boundary

Apply the first-law sign ledger, separate state from path functions, and use total entropy plus the third-law reference without changing the defined system.

ESSENTIAL QUESTIONWhat crosses the system boundary, which quantities depend only on endpoints, and what total-entropy condition selects a direction?

Thermodynamic laws ledger

First-law signs for the named system
Transfer	System sign	Interpretation
Heat enters	q > 0	System gains energy as heat
Heat leaves	q < 0	Surroundings gain that heat
Compression	w > 0	Work is done on the system
Expansion	w < 0	System does work on surroundings

State, path, and entropy boundaries
Decision	Relationship	Boundary
Energy ledger	ΔE = q + w	Keep one system definition
State function	ΔE, ΔH, ΔS, ΔG	Depends on endpoints
Path function	q and w	Depends on process
Second law	ΔS_univ = ΔS_sys + ΔS_surr	Positive for spontaneous direction

ENERGY LEDGER · CAPTIONS INCLUDED

Name the boundary before the sign

For the defined system, heat entering gives q > 0 and work done on the system gives w > 0. Expansion work is negative in the chemistry convention, and ΔE = q + w preserves the complete energy ledger.

System first, sign second
Expansion work is negative

Separate state from path

Internal energy, enthalpy, entropy, and Gibbs energy are state functions fixed by endpoints. Heat and work are path functions, so different routes can divide the same ΔE differently between q and w.

Endpoint change for state functions
Process history for q and w

Apply the second and third laws

A proposed direction is spontaneous when ΔS_univ = ΔS_sys + ΔS_surr is positive and is at equilibrium when that sum is zero. The third-law zero belongs to a perfect crystal at absolute zero—not to every element at ordinary temperature.

Test system plus surroundings
Perfect crystal at 0 K

Worked example

A gas absorbs 125 kJ of heat while doing 40 kJ of expansion work on the surroundings. Find ΔE for the gas.

1
Choose the gas as the system, so absorbed heat gives q = +125 kJ.
2
Work done by the gas is energy leaving the system, so w = −40 kJ.
3
Apply ΔE = q + w = +125 kJ − 40 kJ = +85 kJ.

ConclusionThe internal energy rises by 85 kJ. The sign follows the named system, not whether the surroundings gain part of the energy.

Close the notes first

Retrieve the model.

01What sign does expansion work have for the expanding system in the chemistry convention?

Negative.

The system transfers energy to the surroundings as work.

02Are heat and enthalpy both state functions?

No. Enthalpy is a state function; heat is a path function.

Heat describes energy transfer along a process, while ΔH compares endpoints under stated conditions.

03Can ΔS_sys < 0 occur in a spontaneous process?

Yes, if ΔS_surr is positive by a larger amount.

The second-law test uses ΔS_univ, not the system term alone.

LESSON 2 · 23 MIN

Study + retrieve

Hess cycles and formation enthalpies

Transform thermochemical equations, add an exact Hess cycle, and calculate standard reaction enthalpy from phase-specific formation data.

ESSENTIAL QUESTIONWhat equation operation reaches the target, and was its enthalpy transformed by the identical operation?

Hess-cycle ledger

Transform a thermochemical equation
Equation operation	ΔH operation	Verification
Reverse	Change sign	Initial and final states swap
Multiply by n	Multiply by n	Scale every coefficient
Add equations	Add ΔH values	Cancel only exact intermediates

Standard reaction-property sum
Step	Action	Boundary
Products	Σν(value)	Use stated phases
Reactants	Σν(value)	Use balanced coefficients
Reaction change	Products − reactants	Preserve units and standard conditions
Elemental ΔH°_f	Zero only in reference state	Not zero internal energy

ENERGY LEDGER · CAPTIONS INCLUDED

Transform equation and enthalpy together

Reversing a reaction reverses the sign of ΔH. Multiplying every coefficient by a factor multiplies ΔH by that factor because the written thermochemical equation specifies a reaction extent.

Reverse → reverse sign
Scale coefficients → scale ΔH

Make the net equation exact

Add transformed component equations only after verifying that intermediates cancel and every target species, coefficient, phase, and direction remains. The algebraic sum of component ΔH values then gives the target ΔH.

Cancel intermediates
Audit phases and coefficients

Use formation data product minus reactant

Compute ΔH°rxn = ΣνΔHf°(products) − ΣνΔHf°(reactants). Only an element in its reference standard state has ΔHf° = 0; a different allotrope or phase needs its own value.

Products minus reactants
Zero only for the reference state

Worked example

Use ΔH°f [CO₂(g)] = −393.5, [H₂O(l)] = −285.8, and [CH₄(g)] = −74.8 kJ/mol to find ΔH° for CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(l).

1
Weight products: −393.5 + 2(−285.8) = −965.1 kJ.
2
Weight reactants: −74.8 + 2(0) = −74.8 kJ because O₂(g) is oxygen's reference state.
3
Subtract reactants from products: −965.1 − (−74.8) = −890.3 kJ.

ConclusionThe reaction is exothermic as written. Changing water to vapor would change the endpoint and therefore the numerical enthalpy.

Close the notes first

Retrieve the model.

01What happens to ΔH when a thermochemical equation is reversed?

Its sign reverses.

The initial and final states exchange roles.

02Do Hess component reactions describe a required mechanism?

No.

Hess's law uses state-function path independence, not an asserted elementary pathway.

03Is ΔH°f zero for diamond because diamond is elemental carbon?

Not under the usual reference convention.

Graphite, not diamond, is carbon's reference standard state under ordinary tabulation conditions.

LESSON 3 · 24 MIN

Study + retrieve

Spontaneity, Gibbs energy, and temperature

Keep thermodynamic direction separate from speed, interpret current versus standard Gibbs energy, and solve temperature-dependent favorability with compatible units.

ESSENTIAL QUESTIONIs the claim about direction or rate, and does the Gibbs value describe the current mixture or a standard-state reaction?

Spontaneity and Gibbs ledger

Direction is distinct from rate
Evidence	Supported conclusion	Unsupported leap
ΔG < 0	Forward direction favored	Reaction is fast
ΔG = 0	Current mixture at equilibrium	All concentrations are equal
Catalyst added	Barrier and approach time change	ΔG or K changes
ΔG° supplied	Standard-state reaction value	Current ΔG for every mixture

Temperature regimes from ΔG = ΔH − TΔS
ΔH	ΔS	Favored regime
Negative	Positive	All temperatures
Positive	Negative	No temperatures
Positive	Positive	High temperatures
Negative	Negative	Low temperatures

ENERGY LEDGER · CAPTIONS INCLUDED

Do not turn spontaneous into fast

Spontaneous means thermodynamically favored under the stated conditions. It does not specify activation energy, reaction mechanism, time to equilibrium, irreversibility, or complete conversion.

Direction is not speed
Catalysts do not change ΔG

Use the current Gibbs criterion

At constant temperature and pressure, ΔG < 0 favors the reaction as written, ΔG > 0 favors the reverse, and ΔG = 0 marks equilibrium. ΔG° is the standard-state value and need not equal ΔG for the current composition.

Sign follows reaction direction
Current ΔG differs from ΔG°

Protect temperature and energy units

In ΔG = ΔH − TΔS, use kelvins and convert J to kJ before subtraction. Matching ΔH and ΔS signs can create a threshold: both positive favors high temperature, while both negative favors low temperature under the constant-property model.

Kelvins only
Match J and kJ

Worked example

At 298 K, a process has ΔH = −35.0 kJ/mol and ΔS = −80.0 J/(mol·K). Find ΔG and predict the favored direction.

1
Convert entropy to −0.0800 kJ/(mol·K).
2
Calculate TΔS = 298(−0.0800) = −23.84 kJ/mol.
3
Compute ΔG = −35.0 − (−23.84) = −11.16 kJ/mol, so the process is favored as written.

ConclusionThe negative entropy term does not by itself forbid spontaneity. At this temperature, the favorable enthalpy term is larger in magnitude.

Close the notes first

Retrieve the model.

01What kinetic quantity can be inferred from a negative ΔG?

None by itself.

Thermodynamic favorability does not determine the activation barrier or rate constant.

02What does ΔG = 0 mean at constant temperature and pressure?

The system is at equilibrium for the current composition.

Neither net reaction direction lowers Gibbs energy at that boundary.

03When ΔH > 0 and ΔS > 0, which temperature regime can be spontaneous?

Sufficiently high temperatures.

A large positive TΔS can outweigh the positive enthalpy term.

LESSON 4 · 22 MIN

Study + retrieve

Enthalpy and entropy as reaction evidence

Interpret enthalpy signs at constant pressure, predict bounded entropy trends, and calculate standard reaction changes without confusing formation enthalpy with absolute entropy.

ESSENTIAL QUESTIONWhich thermodynamic property is tabulated, what reference convention applies, and do the species phases support the proposed sign?

Enthalpy and entropy evidence ledger

System enthalpy at constant pressure
Process	System sign	Surroundings response
Exothermic	ΔH < 0	Gain heat
Endothermic	ΔH > 0	Lose heat
Reverse reaction	Sign reverses	Energy transfer reverses
Constant pressure	q_p = ΔH	Only under the stated work model

Qualitative system-entropy evidence
Change	Typical ΔS_sys	Boundary
Solid → liquid → gas	Increases	Compare compatible conditions
Gas expansion	Increases	More accessible volume
More gas coefficients	Often increases	Phase and data outrank shortcut
Reference-state element	S° generally positive	ΔH°_f, not S°, is zero

ENERGY LEDGER · CAPTIONS INCLUDED

Attach enthalpy to the system

At constant pressure with only pressure-volume work, qₚ = ΔH. An exothermic system has ΔH < 0 while its surroundings gain heat; direction, coefficient scale, phase, and conditions remain part of the value.

Exothermic system → negative ΔH
qₚ = ΔH under the model

Use microstate evidence carefully

Heating, expansion, mixing distinguishable particles, and moving solid → liquid → gas commonly increase accessible microstates. Gas-mole count is a useful reaction shortcut only when phases and competing molecular effects do not overturn it.

Phase labels matter
Gas coefficients are a clue, not proof

Keep ΔH°f and S° conventions separate

Both ΔH°rxn and ΔS°rxn use coefficient-weighted products minus reactants. Reference-state elements have ΔH°f = 0, but their standard molar entropies at ordinary temperatures are generally positive.

Same reaction sum
Different zero references

Worked example

For N₂(g) + 3H₂(g) → 2NH₃(g), use ΔH°f[NH₃(g)] = −46.1 kJ/mol and zero for reference-state N₂(g) and H₂(g).

1
Weight the product value: 2(−46.1) = −92.2 kJ.
2
The coefficient-weighted reactant formation enthalpies sum to zero.
3
Subtract reactants from products to obtain ΔH°rxn = −92.2 kJ.

ConclusionThe negative value belongs to the reaction as written. It does not imply that N₂ or H₂ has zero internal energy or zero entropy.

Close the notes first

Retrieve the model.

01What is the sign of ΔH for heat released by the reacting system at constant pressure?

Negative.

Energy leaves the defined system as heat.

02Which generally has greater molar entropy under comparable conditions: a liquid or its vapor?

The vapor.

The gas has many more accessible translational microstates.

03May an element's S° term be deleted because its ΔH°f is zero?

No.

The two properties use different reference conventions.

LESSON 5 · 25 MIN

Study + retrieve

Heat transfer through temperature and phase

Select q = mcΔT or q = CΔT, close a calorimeter energy balance, and sum sensible-heat and phase-change steps with the correct constraint.

ESSENTIAL QUESTIONWhich body gains or loses heat, does temperature change or phase change, and what energy quantity does the apparatus measure?

Heat-transfer ledger

Select the thermal equation
Segment	Relationship	Required data
Sensible heat, specific c	q = mcΔT	Mass and phase-specific c
Sensible heat, object C	q = CΔT	Whole-object heat capacity
Phase change	q = nΔH_phase	Moles and direction
Multistep path	Σq_step	Break at every phase boundary

Close the calorimeter balance
Model	Energy balance	Direct target
Isolated assembly	q_rxn + q_solution + q_cal = 0	Signed heat exchange
Coffee cup	Constant-pressure approximation	q_p = ΔH
Bomb	Constant volume	q_v = ΔE
Ideal phase plateau	Temperature constant	Latent heat, not mcΔT

ENERGY LEDGER · CAPTIONS INCLUDED

Choose the capacity named

Use q = mcΔT with specific heat capacity and q = CΔT with the heat capacity of a whole object. ΔT = Tfinal − Tinitial controls the sign, while a Celsius interval and kelvin interval have the same numerical size.

Specific heat needs mass
Whole-object C already includes amount

Close the calorimeter ledger

For the isolated modeled assembly, qrxn + qsolution + qcal = 0. A coffee cup approximates constant pressure and qₚ = ΔH, while a bomb calorimeter fixes volume and directly measures qᵥ = ΔE.

Signed terms sum to zero
Coffee cup: ΔH; bomb: ΔE

Break a phase path into segments

Use sensible heat within one phase and q = nΔHphase across an ideal phase-change plateau. Split at every boundary, use phase-specific capacities, preserve signs, and add all segment energies.

No ΔT during the ideal plateau
Sum signed segments

Worked example

How much heat warms 18.0 g of ice from −10.0 °C to 0.0 °C, melts it, then warms the water to 20.0 °C? Use cice = 2.09 J/(g·K), cwater = 4.184 J/(g·K), ΔHfus = 6.01 kJ/mol, and 18.0 g/mol.

1
Warm ice: q₁ = (18.0)(2.09)(10.0) = 376 J = 0.376 kJ.
2
Melt 1.00 mol: q₂ = (1.00)(6.01) = 6.01 kJ.
3
Warm water: q₃ = (18.0)(4.184)(20.0) = 1.51 kJ; total q = 7.89 kJ.

ConclusionTemperature stays at 0 °C during the ideal melting step. Applying one mcΔT expression across the full path would omit latent heat.

Close the notes first

Retrieve the model.

01When is mass multiplied by heat capacity?

When the supplied capacity is specific heat capacity.

A whole-object heat capacity already includes the object's amount.

02If the solution warms in an isolated coffee-cup model, what sign does qrxn have?

Negative.

The reaction loses the heat gained by solution and calorimeter.

03What happens to temperature during an ideal equilibrium phase-change plateau?

It remains constant.

Transferred energy changes the phase fraction rather than average kinetic energy.

All twenty Thermodynamics and Thermochemistry problems

Preserve the sign before calculating.

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 boundary, sign, state/path, Hess transformation, Gibbs condition, entropy evidence, or heat-transfer decision.

DAT TRAIN does not claim topic quotas because the official manual does not publish them. Chemical Equilibria owns Q/K composition; Chemical Kinetics owns rate laws and activation-energy calculations.