Worked example 1: solution

Step 2 · Plan the route

One arrow links the units: cal → J. One conversion factor is needed.

Worked example 1: solution

The factor takes cal in the denominator, so cal cancels and J survives:

Worked example 1: solution

Worked example 1: the route on the map

Worked example 2: solution

Step 2 · Plan the route

No single equality links kJ to cal. The route runs through the joule: kJ → J → cal. Two conversion factors are needed.

Worked example 2: solution

The prefix factor cancels kJ; the calorie factor takes J in the denominator:

Worked example 2: solution

Worked example 2: the route on the map

Practice 1 · answer: B

A flipped the 4.184 factor: 6.20 × 10³ × 4.184 = 2.59 × 10⁴, and no unit cancels. C stopped after the prefix factor: 6.20 × 1000 = 6.20 × 10³, a count of joules, not calories. D treated kilojoules as joules: 6.20 / 4.184 = 1.48, a thousand times too small.

Worked example 3: solution

A common first attempt

Worked example 3: solution

A common first attempt

Worked example 3: the correct chain

Step 2 · Plan the route

The capital C marks the food Calorie: 1 Cal = 1000 cal. The route: Cal → cal → J → kJ. Three conversion factors are needed.

Worked example 3: the correct chain

Worked example 3: the correct chain

Worked example 3: the route on the map

Practice 2 · answer: D

A read Cal as cal: 340. × 4.184 / 1000 = 1.42, a thousand times too small. B flipped the 4.184 factor: 340. × 1000 / 4.184 / 1000 = 81.3, and the joule never cancels. C stopped at Calories: 340. is the energy in Cal, never converted to kJ.

Guided example: solution

Step 1 · Sign each quantity

The system is the gas. Heat leaves it, so q = −450 J. The compressor pushes on it, a compression, so w = +310 J.

Guided example: solution

Step 1 · Sign each quantity
Step 2 · Solve ΔE = q + w for the unknown

ΔE is the unknown, so add the two signed values:

Guided example: solution

Step 1 · Sign each quantity
Step 2 · Solve ΔE = q + w for the unknown

ΔE is negative: more energy left the gas than entered it.

Guided example: solution

Step 1 · Sign each quantity
Step 2 · Solve ΔE = q + w for the unknown

Practice 1 · answer: B

A flipped the heat sign: (+165) + (−58) = +107 J. C reversed both signs: 165 + 58 = +223 J, the surroundings' ΔE, not the gas's. D subtracted the work: (−165) − (−58) = −107 J.

Practice 1 · answer: B

A flipped the heat sign: (+165) + (−58) = +107 J. C reversed both signs: 165 + 58 = +223 J, the surroundings' ΔE, not the gas's. D subtracted the work: (−165) − (−58) = −107 J.

Worked example: solution

Step 1 · Sign each quantity

The energy fell: ΔE = −150 J. Work is done on the system: w = +240 J. The unknown is q.

Worked example: solution

Worked example: solution

Worked example: solution

Practice 2 · answer: C

A subtracted in reverse: q − ΔE = (−75) − (+115) = −190 J, the work read from the surroundings' side. B flipped the heat sign: 115 − 75 = +40 J, as if the steam had absorbed the heat. D treated ΔE as zero: w = −q = +75 J, true only for a cycle back to the start.

Practice 2 · answer: C

A subtracted in reverse: q − ΔE = (−75) − (+115) = −190 J, the work read from the surroundings' side. B flipped the heat sign: 115 − 75 = +40 J, as if the steam had absorbed the heat. D treated ΔE as zero: w = −q = +75 J, true only for a cycle back to the start.

Practice 3 · answer: D

A flipped the sign of path 2's heat: 220 + 310 = +530 J, as if the 310 J had left the gas. B stopped at ΔE: +220 J is the energy change both paths share, not path 2's work. C copied path 1's work, +50 J: w is a path function and changes with the route.

Practice 3 · answer: D

A flipped the sign of path 2's heat: 220 + 310 = +530 J, as if the 310 J had left the gas. B stopped at ΔE: +220 J is the energy change both paths share, not path 2's work. C copied path 1's work, +50 J: w is a path function and changes with the route.

Extra example 1: solution

Step 1 · Sign each quantity

Heat flows into the system, so q = +125 J. The system does work on the surroundings (expansion), so w = −78 J.

Extra example 1: solution

Step 1 · Sign each quantity
Step 2 · Solve ΔE = q + w for the unknown

Extra example 1: solution

Step 1 · Sign each quantity
Step 2 · Solve ΔE = q + w for the unknown

Extra example 1: solution

Step 1 · Sign each quantity
Step 2 · Solve ΔE = q + w for the unknown

Extra example 2: solution

Part (a) · expansion against a vacuum

Extra example 2: solution

In liters, ΔV = 0.971 − 0.264 = +0.707 L. The external pressure is constant, so w = −P·ΔV, and the L·atm converts to joules:

Extra example 2: solution

Extra practice · answer: A

B dropped the minus sign in w = −P·ΔV: 3.0 × (−0.631) × 101.325 = −1.9 × 10² J, which would say the squeezed gas lost energy. C stopped at +1.89 L·atm and wrote it as joules; only the 101.325 J per L·atm factor makes joules. D turned the conversion factor upside down: 1.89 / 101.325 = 1.9 × 10⁻² J.

Worked example 1: solution

Step 1 · Name the system

The iron and oxygen are the system. The pouch, the air, your cold hands: surroundings.

Worked example 1: solution

ΔH is negative: −1648 kJ. Negative marks heat leaving the system.

Worked example 1: solution

Worked example 1: solution

Worked example 1: the clue used

Worked example 2: solution

Step 1 · Name the system

The decomposing peroxide is the system; the cut, the skin, the air are surroundings.

Worked example 2: solution

Step 1 · Name the system
Step 2 · Find the heat's direction

The products sit 196 kJ below the reactants. That difference left the system as heat.

Worked example 2: solution

Step 1 · Name the system
Step 2 · Find the heat's direction
Step 3 · State the verdict

Worked example 2: solution

Step 1 · Name the system
Step 2 · Find the heat's direction
Step 3 · State the verdict

Worked example 2: the clue used

Practice 1 · answer: C

A flipped the sign convention: heat absorbed would be counted into the system, +198 kJ, not −198 kJ. B paired the right label with the wrong flow: exothermic means heat exits the system. D paired the right flow with the wrong label: a heat-releasing reaction is exothermic.

Worked example 3: solution

A common first answer

The cold skin is the evidence. Your hand is losing heat, and the hand is surroundings, not system.

Worked example 3: solution

The dissolving salt and water are the system. The pouch, your hand: surroundings.

Worked example 3: solution

Your hand cools: heat is leaving the surroundings and entering the system.

Worked example 3: solution

Worked example 3: solution

Practice 2 · answer: A

B read a product-side heat term as heat gained: it leaves with the products, so it is released, −65 kJ, not +65 kJ. C matched the sign but not the label: a positive ΔH would mean heat in, and a bucket cannot turn hot from that. D matched the label but not the sign: endothermic and ΔH = −65 kJ contradict each other.

Extra practice 1 · answer: C

A took the freezing as the reaction's own change. Freezing does release heat, but the water is surroundings, and its heat went into the flask. B paired the label with the wrong flow: heat into the system is endothermic. D read the sign from the surroundings: the board loses heat, the system gains it, so ΔH is positive.

Extra practice 2 · answer: A

Three conversion factors are needed: kcal → cal → J → kJ.

B read the heat term from the wrong side: it sits with the reactants, so it is absorbed, +182 kJ. C flipped the 4.184 factor: 43.4 ÷ 4.184 = 10.4. D read kcal as cal: 43.4 × 4.184 = 182 J, only 0.182 kJ. E skipped the conversion: 43.4 is the heat in kcal, not kJ.

Extra practice 3 · answer: D

A paired the label with the wrong sign: heat out is exothermic. B signed the work as done on the system: −154 + 2.48 = −152 kJ. The H₂ pushed the air back, so that energy left. C read the heat sign from the surroundings: +154 − 2.48 = +152 kJ. The room gains the heat; the system loses it.

Worked example 1: solution

Step 1 · List the pieces

m = 250 g. c = 4.184 J/g·°C. ΔT = 50.0 − 20.0 = +30.0 °C. The unknown is q.

Worked example 1: solution

q already stands alone on its side of the equation; no rearranging is needed.

Worked example 1: solution

Worked example 1: solution

Worked example 1: solving for q

Worked example 2: solution

Step 1 · List the pieces

q = 1347 J. m = 125 g. ΔT = 46.0 − 22.0 = +24.0 °C. The unknown is c.

Worked example 2: solution

Divide both sides by m·ΔT before any numbers go in.

Worked example 2: solution

No unit cancels here; they assemble into J/g·°C: the unit of a specific heat.

Worked example 2: solution

Worked example 2: identify the metal

Match the property

Only iron matches 0.449 J/g·°C. The block is iron.

Worked example 2: identify the metal

Step 4 · Check the sign

Worked example 2: solving for c

Practice 1 · answer: C

A left out the mass: 0.385 × 20.0 = 7.70 J warms one gram. B left out the temperature change: 75.0 × 0.385 = 28.9 J is one degree's worth. D divided by the specific heat: 75.0 × 20.0 ÷ 0.385 = 3.90 × 10³, with units g²·°C²/J.

Worked example 3: solution

A common first attempt

A lid that is releasing heat cannot end up hotter. The sign of q was dropped.

Worked example 3: solution

Released heat leaves the sample, so q = −8100 J. m = 150 g. c = 0.897 J/g·°C. The unknown is ΔT, then T_final.

Worked example 3: solution

Worked example 3: final temperature

Step 3 · Substitute and cancel units

Worked example 3: final temperature

Worked example 3: solving for ΔT, then T_final

Practice 2 · answer: B

A subtracted the temperatures in the wrong order: 62.0 − 22.0 = +40.0 °C gives +512 J, heat absorbed by a cooling sinker. C stopped at m × c: 100. × 0.128 = 12.8 J, one degree's worth with no sign. D left out the mass: 0.128 × (−40.0) = −5.12 J, the heat for a single gram.

Practice 3 · answer: D

A left q in kilojoules: ΔT = 0.726 °C, so 27.7 °C. B subtracted T_initial: 725.6 − 27.0 = 699 °C; T_final = T_initial + ΔT. C stopped at ΔT: 726 °C is the change, not the final temperature. E multiplied by c: 2340 × 0.129 ÷ 25.0 = 12.07 °C, then 39.1 °C.

Practice 4 · answer: C

c is per gram, so the kilograms convert inside the chain: 2.0 kg is 2000 g. A used 2.0 as if it were grams: 2.0 × 0.89 × 157.0 = 280 J, a thousand times short. B left out the temperature change: 2000 × 0.89 = 1.8 × 10³ J, one degree's worth. D used 180.0 as ΔT: 2000 × 0.89 × 180.0 = 3.2 × 10⁵ J. ΔT is final minus initial: 157.0 °C.

Worked example: solution

Step 1 · Compare the specific heats

Copper's c, 0.385, is smaller than aluminum's, 0.897. But mass is in the denominator too, and the copper sample is four times heavier.

Worked example: solution

Compare the whole sample's m·c: copper 200 × 0.385 = 77.0 J/°C, aluminum 50.0 × 0.897 = 44.85 J/°C. Aluminum's is smaller, so aluminum swings more.

Worked example: solution

Worked example: solution

Worked example: the route on the strip

Practice · answer: A

B compared c alone: "smaller c wins" holds only at equal masses, and the lead block is heavier. C equal heat does not mean equal ΔT: the two samples' m·c differ. D computed both rises and pinned them on the wrong metals.

Extra practice 1 · answer: B

A took equal heat as an equal rise: 36.0 °C holds only when the specific heats match. C flipped the ratio: 36.0 × 0.385 / 0.128 = 108, but the larger c gives the smaller rise. D stopped halfway: 0.128 × 36.0 = 4.61 is the heat per gram, in J/g, not copper's rise.

Extra practice 2 · answer: A

B stopped in joules: 321 J is the right gap in the wrong unit. C put the factor upside down: 321 × 4.184 = 1.34 × 10³. D never subtracted silver's heat: 673.5 / 4.184 = 161 cal is iron's whole heat.

Extra practice 3 · answer: D

A took equal heat as an equal rise: that holds only when m·c matches, and 62.86 is not 35.88. B compared c alone: iron's 140. g makes its m·c the larger, so it rises only 23.9 °C. C stopped at the rise: 41.8 °C is how far aluminum climbs from 22.0 °C, not where it ends.

Worked example 1: solution

Step 1 · Name the phase change

The ice is melting: solid water turns to liquid, all at 0 °C.

Worked example 1: solution

Melting uses the heat of fusion, ΔH_fus = 335 J/g.

Worked example 1: solution

Write the factor so grams cancel. Only one orientation does:

Worked example 1: solution

Worked example 1: the route on the map

Worked example 2: solution

Step 1 · Name the phase change

The water is boiling: liquid turns to gas, all at 100 °C.

Worked example 2: solution

Boiling uses the heat of vaporization, ΔH_vap = 2259 J/g.

Worked example 2: solution

Grams cancel; joules remain.

Worked example 2: solution

Worked example 2: the route on the map

Practice 1 answer: A

The heat is given, so ΔH_vap enters upside down: J on the bottom, so joules cancel. B used the heat of fusion: 5.00 × 10⁴ ÷ 335 = 149 g is the ice this heat would melt, not the water it boils. C used water's specific heat: 5.00 × 10⁴ ÷ 4.184 = 1.20 × 10⁴, but a plateau has no ΔT, so c does not apply. D kept the factor right side up: 5.00 × 10⁴ × 2259 = 1.13 × 10⁸, in J²/g, nothing cancels.

Worked example 3: solution

A common first attempt

The steam condenses at 100 °C into water at 100 °C. ΔT = 0, so the m·c·ΔT term adds nothing.

Worked example 3: solution

A common first attempt

The steam condenses at 100 °C into water at 100 °C. ΔT = 0, so the m·c·ΔT term adds nothing.
Step 1 · Name the phase change

The steam is condensing: gas turns to liquid, all at 100 °C.

Worked example 3: solution

A common first attempt

The steam condenses at 100 °C into water at 100 °C. ΔT = 0, so the m·c·ΔT term adds nothing.
Step 1 · Name the phase change
Step 2 · Pick its ΔH

Condensing uses the heat of vaporization, ΔH_vap = 2259 J/g, the same value as boiling.

Worked example 3: heat released

Step 3 · Multiply

Worked example 3: heat released

Heat leaves the steam as it condenses, so q is negative: q = −18,072 J.

Worked example 3: the route on the map

Worked example 4: solution

Step 1 · Name the phase change

The ice is melting at 0 °C, the same change as ever. Only the units of its heat are new: kilojoules per mole.

Worked example 4: solution

Melting uses the heat of fusion: ΔH_fus = 6.02 kJ/mol.

Worked example 4: solution

Grams cancel into moles, moles cancel into kilojoules.

Worked example 4: solution

The per-gram route on the same sample: 60.0 g × 335 J/g = 20,100 J = 20.1 kJ. Two routes, one heat, agreeing to the rounding of the constants.

Worked example 4: the route on the map

Practice 2 answer: B

The heat is given, so ΔH_fus enters upside down and kJ cancels. A picked the vaporization heat: 15.0 ÷ 40.7 × 18.02 = 6.64 g, but no boiling or condensing happens here. C divided by the molar mass: 2.49 ÷ 18.02 = 0.138 g. D stopped at moles: 2.49 mol of water, one factor short of grams.

Practice 3 answer: D

A added the two heats: 63,252 + 9,380 = 7.26 × 10⁴ J is the cost of both changes, not the gap. B divided the heats: 2259 ÷ 335 = 6.74 says how many times more, not how many joules more. C stopped at the boiling heat: 28.0 × 2259 = 6.33 × 10⁴ J, with the melt never subtracted.

Practice 4 answer: A

Two conversion factors are needed, and the first one is ethanol's molar mass: 2(12.01) + 6(1.008) + 16.00 = 46.07 g/mol.

B skipped the molar mass: 62.0 × 38.6 = 2.39 × 10³, as if each gram were a mole. C used water's molar mass: 62.0 ÷ 18.02 × 38.6 = 133, but the liquid is ethanol. D put the molar mass upside down: 62.0 × 46.07 × 38.6 = 1.10 × 10⁵, and grams do not cancel.

Practice 5 answer: C

A stopped at ΔT: 6.41 °C is the rise, not the final temperature. B warmed the ice's 32.0 g: 21.0 + 10,720 ÷ (32.0 × 4.184) = 101.1, but the 400. g of water takes the heat. D used the heat of vaporization: 32.0 × 2259 = 72,288 J, giving 64.2, but nothing boils.

Worked example 1: solution

Step 1 · Identify each segment

The ice sits at its melting point. Two pieces follow: melt at a constant 0 °C, then warm the liquid to 25 °C.

Worked example 1: solution

Worked example 1: solution

Worked example 1: the route on the curve

Worked example 2: the melting step

A common first attempt

Melting happens at a constant 0 °C. There is no temperature change to multiply, so ΔHfus already gives the whole melting heat.

Worked example 2: the melting step

Two pieces: melt at 0 °C, then warm the liquid from 0 °C to 40 °C. Only the warming piece has a ΔT.

Worked example 2: the two pieces

Step 2 · Compute each piece

Worked example 2: the two pieces

Worked example 2: the route on the curve

Worked example 3: warming and melting

Step 1 · Identify each segment

Five pieces: warm ice (−20 → 0), melt at 0 °C, warm water (0 → 100), boil at 100 °C, warm steam (100 → 120). Three slopes and two plateaus.

Worked example 3: warming and melting

Worked example 3: boiling, then the total

Step 2 · Compute each piece

Worked example 3: boiling, then the total

Worked example 3: the route on the curve

Practice 1 answer: C

A skipped the melting plateau: 20.0 × 4.184 × 30.0 = 2.51 × 10³ J warms the liquid but never melts the ice. B stopped at the melting point: 20.0 × 335 = 6.70 × 10³ J leaves the water at 0 °C. D multiplied the plateau by a ΔT: 20.0 × 335 × 30.0 = 2.01 × 10⁵ J.

Practice 2 answer: D

A lost the sign: a cooling path releases heat, so q is negative. B skipped the condensation plateau: −285 − 1255 = −1.54 × 10³ J. C stopped at 100 °C: −285 − 22,590 = −2.29 × 10⁴ J.

Practice 3 answer: D

A skipped the cooling leg: −16,750 J = −1.68 × 10⁴ J. B lost the sign: the water releases heat, so q is negative. C skipped the freezing plateau: −4.60 × 10³ J. E cooled the liquid with ice's c: 50.0 × 2.03 × (−22.0) − 16,750 = −1.90 × 10⁴ J.

Practice 4 answer: C

A skipped the melt: 107,875 − 11,725 = 96.2 kJ. B used 110.0 °C as the steam's ΔT: 35.0 × 1.9 × 110.0 = 7315 J, total 115 kJ. D left the total in joules: 1.08 × 10⁵. E stopped before boiling: 1776 + 11,725 + 14,644 = 28.1 kJ.

Practice 5 answer: B

A skipped the ice leg: 6625 ÷ 104.6 = 63.3. C skipped the melt: 14,492.5 ÷ 104.6 = 139, past the boiling point. D started the water at −10.0 °C: −10.0 + 58.5 = 48.5. E warmed the ice with water's c: 5579 ÷ 104.6 = 53.3.

Worked example 1: solution

Step 1 · Set the balance

Each body keeps its own ΔT = T_f − T_i: 150 · (T_f − 70) = −100 · (T_f − 20). The hot side's ΔT will come out negative; the minus sign hands its released heat to the cold side.

Worked example 1: solution

The balance point is the mass-weighted average of the two starting temperatures. No distributing, no collecting terms: the masses weight the two starts directly.

Worked example 1: solution

Worked example 1: the route on the map

Worked example 2: solution

A common first attempt

Same signs on both heats is the slip. One body loses, the other gains.

Worked example 2: solution

Worked example 2: the final temperature

Step 2 · Solve without expanding

Worked example 2: the route on the map

Worked example 3: solution

Two moves are needed.

Step 1 · Heat gained by the solution

Worked example 3: solution

Worked example 3: the route on the map

Worked example 4: solution

One heat capacity for the whole rig

The bomb, the water, the walls: C_cal covers all of it at once, 9.90 kJ per degree. No mass and no specific heat are needed; C_cal already contains them.

Worked example 4: solution

Worked example 4: solution

Practice 1 · answer: A

B swapped the masses onto the wrong specific heats: 40.3 °C. C dropped the specific heats and weighted by mass alone: (60.0 × 120.0 + 125 × 25.0) ÷ 185 = 55.8 °C, a shortcut that works only when both bodies are water. D took the plain average of the starts: (120.0 + 25.0) ÷ 2 = 72.5 °C, as if the two bodies weighed the same and shared one c.

Practice 2 · answer: D

A dropped the minus in q_metal = −q_water: 2761 ÷ (95.0 × (−64.5)) = −0.451, a negative specific heat. B stopped at q_water: 2761 J, before dividing by the metal's m·ΔT. C used the final temperature as the metal's ΔT: 2761 ÷ (95.0 × 34.5) = 0.843. E swapped the two masses: −(95.0 × 4.184 × 16.5) ÷ (40.0 × (−64.5)) = 2.54.

Practice 3 · answer: A

B stopped at the heat: 18.4 kJ is q, not a calorimeter constant. C skipped the mole conversion: 0.562 × 393.5 ÷ 1.19 = 186 treats the grams as moles. D divided by the final temperature: 18.4 ÷ 27.93 = 0.659; the divisor is ΔT = 1.19 °C.

Worked example 1: solution

Step 1 · Match the target to the given

The target runs the other way: CO₂ breaks apart into C and O₂. It is the reverse.

Worked example 1: solution

Worked example 2: solution

Step 1 · Match the target to the given

The target is reversed and every coefficient is doubled. Two changes, so ΔH takes two steps.

Worked example 2: solution

Worked example 3: solution

Two conversion factors are needed.

Step 1 · Pick the ΔH orientation

Two orientations exist. Only the one canceling mol CH₄ is used: −890 kJ / 1 mol CH₄, not 1 mol CH₄ / −890 kJ.

Worked example 3: solution

Worked example 3: the route on the map

Worked example 4: solution

Step 1 · Write ΔH per mole of SO₂

−198.2 kJ is the heat for the equation as written, which burns 2 mol of SO₂. Per mole of SO₂ that is −198.2/2 = −99.1 kJ. Keeping −198.2 per mole doubles the answer to −272 kJ: wrong.

Worked example 4: solution

Worked example 4: the route on the map

Practice 1 · answer: B

A scaled but kept the sign: 2 × 180 = +360 kJ. C flipped the sign but did not scale: −180 kJ. D left ΔH untouched at +180 kJ.

Practice 2 · answer: A

B kept 1203 kJ per 1 mol Mg: 2.50 × 10² ÷ 1203 × 24.31 = 5.05 g, half the true mass, because the equation burns 2 mol. C stopped at moles: 2.50 × 10² × 2/1203 = 0.416 mol Mg, one factor short of grams. D divided by the molar mass instead of multiplying: 0.416 ÷ 24.31 = 0.0171 g.

Practice 3 · answer: D

A flipped the ΔH factor: 0.306 mol × (2 mol / −10,941 kJ) = −5.60 × 10⁻⁵, and mol never cancels. B stopped at moles: 35.0 ÷ 114.22 = 0.306 mol, one factor short of kilojoules. C dropped the sign: combustion releases heat, so ΔH is negative. E used −10,941 kJ per 1 mol: 0.306 × (−10,941) = −3.35 × 10³ kJ, double the answer, because the equation burns 2 mol.

Worked example 1: solution

Step 1 · Match each given to the target

C(s) is a reactant in (1): keep (1) as written. CO(g) must end up a product, but (2) consumes it: reverse (2).

Worked example 1: solution

Worked example 1: solution

Worked example 1: the route on the map

Match on the map

(1) has C(s) on the reactant side, one mole: it passes both questions. (2) has CO(g) on the wrong side: it reverses.

Worked example 1: the route on the map

Only (2) is reversed: −283.0 kJ becomes +283.0 kJ. Every coefficient already matches, so nothing is scaled.

Worked example 1: the route on the map

Only (2) is reversed: −283.0 kJ becomes +283.0 kJ. Every coefficient already matches, so nothing is scaled.

Worked example 2: solution

Step 1 · Match each given to the target

The target needs 2 CO as reactants: (2) makes CO, so reverse it and double it. The target needs 2 CO₂ as products: double (1).

Worked example 2: solution

Worked example 2: solution

Worked example 2: the route on the map

Practice 1 · answer: B

Neither given needs reversing or scaling. A dropped the minus on (2): 180.5 + 114.1 = +294.6. C stopped at the first given: +180.5 kJ never adds step (2). D flipped (1) instead: −180.5 − 114.1 = −294.6.

Worked example 3: rearranging the givens

Step 1 · Match each given to the target

C(s): reactant in (1), keep it. 2 H₂: double (2). CH₄ must end up a product: reverse (3).

Worked example 3: rearranging the givens

Worked example 3: adding the three equations

Step 3 · Add and cancel

Worked example 3: adding the three equations

Step 3 · Add and cancel

Worked example 4: solution

Step 1 · Match each given to the target

The target needs 2 C as reactants: double (1). One H₂: keep (2). C₂H₂ must end up a product, one mole of it: reverse (3) and halve it.

Worked example 4: solution

Worked example 4: solution

Worked example 4: the route on the map

Practice 2 · answer: A

B stopped after (1): +534.2 kJ never brings in the hydrogen. C reversed (1) but never halved it: 1068.4 − 483.6 = +584.8, and 2 H₂O and O₂ are left over. D halved (1) but never flipped its sign: −534.2 + (−483.6) = −1017.8. E never doubled (2): 534.2 − 241.8 = +292.4.

Practice 2 · answer: A

Practice 3 · answer: C

A never halved (1): −4116.0 + 2219.9 − 285.8 = −2181.9. B reversed (2) but kept its sign: −2058.0 − 2219.9 − 285.8 = −4563.7. D reversed the propene equation instead of the propane one: +2058.0 − 2219.9 − 285.8 = −447.7. E stopped before equation (3): −2058.0 + 2219.9 = +161.9, and H₂ never enters.

Practice 3 · answer: C

Practice 4 · answer: B

A added the givens unscaled, then divided: −907 − 113 − 139 = −1159 kJ, and −1159 kJ ÷ 4 = −290. kJ/mol. C divided by the 2 in equation (3) instead of the 4 in the sum: −762 kJ/mol. D never divided: −1524 kJ, or −1,520 to three figures, is ΔH for four moles.

Worked example 1: solution

Elements in their most stable forms

The reactants are N₂(g) and H₂(g), never lone N or H atoms.

Worked example 1: solution

Guided example: solution

Step 1 · Balance the equation with states

Four Al and six O on each side. The (s) on Al₂O₃ picks its row.

Guided example: solution

Al(s) and O₂(g) are elements in their standard states: zero. Only Al₂O₃(s) carries a value.

Guided example: solution

Guided example: solution

Worked example 2: solution

Step 1 · Balance the equation with states
Step 2 · Look up every ΔH°f

The equation is balanced, and liquid water picks the −285.8 row. O₂ is an element in its standard state: zero.

Worked example 2: solution

Worked example 2: solution

Practice 1 · answer: D

A starts from free gaseous atoms, not C(s, graphite), H₂(g) and O₂(g). B starts from CO, a compound: industry makes methanol this way, but its ΔH is not ΔH°f. C makes 2 mol, so its ΔH is 2 × (−238.7) = −477.4 kJ. E uses diamond, not carbon's most stable form.

Practice 2 · answer: C

A dropped the coefficient 2 on NaHCO₃: −1766.0 − (−950.8) = −815.2. B stopped at the products' sum: −1766.0, before subtracting the reactants. D subtracted reactants minus products: −135.6.

Worked example 3: solution

Write the formula with x

Worked example 3: solution

Worked example 4: solution

Step 1 · Balance the equation with states
Step 2 · Look up every ΔH°f

The states pick the rows. Al(s) is aluminum's standard state: zero. Fe(l) is not iron's: melting costs energy, so it carries +12.40. Calling it zero gives −847.6 kJ, wrong.

Worked example 4: solution

Worked example 4: solution

Worked example 4: per gram of aluminum

Convert the reaction total to a per-gram figure

Practice 3 · answer: B

A wrote reactants minus products: −3119.4 = 2x − (−3288.8) gives x = −3204.1. C stopped at 2x: −169.4 kJ belongs to 2 mol of ethane. D multiplied by the 2 instead of dividing: 2 × (−169.4) = −338.8.

Practice 4 · answer: C

A took the liquid-water row against the equation's (g): −3535.5. B stopped at the products' sum: 5 × (−393.5) + 6 × (−241.8) = −3418.3, before subtracting the reactant. D subtracted reactants minus products: +3271.5.