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AP02 · MIXED UNIT 6 PRACTICE

Connect heat, phase changes and reaction-energy evidence

Solve eight unfamiliar thermochemistry cases, compare competing models and distinguish calculation agreement from experimental support.

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Infer the apparatus capacity from independent pulses

A stipulated 120 g solution has specific heat 3.0 J/(g K). A known 1260 J electrical pulse gives a 3.0 K rise; a second 840 J pulse gives 2.0 K. With no external loss and constant capacities, enter the apparatus heat capacity in J/K. Explain why two agreeing pulses do not prove all later reaction runs are isolated.

Quantity / representation	Packet value
Solution	120 g; 3.0 J/(g K)
Pulses	1260 J / 3 K; 840 J / 2 K
Boundary	Exact teaching values; calibrated receiver includes solution and apparatus

Numerical response (units specified in the question) Your reasoning for Infer the apparatus capacity from independent pulses

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Keep reaction heat opposite to its receiving surroundings

A calibrated receiver has capacity 800 J/K. A reaction run changes temperature by −1.5 K; a matched additive blank changes it by +0.5 K. Reaction extent is 0.040 mol. Under constant pressure, only pressure-volume work and no uncharacterized heat exchange, enter molar reaction enthalpy in kJ/mol reaction. Explain the sign and why a mismatched blank would prevent this inference.

Quantity / representation	Packet value
Run / blank	−1.5 K / +0.5 K
Receiver capacity	800 J/K
Extent	0.040 mol of reaction as written

Numerical response (units specified in the question) Your reasoning for Keep reaction heat opposite to its receiving surroundings

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Account for material already in the vapor phase

An original pure-substance model has 2.5 mol total, initially 20% vapor, at phase equilibrium under the appropriate constant pressure. Stipulated molar vaporization enthalpy is 30 kJ/mol. Adding 22.5 kJ stays inside the phase interval. Enter the final vapor mole fraction and explain the final liquid/vapor amounts without interpreting the fraction as container volume.

Quantity / representation	Packet value
Initial inventory	2.5 mol; vapor fraction 0.20
Heat / molar phase energy	+22.5 kJ / 30 kJ/mol
Boundary	Closed variable-volume phase model; heat and amount stipulated exact

Numerical response (units specified in the question) Your reasoning for Account for material already in the vapor phase

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Distinguish a nominal phase endpoint from supported completion

One mole starts as liquid. Supplied heat is exactly 20 kJ and the molar vaporization reference is 20±1 kJ/mol, treated as an endpoint range. Which conclusion is supported? Calculate the possible algebraic fractions and name the information needed if all liquid is consumed before the heat input ends.

Quantity / representation	Packet value
Nominal result	20/(1×20)=1
Reference range	19–21 kJ/mol; no probability interpretation
Missing model	No single-phase heat capacity or post-transition state model

Your response

All material is certainly vapor because the central calculation equals oneReplace every fraction above one by one and discard the excess heatThe source range straddles completion; a phase-only model cannot certify the final state

Your reasoning for Distinguish a nominal phase endpoint from supported completion

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Reverse and scale an unfamiliar reaction-energy chain

Three hypothetical isomers X,Y,Z share an elemental formula. Supplied transformations are X→Y, ΔH=+25±2 kJ, and Y→Z, ΔH=−70±3 kJ, each per equation as written. Enter ΔH for 2Z→2X. Explain intermediate cancellation and the absolute endpoint envelope without assuming Gaussian independence.

Quantity / representation	Packet value
Source 1	X→Y: +25±2 kJ
Source 2	Y→Z: −70±3 kJ
Target	2Z→2X

Numerical response (units specified in the question) Your reasoning for Reverse and scale an unfamiliar reaction-energy chain

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Do not assign zero to every elemental form

Use supplied standard formation values CO₂(g)=−393.51 and CO(g)=−110.53 kJ/mol. A separate stipulated conversion C(graphite)→C(alternative solid) costs +1.90 kJ/mol. Enter the energy for 2CO(g)→CO₂(g)+C(alternative solid). Explain why elemental composition alone cannot justify replacing the alternative solid formation value by zero.

Quantity / representation	Packet value
Reference forms	Graphite and O₂(g) use the elemental zero convention
Gas formation inputs	CO₂ −393.51; CO −110.53 kJ/mol
Additional stipulated state change	Graphite→alternative solid: +1.90 kJ/mol

Numerical response (units specified in the question) Your reasoning for Do not assign zero to every elemental form

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Separate common temperature, molecular speed and total energy

Two separate ideal monatomic gas samples are at the same temperature and contact through a heat-conducting wall. One has particle mass 4 u and twice as many particles as the other, whose particle mass is 40 u. Which statement follows? Explain the net heat-transfer direction and distinguish temperature from a stored quantity called heat.

Quantity / representation	Packet value
Temperature	Same equilibrium temperature
Masses / counts	4 u with 2N particles; 40 u with N particles
Model	Ideal monatomic gases; equilibrium, no bulk motion

Your response

Equal temperature means identical particle speeds and identical total energy regardless of amountMean translational kinetic energy is equal; lighter particles move faster on average, and twice as many monatomic particles carry twice the ideal-gas internal energyThe gas with heavier particles must transfer net heat to the lighter gas at equal temperature

Your reasoning for Separate common temperature, molecular speed and total energy

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Limit a conclusion to the energy evidence actually supplied

A valid constant-pressure measurement for a stated reaction and conditions gives ΔH=−40 kJ per mole of reaction. No rate, activation-barrier, entropy or equilibrium data are supplied. Which conclusion is supported? Explain the sign without extending the claim to every temperature or pathway.

Quantity / representation	Packet value
Reported property	ΔH=−40 kJ/mol reaction under stated conditions
Work boundary	Only pressure-volume work
Absent evidence	Rate, barrier, entropy and equilibrium composition

Your response

The specified constant-pressure process releases energy to the receiving surroundings; its rate and universal spontaneity are not establishedNegative enthalpy proves rapid completion under every conditionThe negative value is the activation barrier and fixes the equilibrium composition

Your reasoning for Limit a conclusion to the energy evidence actually supplied

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Retain a snapshot for a human review

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Prepare Unit 6 text snapshotCurrent Unit 6 report textscience:ap-chemistry-unit6-review@1 PUBLIC PRACTICE REPORT — not an independent assessment or course-completion certificate. First/latest numerical or choice responses are retained; explanation text is the CURRENT EDITABLE DRAFT, not a locked first explanation. Keys are public. No automatic writing grade is assigned. No unrecorded key or reasoning is included below. unit6-calibration-inventory — Infer the apparatus capacity from independent pulses A stipulated 120 g solution has specific heat 3.0 J/(g K). A known 1260 J electrical pulse gives a 3.0 K rise; a second 840 J pulse gives 2.0 K. With no external loss and constant capacities, enter the apparatus heat capacity in J/K. Explain why two agreeing pulses do not prove all later reaction runs are isolated. Solution: 120 g; 3.0 J/(g K) Pulses: 1260 J / 3 K; 840 J / 2 K Boundary: Exact teaching values; calibrated receiver includes solution and apparatus Attempts: 0/3; known feedback exposure: no First response: No response Latest response: No response CURRENT EDITABLE EXPLANATION: unit6-cooling-blank-extent — Keep reaction heat opposite to its receiving surroundings A calibrated receiver has capacity 800 J/K. A reaction run changes temperature by −1.5 K; a matched additive blank changes it by +0.5 K. Reaction extent is 0.040 mol. Under constant pressure, only pressure-volume work and no uncharacterized heat exchange, enter molar reaction enthalpy in kJ/mol reaction. Explain the sign and why a mismatched blank would prevent this inference. Run / blank: −1.5 K / +0.5 K Receiver capacity: 800 J/K Extent: 0.040 mol of reaction as written Attempts: 0/3; known feedback exposure: no First response: No response Latest response: No response CURRENT EDITABLE EXPLANATION: unit6-initial-phase-inventory — Account for material already in the vapor phase An original pure-substance model has 2.5 mol total, initially 20% vapor, at phase equilibrium under the appropriate constant pressure. Stipulated molar vaporization enthalpy is 30 kJ/mol. Adding 22.5 kJ stays inside the phase interval. Enter the final vapor mole fraction and explain the final liquid/vapor amounts without interpreting the fraction as container volume. Initial inventory: 2.5 mol; vapor fraction 0.20 Heat / molar phase energy: +22.5 kJ / 30 kJ/mol Boundary: Closed variable-volume phase model; heat and amount stipulated exact Attempts: 0/3; known feedback exposure: no First response: No response Latest response: No response CURRENT EDITABLE EXPLANATION: unit6-boundary-uncertainty — Distinguish a nominal phase endpoint from supported completion One mole starts as liquid. Supplied heat is exactly 20 kJ and the molar vaporization reference is 20±1 kJ/mol, treated as an endpoint range. Which conclusion is supported? Calculate the possible algebraic fractions and name the information needed if all liquid is consumed before the heat input ends. Nominal result: 20/(1×20)=1 Reference range: 19–21 kJ/mol; no probability interpretation Missing model: No single-phase heat capacity or post-transition state model Attempts: 0/3; known feedback exposure: no First response: No response Latest response: No response CURRENT EDITABLE EXPLANATION: unit6-reverse-double-hess — Reverse and scale an unfamiliar reaction-energy chain Three hypothetical isomers X,Y,Z share an elemental formula. Supplied transformations are X→Y, ΔH=+25±2 kJ, and Y→Z, ΔH=−70±3 kJ, each per equation as written. Enter ΔH for 2Z→2X. Explain intermediate cancellation and the absolute endpoint envelope without assuming Gaussian independence. Source 1: X→Y: +25±2 kJ Source 2: Y→Z: −70±3 kJ Target: 2Z→2X Attempts: 0/3; known feedback exposure: no First response: No response Latest response: No response CURRENT EDITABLE EXPLANATION: unit6-nonreference-element — Do not assign zero to every elemental form Use supplied standard formation values CO₂(g)=−393.51 and CO(g)=−110.53 kJ/mol. A separate stipulated conversion C(graphite)→C(alternative solid) costs +1.90 kJ/mol. Enter the energy for 2CO(g)→CO₂(g)+C(alternative solid). Explain why elemental composition alone cannot justify replacing the alternative solid formation value by zero. Reference forms: Graphite and O₂(g) use the elemental zero convention Gas formation inputs: CO₂ −393.51; CO −110.53 kJ/mol Additional stipulated state change: Graphite→alternative solid: +1.90 kJ/mol Attempts: 0/3; known feedback exposure: no First response: No response Latest response: No response CURRENT EDITABLE EXPLANATION: unit6-temperature-particles — Separate common temperature, molecular speed and total energy Two separate ideal monatomic gas samples are at the same temperature and contact through a heat-conducting wall. One has particle mass 4 u and twice as many particles as the other, whose particle mass is 40 u. Which statement follows? Explain the net heat-transfer direction and distinguish temperature from a stored quantity called heat. Temperature: Same equilibrium temperature Masses / counts: 4 u with 2N particles; 40 u with N particles Model: Ideal monatomic gases; equilibrium, no bulk motion Attempts: 0/3; known feedback exposure: no First response: No response Latest response: No response CURRENT EDITABLE EXPLANATION: unit6-heat-claim-limits — Limit a conclusion to the energy evidence actually supplied A valid constant-pressure measurement for a stated reaction and conditions gives ΔH=−40 kJ per mole of reaction. No rate, activation-barrier, entropy or equilibrium data are supplied. Which conclusion is supported? Explain the sign without extending the claim to every temperature or pathway. Reported property: ΔH=−40 kJ/mol reaction under stated conditions Work boundary: Only pressure-volume work Absent evidence: Rate, barrier, entropy and equilibrium composition Attempts: 0/3; known feedback exposure: no First response: No response Latest response: No response CURRENT EDITABLE EXPLANATION: Reviewer follow-up: request a new case, inspect assumptions and representations, and assess the explanation separately. This text is an exported snapshot; later edits do not update it.

Source and model boundaries

All eight cases are original public transfer packets, using explicitly supplied values. The College Board CED, Unit 6 supplies the capability framework. The calibration, Hess construction and phase-energy activities explain source and model limits. Existing HS thermal lessons supply contextual support. No restricted exam content or newly measured experimental dataset is claimed.

A reviewer should request fresh work, inspect units, representations and assumptions, and assess the explanation separately. Fusion/sublimation breadth, authentic measurement and independent practical acceptance remain open. This public review does not certify full-unit mastery.

Return to the separate Unit 5 thermochemistry review.

Continue to the separate Unit 7 equilibrium review.

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