Chemistry · Chapter 5

Thermochemistry

Every chemical change moves energy. Burn methane and the room gets warm. Open a cold pack and your hand gets cold. Thermochemistry is the part of chemistry that tracks that energy. In this chapter you learn to say how much heat a reaction gives up or takes in. You learn to measure it with a cup of water and a thermometer. Then you learn three ways to predict it on paper. One adds up known steps. One looks up each compound in a table. One counts the bonds that break and form. By the end, a number like −890.5 kJ will tell you a whole story.

By the end of this chapter you can

Tell heat apart from work, and give each one the right sign.
Use the first law, ΔE = q + w, on a system you choose yourself.
Find heat with q = mcΔT, and read the result from a calorimeter.
Write a thermochemical equation, then scale its ΔH to any mass.
Flip, scale and add known steps with Hess's law to reach a target.
Apply products minus reactants to standard formation enthalpies.
Estimate a reaction enthalpy from average bond energies, and say why it is only an estimate.

Section 1Energy, heat and work

Two kinds of energy

Energy is the ability to do work or to warm something up. It shows up in two broad kinds. Kinetic energy is the energy of motion. A rolling ball has it. So does a single molecule darting across a flask. You find it with KE = ½mv², so speed counts twice over. Potential energy is stored energy. It depends on where a thing sits, or on what it is made of. A book on a high shelf holds it. So do the bonds in a tank of gasoline.

Heat is not a third kind of energy that a thing owns. Thermal energy is the energy of random particle motion inside a sample. Temperature reports how fast those particles move, on average. Heat is thermal energy on the move, from a warm place to a cooler one. A hot pan does not hold heat. It holds thermal energy, and it gives off heat.

Draw a box around the problem

Before you can count energy, you must say what you are counting. The system is the part you care about. The surroundings are everything else in the universe. For a reaction in a beaker, the chemicals are the system. The glass, the air and your hand are the surroundings. A closed system can trade energy but not matter. An isolated system trades neither one.

Energy is never destroyed. The law of conservation of energy says it only moves or changes form. Whatever the system gains, the surroundings gave up. Written for a chemical system, that idea is the first law of thermodynamics.

ΔE = q + w

Here ΔE is the change in the system's internal energy, the total of all its kinetic and potential energy. The letter q stands for heat. The letter w stands for work, which in chemistry is usually a gas pushing on its container or getting squeezed.

Getting the signs right

Both q and w carry a sign, and the sign always describes the system. Think of a bank account. Money coming in is positive. Money going out is negative.

q is positive when heat flows into the system. That change is endothermic.
q is negative when heat flows out of the system. That change is exothermic.
w is positive when the surroundings do work on the system, such as squeezing a gas.
w is negative when the system does work on the surroundings, such as a gas expanding.

The SI unit of energy is the joule, J. One joule is one kg·m²/s², which is tiny. A calorie is larger, and it equals 4.184 J by definition. The Calorie on a food label is really a kilocalorie, so one of them is 4184 J.

Worked example · signs and the first law

A gas takes in 55 J of heat. It expands and does 25 J of work on the air outside. What is ΔE for the gas?

Step 1 · Name the system.The gas is the system. The air outside is the surroundings.

Step 2 · Set each sign.Heat flows in, so q = +55 J. The gas pushes outward, so work leaves it and w = −25 J.

Step 3 · Add them.ΔE = (+55 J) + (−25 J) = +30 J.

Step 4 · Check it.The gas took in more than it spent. Its internal energy should rise, and +30 J says it rose by 30 J.

Figure 5.1 Heat and work cross the boundary in both directions. Arrows pointing into the system are positive. Arrows pointing out are negative. Add the two signed amounts and you have ΔE.

Worked example · food Calories to joules

A snack bar lists 240 Calories. How many joules is that?

Step 1 · Fix the unit.A food Calorie is a kilocalorie. So the bar holds 240 kcal.

Step 2 · Convert.240 kcal × 4184 J/kcal = 1,004,160 J.

Step 3 · Round.Three significant figures gives 1.00 × 10⁶ J, or 1.00 × 10³ kJ.

Key points

Heat is energy in transit. No sample stores heat, it stores thermal energy.
Choose the system first. Every sign in the problem hangs on that choice.
ΔE = q + w. Energy in is positive, energy out is negative.
1 cal = 4.184 J exactly, and one food Calorie equals 1 kcal, or 4184 J.

Interactive Move energy between system and surroundings and track the signs. Open full ↑

Section 2Heat capacity and calorimetry

Why water fights back

Put the same amount of heat into a steel pan and into a pot of water. The pan gets hot fast. The water barely budges. That difference has a name. Heat capacity, C, is the heat needed to raise a whole object by one degree. Its unit is J/°C. It depends on how much stuff you have, so a big pot has a larger value than a small one.

Specific heat capacity, c, is the heat needed to raise one gram by one degree. Its unit is J/g·°C. This one is a property of the substance itself, so you can look it up. Molar heat capacity is the same idea per mole instead of per gram.

Water has a specific heat of 4.184 J/g·°C. That is high for a liquid. Hydrogen bonds soak up energy before the molecules speed up much. The result shapes daily life. Lakes warm and cool slowly. Your blood holds your body near 37 °C. Car engines use water to carry heat away.

Substance	c (J/g·°C)	What that means
Water (liquid)	4.184	Hard to heat, hard to cool
Ethanol	2.44	Warms about twice as fast as water
Aluminum	0.897	Light metal, moderate for a metal
Iron	0.449	A pan heats quickly
Copper	0.385	Good at moving heat along
Silver	0.233	Heats fast
Lead	0.130	Very small, so it heats very fast
Gold	0.129	The smallest on this list

The equation that does the work

Heat, mass, specific heat and temperature change all tie together in one line.

q = mcΔT, where ΔT = T_final − T_initial

Write ΔT as final minus initial every time. Then the sign takes care of itself. A sample that warms up has a positive ΔT and a positive q, so it gained heat. A sample that cools has a negative ΔT and a negative q. One more bonus: a change of one degree Celsius is the same size as a change of one kelvin. So you may use either unit for ΔT.

Measuring heat in a cup

Calorimetry is the practice of measuring heat by watching a temperature change. The device is a calorimeter. The simplest one is a foam cup of water with a lid and a thermometer. Foam is a poor conductor, so very little heat escapes. The cup sits open to the room, so pressure stays constant.

A bomb calorimeter is the heavy-duty version. A sealed steel vessel holds the sample and pure oxygen. The whole vessel sits in a water bath. Because the volume cannot change, it measures heat at constant volume instead.

Either way the logic is the same. If no heat leaks out, the heat one part loses is the heat the other part gains.

q_hot = −q_cold

Worked example · heating water

How much heat warms 125 g of water from 22.0 °C to 95.0 °C?

Step 1 · Find ΔT.ΔT = 95.0 °C − 22.0 °C = 73.0 °C.

Step 2 · Build the product.mc = (125 g)(4.184 J/g·°C) = 523.0 J/°C.

Step 3 · Multiply.q = (523.0 J/°C)(73.0 °C) = 38,179 J.

Step 4 · Round and report.Three significant figures gives q = 3.82 × 10⁴ J, or 38.2 kJ. The grams and degrees cancel, so joules are left. The value is positive, which fits a sample that warmed up.

Figure 5.2 A coffee cup calorimeter. The hot metal cools, the water warms, and the thermometer settles at one shared temperature. Heat lost by the metal equals heat gained by the water, with the sign flipped.

Worked example · finding a specific heat

A 50.0 g block of metal is heated to 98.0 °C. It is dropped into 100.0 g of water at 21.0 °C. The water settles at 24.4 °C. What is the metal's specific heat?

Step 1 · Heat gained by the water.ΔT = 24.4 − 21.0 = 3.4 °C. q = (100.0 g)(4.184 J/g·°C)(3.4 °C) = 1423 J.

Step 2 · Heat lost by the metal.q_metal = −1423 J. Its ΔT = 24.4 − 98.0 = −73.6 °C.

Step 3 · Solve q = mcΔT for c.c = q ÷ (m ΔT) = (−1423 J) ÷ [(50.0 g)(−73.6 °C)] = 1423 ÷ 3680 = 0.387 J/g·°C.

Step 4 · Round honestly.The water's ΔT of 3.4 °C has only two significant figures, so the answer does too. c = 0.39 J/g·°C. Copper is 0.385 J/g·°C, so the block is very likely copper.

Worth knowing. Small temperature changes are the weak link in calorimetry. A 3.4 °C rise read on a cheap thermometer limits you to two significant figures, no matter how carefully you weighed the sample.

Key points

Heat capacity belongs to an object. Specific heat belongs to a substance.
q = mcΔT, and ΔT is always final minus initial.
Water's 4.184 J/g·°C is high, which is why water steadies temperatures.
In a calorimeter, heat lost by one part equals heat gained by the other, with opposite signs.

Interactive Drop a hot metal into water and solve q = mcΔT. Open full ↑

Section 3Enthalpy and thermochemical equations

A quantity built for the lab bench

Most reactions you run sit in an open container. The pressure stays near 1 atm the whole time. If a gas forms, it has to push the air back, and that costs energy as work. So ΔE is awkward to measure in an open cup, because part of the energy slipped away as work you never saw. Chemists get around this with a quantity called enthalpy, H.

Enthalpy is defined as H = E + PV. You will never need to find H itself, and that is fine. What matters is the change. At constant pressure, the enthalpy change is exactly the heat you measure.

ΔH = H_products − H_reactants = q_p

That is a gift. The thermometer reading from an open cup gives you ΔH with no extra work term to chase. The sign rules match the ones you already know. A negative ΔH means heat left the system, so the reaction is exothermic. A positive ΔH means heat went in, so the reaction is endothermic. Enthalpy is an extensive property, which means the amount of substance changes the size of the answer.

Thermochemical equations

A thermochemical equation is a balanced equation with two extra pieces of care. Every substance carries its physical state. And a value of ΔH is written beside it for the equation exactly as balanced.

CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(l) ΔH = −890.5 kJ

Read that as a package deal. Burn one mole of methane with two moles of oxygen, and make liquid water, and 890.5 kJ of heat leaves. States are not decoration. If the water comes off as vapor instead, the same reaction gives only 802.5 kJ. The missing 88 kJ stayed in the vapor, because turning 2 mol of liquid water into vapor costs about 44.0 kJ per mole.

A value marked ΔH° is a standard enthalpy change. The little circle means every substance was in its standard state: a pressure of 1 bar for gases, a concentration of 1 M for solutions, and a stated temperature that is almost always 25 °C.

Scaling and reversing

Two rules let one equation answer many questions. Multiply the whole equation by a number, and ΔH gets multiplied by the same number. Reverse the equation, and ΔH keeps its size but flips its sign. Watch all three forms below say the same thing.

2H₂(g) + O₂(g) → 2H₂O(l)    ΔH = −571.6 kJ
H₂(g) + ½O₂(g) → H₂O(l)    ΔH = −285.8 kJ
2H₂O(l) → 2H₂(g) + O₂(g)    ΔH = +571.6 kJ

The last line is worth a pause. Splitting water back into its elements takes in exactly the heat that forming it gave off. Nature keeps a tidy ledger.

Figure 5.3 Two enthalpy diagrams. On the left the products sit lower, so heat leaves and ΔH is negative. On the right the products sit higher, so heat must go in and ΔH is positive. The height of the arrow is the size of ΔH.

Worked example · heat from a given mass

How much heat is released when 25.0 g of methane burns to carbon dioxide and liquid water?

Step 1 · Find the molar mass.CH₄ is 12.011 + 4(1.008) = 16.043, so 16.04 g/mol.

Step 2 · Convert grams to moles.n = 25.0 g ÷ 16.04 g/mol = 1.559 mol.

Step 3 · Use ΔH as a conversion factor.The equation gives −890.5 kJ per mole of CH₄. So q = (1.559 mol)(−890.5 kJ/mol) = −1388 kJ.

Step 4 · Round and read the sign.q = −1.39 × 10³ kJ. The minus sign says the heat left the system, so 1.39 × 10³ kJ warmed the room.

Key points

At constant pressure, ΔH equals the heat you measure, so ΔH = q_p.
Negative ΔH is exothermic. Positive ΔH is endothermic.
A thermochemical equation needs states and a ΔH for the equation as written.
Scale the equation and ΔH scales with it. Reverse the equation and ΔH flips sign.

Interactive Read endothermic and exothermic energy diagrams. Open full ↑

Section 4Hess's law

Enthalpy does not care how you got there

Imagine hiking to a lookout 300 m above the car park. You could go straight up the steep face. You could take a long switchback trail instead. Either way you climbed 300 m. Altitude depends only on where you start and where you stop. Enthalpy behaves the same way, which makes it a state function. Its change depends on the starting state and the final state, never on the route between them.

That single fact is Hess's law. If a reaction can be written as the sum of two or more steps, then the reaction's ΔH is the sum of the steps' ΔH values. The steps do not even have to be real. They only have to add up on paper.

Hess's law is how chemists get numbers they cannot measure. Take the reaction of carbon with oxygen to make carbon monoxide. You cannot measure it cleanly, because some of the CO always burns on to CO₂. So you measure two reactions you can control, then combine them.

Three moves and nothing else

Every Hess's law problem uses the same three moves on the steps you were given.

Use a step as written, and use its ΔH as given.
Reverse a step, and change the sign of its ΔH.
Multiply a step by any factor, and multiply its ΔH by that same factor.

Then stack the adjusted steps and add. Anything that appears on the left in one step and on the right in another will cancel. What is left should be the target equation. If it is not, go back and adjust a factor.

A good habit is to work from the target, not from the steps. Find a substance that appears in the target. Hunt for it in the given steps. Ask two questions. Is it on the correct side? Is the coefficient correct? Flip or scale until both answers are yes, then move to the next substance.

Worked example · building a reaction from two steps

Target: C(s) + ½O₂(g) → CO(g). Find ΔH from these two measured reactions.

(1) C(s) + O₂(g) → CO₂(g) ΔH₁ = −393.5 kJ
(2) 2CO(g) + O₂(g) → 2CO₂(g) ΔH₂ = −566.0 kJ

Step 1 · Place the carbon.The target has one C(s) on the left. Step (1) already has it there, with a coefficient of 1. Keep step (1) as written.

Step 2 · Place the CO.The target has CO on the right. Step (2) has CO on the left, and two of them. Reverse step (2) and halve it.

Step 3 · Adjust the enthalpy with it.Reversing flips the sign to +566.0 kJ. Halving gives +283.0 kJ. The adjusted step is CO₂(g) → CO(g) + ½O₂(g).

Step 4 · Stack and cancel.C(s) + O₂(g) → CO₂(g) sits above CO₂(g) → CO(g) + ½O₂(g). One CO₂ cancels. One O₂ on the left meets ½O₂ on the right, leaving ½O₂ on the left. The sum is C(s) + ½O₂(g) → CO(g).

Step 5 · Add the enthalpies.ΔH = (−393.5 kJ) + (+283.0 kJ) = −110.5 kJ.

Step 6 · Check it.Tables list the formation enthalpy of CO(g) as −110.5 kJ/mol. The two agree exactly.

Figure 5.4 Two routes from carbon and oxygen down to carbon dioxide. The direct drop is −393.5 kJ. The two step route stops at carbon monoxide, falling −110.5 kJ and then −283.0 kJ. Both routes total the same, so the first leg must be −110.5 kJ.

Worth knowing. Fractional coefficients look odd in a balanced equation, but Hess's law allows them. A half mole of O₂ is a perfectly real amount, and it keeps the target equation from being doubled for no reason.

Key points

Enthalpy is a state function, so ΔH ignores the path taken.
Hess's law: if steps add up to the target, their ΔH values add up too.
Reversing a step flips the sign. Scaling a step scales the value.
Work from the target substance by substance, checking side and coefficient.

Interactive Stack reaction steps and add their enthalpies to a target. Open full ↑

Section 5Standard enthalpies of formation

One table, thousands of reactions

Hess's law is powerful, but hunting for the right steps gets old. There is a shortcut that works for almost any reaction. Pick one special step and tabulate it for every compound. That step is formation from the elements.

The standard enthalpy of formation, written ΔH_f°, is the enthalpy change when one mole of a substance forms from its elements, with everything in its standard state. The standard state of an element is its most stable form at 1 bar and the stated temperature, usually 25 °C. For carbon that is graphite, not diamond. For oxygen it is O₂(g), not ozone. For bromine it is a liquid.

Two details follow right away. First, the equation must make exactly one mole of the product, so fractional coefficients are normal. Second, an element already in its standard state has nothing to form, so its ΔH_f° is zero.

H₂(g) + ½O₂(g) → H₂O(l) ΔH_f° = −285.8 kJ/mol

Products minus reactants

Once every compound has a number, any reaction enthalpy is one subtraction.

ΔH°_rxn = Σ nΔH_f°(products) − Σ nΔH_f°(reactants)

Hess's law is why this works. Think of a two leg route. First take the reactants apart into their elements, which is the reverse of forming them, so you subtract their values. Then build the products from those elements, which adds their values. The elements act like sea level, a shared zero that both sides are measured from.

Three cautions save most of the lost points. Multiply each value by the coefficient from the balanced equation. Match the state in the table to the state in your equation, since H₂O(l) and H₂O(g) differ by 44.0 kJ/mol. And subtract the whole reactant sum, not just the first term.

Substance	ΔH_f° (kJ/mol)	Substance	ΔH_f° (kJ/mol)
H₂O(l)	−285.8	NH₃(g)	−45.9
H₂O(g)	−241.8	NO(g)	+90.25
CO₂(g)	−393.5	NO₂(g)	+33.2
CO(g)	−110.5	SO₂(g)	−296.8
CH₄(g)	−74.6	SO₃(g)	−395.7
C₂H₂(g)	+227.4	NaCl(s)	−411.2
C₂H₅OH(l)	−277.6	CaCO₃(s)	−1207.6
C₆H₁₂O₆(s)	−1273.3	CaO(s)	−634.9
H₂O₂(l)	−187.8	Fe₂O₃(s)	−824.2
Any element in its standard state	0	Al₂O₃(s)	−1675.7

Figure 5.5 The elements set a shared zero. Limestone sits 1207.6 kJ below it. Lime plus carbon dioxide sit 1028.4 kJ below it. The gap between the two shelves is the reaction enthalpy, +179.2 kJ.

Worked example · breaking down limestone

A lime kiln runs CaCO₃(s) → CaO(s) + CO₂(g). Find ΔH°.

Step 1 · Sum the products.Σ products = (1)(−634.9) + (1)(−393.5) = −1028.4 kJ.

Step 2 · Sum the reactants.Σ reactants = (1)(−1207.6) = −1207.6 kJ.

Step 3 · Subtract.ΔH° = (−1028.4 kJ) − (−1207.6 kJ) = +179.2 kJ.

Step 4 · Read the sign.Positive, so the reaction is endothermic. That is why a kiln has to burn fuel all day to keep making lime.

Worked example · burning ethanol

Find ΔH° for C₂H₅OH(l) + 3O₂(g) → 2CO₂(g) + 3H₂O(l).

Step 1 · Sum the products with coefficients.2(−393.5) + 3(−285.8) = −787.0 + (−857.4) = −1644.4 kJ.

Step 2 · Sum the reactants.1(−277.6) + 3(0) = −277.6 kJ. Oxygen gas is an element in its standard state, so it contributes nothing.

Step 3 · Subtract.ΔH° = (−1644.4 kJ) − (−277.6 kJ) = −1366.8 kJ.

Step 4 · Sanity check.Burning a fuel should be exothermic, and the answer is strongly negative. One mole of ethanol weighs 46.07 g, so that is about 29.7 kJ per gram.

Key points

ΔH_f° is for one mole of a substance made from its elements in standard states.
An element in its own standard state has ΔH_f° = 0.
ΔH°_rxn = sum of products minus sum of reactants, each value times its coefficient.
States matter. Liquid and gaseous water sit 44.0 kJ/mol apart.

Interactive Use products minus reactants to get a reaction enthalpy. Open full ↑

Section 6Bond energies

Breaking costs, making pays

Formation enthalpies tell you about whole compounds. Bond energies let you look inside them. A bond dissociation energy, D, is the energy needed to break one mole of a given bond in the gas phase, pulling the pieces apart into free atoms. Breaking a bond always takes energy in, so D is always positive. Forming that same bond gives back the same amount, so it always counts as negative.

There is a wrinkle. The strength of a bond shifts a little depending on what sits nearby. The first C–H bond in methane takes about 439 kJ/mol to break. The four of them average out to about 415 kJ/mol. Tables list that kind of average bond energy, pooled across many molecules. So answers from bond energies are good estimates, not exact values. Expect to land within a few percent.

Bond order matters a lot. More shared pairs pull the atoms closer and hold them tighter. A C–C single bond is 347 kJ/mol. The double bond is 614, and the triple is 839. Notice the double bond is not twice the single, and the triple is not three times it. Each extra pair adds less than the one before.

Bond	D (kJ/mol)	Bond	D (kJ/mol)
H–H	436	C–C	347
C–H	415	C=C	614
N–H	390	C≡C	839
O–H	464	C–O	350
H–F	569	C=O (in CO₂)	799
H–Cl	431	C–Cl	339
F–F	159	N–N	160
Cl–Cl	242	N≡N	945
O–O	140	O=O	498

Counting a whole reaction

Picture a reaction taking the long way around. First tear every reactant apart into loose atoms. That climb costs the sum of all the bonds you broke. Then let those atoms snap together into the products. That drop pays back the sum of all the bonds you made. The difference between the climb and the drop is the reaction enthalpy.

ΔH ≈ Σ D(bonds broken) − Σ D(bonds formed)

This is Hess's law again, with free atoms as the stopping point. If the new bonds are stronger than the old ones, the drop beats the climb and ΔH comes out negative. That is the real reason fuels release heat. Carbon dioxide and water hold unusually strong bonds, stronger than the ones in the fuel and in O₂.

One shortcut helps on paper. Bonds that appear unchanged on both sides cancel out. Still, counting every bond is safer while you are learning, because a missed bond is easy to spot in a full list.

Figure 5.6 The long way around. Breaking H–H and Cl–Cl lifts the system 678 kJ to free atoms. Forming two H–Cl bonds drops it 862 kJ. The net change is −184 kJ, so the reaction gives off heat.

Worked example · hydrogen plus chlorine

Estimate ΔH for H₂(g) + Cl₂(g) → 2HCl(g).

Step 1 · List the bonds broken.One H–H at 436 kJ/mol and one Cl–Cl at 242 kJ/mol. Sum = 436 + 242 = 678 kJ.

Step 2 · List the bonds formed.Each HCl has one H–Cl bond, and there are two molecules. Sum = 2(431) = 862 kJ.

Step 3 · Subtract.ΔH ≈ 678 kJ − 862 kJ = −184 kJ.

Step 4 · Compare.Formation data give 2(−92.3) = −184.6 kJ. The estimate is off by less than 1 kJ, which is better than usual.

Worked example · burning methane the long way

Estimate ΔH for CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(g).

Step 1 · Bonds broken.Four C–H bonds at 415 and two O=O bonds at 498. Sum = 4(415) + 2(498) = 1660 + 996 = 2656 kJ.

Step 2 · Bonds formed.CO₂ has two C=O bonds at 799. Each water has two O–H bonds, so four at 464. Sum = 2(799) + 4(464) = 1598 + 1856 = 3454 kJ.

Step 3 · Subtract.ΔH ≈ 2656 kJ − 3454 kJ = −798 kJ.

Step 4 · Compare.Formation data give −802.5 kJ for this version with water vapor. The estimate is within 5 kJ, or about half a percent.

Key points

Breaking bonds takes energy in. Forming bonds gives energy out.
ΔH ≈ sum of bonds broken minus sum of bonds formed.
Stronger bonds in the products mean a negative ΔH, which is why fuels burn hot.
Tabulated values are averages, so bond energy answers are estimates.

Interactive Break bonds, make bonds and total the energy change. Open full ↑

The chapter in one breath

Pick the system first. Every sign in a thermochemistry problem is written from the system's point of view.
ΔE = q + w. Energy entering the system is positive. Energy leaving it is negative.
q = mcΔT turns a thermometer reading into joules, and ΔT is always final minus initial.
Water's specific heat of 4.184 J/g·°C is high, which is why water makes such a steady heat sink.
At constant pressure, ΔH is the heat. Negative means exothermic, positive means endothermic.
Thermochemical equations scale. Double the equation and double ΔH. Reverse it and flip the sign.
Hess's law adds steps. Flip, scale and stack known reactions until they equal the target.
Formation enthalpies give a shortcut: products minus reactants, each value times its coefficient.
Bond energies explain the heat. Break bonds, form stronger ones, and the difference comes out as heat.

ReferenceWords worth knowing

These are the words this chapter leans on. Come back to the list whenever a problem uses one you are not sure of.

Bomb calorimeter: A sealed steel vessel used to burn a sample in oxygen at constant volume, sitting in a measured water bath.
Bond dissociation energy (D): The energy needed to break one mole of a given bond in the gas phase. Always a positive number.
Calorimetry: Measuring heat by tracking the temperature change of a known mass.
Endothermic: A change that takes heat in from the surroundings, so q and ΔH are positive.
Enthalpy (H): A thermodynamic quantity defined as H = E + PV. Its change is what chemists measure at constant pressure.
Enthalpy change (ΔH): The enthalpy of the products minus the enthalpy of the reactants. At constant pressure it equals the heat exchanged.
Exothermic: A change that gives heat out to the surroundings, so q and ΔH are negative.
First law of thermodynamics: Energy is conserved. For a system, ΔE = q + w.
Heat (q): Thermal energy moving because of a temperature difference. It flows from warmer to cooler.
Heat capacity (C): The heat needed to raise a whole object by one degree. Measured in J/°C, and it depends on size.
Hess's law: If a reaction is the sum of several steps, its ΔH is the sum of those steps' ΔH values.
Internal energy (E): The total kinetic and potential energy stored inside a system.
Joule (J): The SI unit of energy, equal to 1 kg·m²/s². One calorie is 4.184 J.
Specific heat capacity (c): The heat needed to raise one gram of a substance by one degree, in J/g·°C. A property of the substance.
Standard enthalpy of formation (ΔH_f°): The enthalpy change for making one mole of a substance from its elements, all in standard states.
Standard state: The reference condition for tabulated values: 1 bar for gases, 1 M for solutions, and the element's most stable form, at a stated temperature.
State function: A property whose change depends only on the starting and final states, never on the path taken.
Surroundings: Everything outside the system that can trade energy or matter with it.
System: The part of the universe you choose to study, such as the contents of a beaker.
Work (w): Energy transferred by a force acting through a distance. In chemistry it is usually a gas expanding or being squeezed.

Check yourself

1. A system gives off 45 J of heat. At the same time the surroundings do 12 J of work on it. What is ΔE?

ΔE = −33 J. Heat out makes q = −45 J. Work done on the system makes w = +12 J. So ΔE = (−45 J) + (+12 J) = −33 J. The system ends up with 33 J less than it started with.

2. How much heat is needed to warm 250.0 g of water from 20.0 °C to 80.0 °C?

62.8 kJ. ΔT = 80.0 − 20.0 = 60.0 °C. Then mc = (250.0 g)(4.184 J/g·°C) = 1046.0 J/°C, and q = (1046.0)(60.0) = 62,760 J. Rounded to three significant figures that is 6.28 × 10⁴ J, or 62.8 kJ.

3. You add 500. J of heat to 100.0 g of copper, and the same 500. J to 100.0 g of water. How much does each one warm up?

Copper rises 13.0 °C. Water rises only 1.20 °C. Rearrange to ΔT = q ÷ mc. For copper, ΔT = 500. ÷ [(100.0)(0.385)] = 500. ÷ 38.5 = 13.0 °C. For water, ΔT = 500. ÷ [(100.0)(4.184)] = 500. ÷ 418.4 = 1.20 °C. Water's larger specific heat means a smaller temperature change for the same heat.

4. A 15.0 g piece of aluminum at 85.0 °C is dropped into 75.0 g of water at 22.0 °C. What final temperature do they reach? Use c = 0.897 J/g·°C for aluminum.

About 24.6 °C. Set heat lost equal to heat gained. For aluminum, mc = (15.0)(0.897) = 13.455 J/°C. For water, mc = (75.0)(4.184) = 313.80 J/°C. Solving 13.455(T − 85.0) + 313.80(T − 22.0) = 0 gives T = (1143.675 + 6903.6) ÷ 327.255 = 24.6 °C. The water barely moves because it has the far bigger mc.

5. Given 2H₂(g) + O₂(g) → 2H₂O(l) with ΔH = −571.6 kJ, what is ΔH for H₂O(l) → H₂(g) + ½O₂(g)?

ΔH = +285.8 kJ. Two changes apply. Halving the equation halves the value to −285.8 kJ. Reversing it flips the sign to +285.8 kJ. Splitting water takes in exactly the heat that forming it gave off.

6. Methane burns by CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(l), ΔH = −890.5 kJ. How much heat comes from burning 5.00 g of methane?

278 kJ is released. The molar mass of CH₄ is 16.04 g/mol, so n = 5.00 ÷ 16.04 = 0.3117 mol. Then q = (0.3117 mol)(−890.5 kJ/mol) = −277.6 kJ. To three significant figures, 278 kJ leaves the system.

7. Use these two steps to find ΔH for N₂(g) + 2O₂(g) → 2NO₂(g). Step A: N₂(g) + O₂(g) → 2NO(g), ΔH = +180.5 kJ. Step B: 2NO(g) + O₂(g) → 2NO₂(g), ΔH = −114.1 kJ.

ΔH = +66.4 kJ. Both steps already point the right way with the right coefficients, so just add them. The 2NO cancels, one O₂ from each step gives 2O₂ on the left, and ΔH = (+180.5) + (−114.1) = +66.4 kJ.

8. Why is the standard enthalpy of formation of O₂(g) equal to zero, while the value for O₃(g) is not?

O₂(g) is oxygen's standard state, so there is nothing to form. Formation enthalpy is measured from the elements in their most stable forms. At 1 bar and 25 °C that form is O₂(g), so making O₂ from O₂ involves no change. Ozone is a different, less stable form, so building it from O₂ does cost energy.

9. Find ΔH° for 2SO₂(g) + O₂(g) → 2SO₃(g). Use ΔH_f° = −296.8 kJ/mol for SO₂(g) and −395.7 kJ/mol for SO₃(g).

ΔH° = −197.8 kJ. Products: 2(−395.7) = −791.4 kJ. Reactants: 2(−296.8) + 1(0) = −593.6 kJ, since O₂ is an element in its standard state. Subtracting gives (−791.4) − (−593.6) = −197.8 kJ.

10. Estimate ΔH for N₂(g) + 3H₂(g) → 2NH₃(g) from bond energies. Use N≡N = 945, H–H = 436 and N–H = 390 kJ/mol.

About −87 kJ. Bonds broken: one N≡N plus three H–H, so 945 + 3(436) = 945 + 1308 = 2253 kJ. Bonds formed: each NH₃ has three N–H bonds, so six in all, 6(390) = 2340 kJ. Then ΔH ≈ 2253 − 2340 = −87 kJ. Tables give −91.8 kJ, so the estimate is close.

11. A cold pack gets cold when ammonium nitrate dissolves in water. Which way does heat flow, and what is the sign of ΔH?

Heat flows from your hand into the pack, and ΔH is positive. The dissolving is the system. It takes heat in, which makes it endothermic with a positive ΔH of about +25.7 kJ/mol. Your hand is part of the surroundings, and it loses that heat, which is why it feels cold.

12. Why does a bond energy calculation usually miss the tabulated ΔH by a few kilojoules?

Because tabulated bond energies are averages. A bond's real strength shifts with the atoms around it. The C–H bond in methane is about 439 kJ/mol, but the table lists 415 kJ/mol as an average over many compounds. Bond energies also assume gas phase species, so any liquid or solid in the equation adds more error.

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