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Working an enthalpy change out from the bonds that broke and the bonds that formed, and why the answer is an estimate rather than a truth.
Paper packet. Every task here also exists on screen, where it is checked automatically; answers written on paper are not assessed by Nydus. When you are back at a device, enter your answers there.
By the end of this lesson you will be able to estimate the enthalpy change of a reaction from mean bond enthalpies: add up every bond broken on the left, add up every bond made on the right, and subtract the second from the first. You will be able to say why breaking always costs energy and making always releases it, why a reaction is exothermic exactly when the bonds it ends with are stronger than the ones it started with, and why the answer is expected to disagree with a measured value.
You can say whether a change is exothermic or endothermic, you can read an enthalpy change off a temperature rise, and you know what a covalent bond is: a shared pair of electrons that two nuclei both attract. This lesson works the enthalpy change out from the bonds themselves, without a thermometer anywhere in the room.
| Term | What it means |
|---|---|
| Bond enthalpy | The energy to break one mole of a particular bond, all in the gas state. |
| Mean bond enthalpy | That value averaged over every compound the bond appears in. |
| Break | Pull two bonded atoms apart against their attraction. |
| Form | Let two atoms come together into a bond. |
| Estimate | A value with a known reason to be a little wrong. |
Every reaction does the same two things in the same order. Old bonds come apart, and new ones form.
Breaking a bond always takes energy in. The two atoms are attracted to each other, so pulling them apart is work, exactly as lifting something is work. There is no such thing as a bond that releases energy when it breaks, and the phrase energy stored in a bond is a trap for this reason: a bond is not a battery, it is an attraction, and it costs to undo.
Forming a bond always gives energy out. Two atoms falling together release energy, the same amount it would take to separate them again.
So:
$$\Delta H = \sum(\text{bonds broken}) - \sum(\text{bonds made})$$
The order matters and is not arbitrary. The first sum is money spent and the second is money refunded, and $\Delta H$ is the change in what the reaction holds, which is spending minus refund.
Read the sign the same way as before. Negative means the refund beat the cost, which means the new bonds are stronger overall than the old ones, which means the reaction gives energy out. Exothermic reactions are exothermic because they end up with stronger bonds, and that is the most useful single sentence in this unit.
Another way: picture
Think of a demolition and a rebuild. Taking the old building down costs money and the rubble is worth nothing; putting the new one up earns you a building. Whether the project made money depends entirely on whether what you built is worth more than what it cost to clear the site. A reaction is exothermic when the new bonds are worth more than the old ones cost.
Another way: steps
To estimate any enthalpy change from bond enthalpies:
Balance the equation. The coefficients tell you how many molecules of each substance, and so how many of each bond.
List every bond on the left. Draw the molecules if it helps. Count repeats: methane has four carbon-to-hydrogen bonds, and two molecules of it have eight.
Total the cost. Multiply each kind of bond by its count and its mean enthalpy, and add.
List and total every bond on the right the same way. That is the refund.
Subtract. Broken minus made gives $\Delta H$ for the equation as written.
Check the work. First, the sign: a combustion, a neutralization or any reaction you know gives out heat must come out negative; if it is positive, the subtraction is backwards. Second, the counts: did every atom on the left end up in a bond on the right? Third, the size: compare with a measured value if you have one. Expect to be off by up to a hundred kilojoules for a reaction that makes liquid water, and close for one that is all gases.
Counting every bond on both sides is allowed because the method imagines the reaction in two stages: every reactant torn into separate atoms, then every product assembled from those atoms. The real reaction does not go that way, but the energy change does not depend on the route — only on where it starts and ends.
Adding the costs is allowed because energy is additive. Breaking four carbon-to-hydrogen bonds costs four times what breaking one does.
Subtracting made from broken is allowed because the made bonds return energy to the surroundings, and $\Delta H$ is the system's net change: what went in minus what came out.
Leaving out bonds that survive unchanged is allowed because they would appear on both sides with the same value and cancel.
Treating the result as an estimate is required because mean values are averages; the exact strength of a bond depends on what else is in the molecule.
A bond enthalpy is an average over every molecule that bond appears in, so an enthalpy change worked out from bond enthalpies is an estimate and is expected to disagree with a measured value. It is also only ever right for substances that are all gases, because it accounts for nothing that happens when a liquid boils.
Here is what that costs in practice. Each estimate below was worked out exactly as this lesson works them out; each measured value is the one a calorimeter gives.
| Reaction | Estimate | Measured | Difference |
|---|---|---|---|
| methane burning | $-818$ kJ | $-890$ kJ | $72$ kJ |
| nitrogen and hydrogen making ammonia | $-93$ kJ | $-92$ kJ | $1$ kJ |
| ethene burning | $-1318$ kJ | $-1411$ kJ | $93$ kJ |
The middle row is nearly exact and the other two are out by the better part of a hundred kilojoules, and the pattern is not an accident. Ammonia, nitrogen and hydrogen are all gases, which is the condition a bond enthalpy is quoted under. The two combustions make liquid water, and condensing that water releases energy that the bond-enthalpy method knows nothing about, because as far as it is concerned the reaction stopped when the last bond formed.
So the two reasons an estimate misses are worth naming separately:
Neither is a mistake in the arithmetic, and neither can be fixed by doing the arithmetic more carefully.
A mean bond enthalpy is measured, not deduced, and every value in this course comes from one table so that two items cannot disagree about the strength of a carbon to hydrogen bond.
Three patterns in that table are worth noticing, because they are consequences of structure rather than a list to learn:
The method's real use is not getting an enthalpy change to the nearest kilojoule — a calorimeter does that better. It is answering why: which bonds were expensive, which were cheap, and what a chemist would have to change to shift the balance.
The engines that lifted the Space Shuttle, and the RS-25 engines on NASA's Space Launch System today, burn liquid hydrogen with liquid oxygen: $\mathrm{2H_2 + O_2 \rightarrow 2H_2O}$. A bond-enthalpy ledger explains why engineers put up with the difficulty of storing hydrogen at −423 °F.
Breaking two hydrogen bonds and one oxygen double bond costs $2 \times 436 + 498 = 1370$ kJ. Making four oxygen-to-hydrogen bonds refunds $4 \times 464 = 1856$ kJ. The estimate is $1370 - 1856 = -486$ kJ for every 4 g of hydrogen burned — about $-121$ kJ per gram of hydrogen, more energy per gram than any other chemical fuel. For comparison, burning gasoline releases about 45 kJ per gram.
That is exactly what a rocket needs, because every gram it carries must itself be lifted. The ledger also says why hydrogen does so well: the oxygen-to-hydrogen bond is strong, so the refund is large, while hydrogen's own bond is cheap to break and the hydrogen atom is the lightest there is. The cost is practical rather than chemical. Liquid hydrogen is so light that its tank must be enormous — the giant orange tank of the Shuttle held mostly hydrogen by volume — and so cold that it boils away through any gap in the insulation.
Fixing nitrogen means breaking the triple bond at 945 kJ per mole. Even though making ammonia is slightly exothermic overall, that huge cost up front is why the Haber process needs high temperature, high pressure and a catalyst, and why it uses about one percent of the world's energy supply.
Thinking energy is stored in a bond and released when it breaks. Breaking always costs. The energy a fuel releases comes from the bonds that form afterwards being stronger than the ones that broke, not from the fuel's own bonds letting go of something.
Subtracting the wrong way round. Broken minus made. If a combustion comes out positive, the two totals have been swapped.
Forgetting the bonds that did not change. In many reactions some bonds survive untouched, and those can be left out of both totals because they cancel. The mistake is counting them on one side and not the other.
Treating the answer as exact. A bond enthalpy is an average over every molecule that bond appears in, so an enthalpy change worked out from bond enthalpies is an estimate and is expected to disagree with a measured value. It is also only ever right for substances that are all gases, because it accounts for nothing that happens when a liquid boils.
Expecting it to work for ionic compounds. A bond enthalpy describes a covalent bond between two atoms in a gas. A salt has no such bonds, and its lattice has to be accounted for a different way.
List the bonds broken.
$\mathrm{H-H} + \mathrm{Cl-Cl}$
From $\mathrm{H_2 + Cl_2 \rightarrow 2HCl}$.
Total the cost.
$436 + 243 = 679 \text{ kJ}$
Each value from the table.
List the bonds made.
$2 \times \mathrm{H-Cl}$
Two molecules, two bonds.
Total the refund.
$2 \times 432 = 864 \text{ kJ}$
Count times energy.
Subtract made from broken.
$679 - 864 = -185 \text{ kJ}$
Exothermic: the new bonds are stronger.
List the bonds broken.
$4 \ \mathrm{C-H} + 2 \ \mathrm{O=O}$
From $\mathrm{CH_4 + 2O_2 \rightarrow CO_2 + 2H_2O}$.
Total the cost.
$4 \times 413 + 2 \times 498 = 2648 \text{ kJ}$
Counting repeats.
List the bonds made.
$2 \ \mathrm{C=O} + 4 \ \mathrm{O-H}$
One carbon dioxide, two waters.
Total the refund.
$2 \times 805 + 4 \times 464 = 3466 \text{ kJ}$
Counting repeats.
Subtract made from broken.
$2648 - 3466 = -818 \text{ kJ}$
The estimate.
Compare with measurement.
$-818 \text{ against } -890 \text{ kJ}$
Condensing water adds energy the ledger cannot see.
List the bonds broken.
$\mathrm{C-C} + 6 \ \mathrm{C-H}$
From $\mathrm{C_2H_6 \rightarrow C_2H_4 + H_2}$.
Total the cost.
$347 + 6 \times 413 = 2825 \text{ kJ}$
Every bond in ethane.
List the bonds made.
$\mathrm{C=C} + 4 \ \mathrm{C-H} + \mathrm{H-H}$
Ethene and hydrogen.
Total the refund.
$612 + 4 \times 413 + 436 = 2700 \text{ kJ}$
Four C–H bonds survive on both sides.
Subtract made from broken.
$2825 - 2700 = +125 \text{ kJ}$
Positive: endothermic.
Check by cancelling survivors.
$(347 + 2 \times 413) - (612 + 436) = 125$
The same answer, shorter.
Read the engineering fact.
$\text{a cracker must be a furnace}$
An endothermic process needs a steady supply.
Total the bonds broken.
$436 + 193 = 629 \text{ kJ}$
The cost side first.
Total the bonds made.
$2 \times 366 = 732 \text{ kJ}$
Two hydrogen bromide bonds.
Subtract made from broken.
Match each bond to the mean energy it takes to break one mole of it.
| $464$ kJ per mole | $366$ kJ per mole | $243$ kJ per mole | $391$ kJ per mole | |
|---|---|---|---|---|
| $\mathrm{O-H}$, the oxygen to hydrogen bond | ||||
| $\mathrm{H-Br}$, the hydrogen to bromine bond | ||||
| $\mathrm{Cl-Cl}$, the chlorine to chlorine bond | ||||
| $\mathrm{N-H}$, the nitrogen to hydrogen bond |
Complete the worked solution: estimate the enthalpy change of $\mathrm{H_2 + F_2 \rightarrow 2HF}$. The hydrogen bond takes four hundred thirty-six kilojoules per mole, the fluorine bond one hundred fifty-eight, and each hydrogen-to-fluorine bond releases five hundred sixty-eight.
Total the bonds broken.
$\text{four thirty-six} + \text{one fifty-eight} =$ b
The cost side.
Total the bonds made.
$\text{two} \times \text{five sixty-eight} =$ m
Two molecules, two bonds.
Subtract made from broken.
$(\text{broken}) - (\text{made}) =$ h
Negative, so strongly exothermic.
In the Haber process making ammonia, $\mathrm{N_2 + 3H_2 \rightarrow 2NH_3}$, breaking every bond on the left — one nitrogen-nitrogen triple bond and three hydrogen-hydrogen bonds — takes $2253$ kJ in total, and forming every bond on the right — six nitrogen-hydrogen bonds — gives out $2346$ kJ in total. What is the enthalpy change for the equation as written? Give the sign.
Answer: unit: J / MJ / kJ
For hydrogen reacting with bromine vapor, $\mathrm{H_2 + Br_2 \rightarrow 2HBr}$. The number of bonds of each kind and the mean enthalpy of each are given. Work out the total for each kind, and then the enthalpy change for the equation as written.
| how many | energy each, in kJ | total, in kJ | |
|---|---|---|---|
| $\mathrm{H-H}$ broken | 1 | 436 | |
| $\mathrm{Br-Br}$ broken | 1 | 193 | |
| $\mathrm{H-Br}$ made | 2 | 366 | |
| the enthalpy change for the equation as written | — | — |
NASA's large rocket engines burn liquid hydrogen with oxygen: $\mathrm{2H_2 + O_2 \rightarrow 2H_2O}$. Using $\mathrm{H-H}$ at $436$, $\mathrm{O=O}$ at $498$ and $\mathrm{O-H}$ at $464$ kJ per mole, estimate the enthalpy change, in kJ, for burning $12$ mol of hydrogen.
The answer: a kJ.
A process engineer wants to run hydrogen burning in oxygen the other way round, turning the products back into the reactants. In the forward direction, breaking two hydrogen-hydrogen bonds and one oxygen-oxygen double bond takes $1370$ kJ and making four oxygen-hydrogen bonds gives out $1856$ kJ. What is the enthalpy change of the **reverse** reaction? Give the sign.
Answer: unit: J / MJ / kJ
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
For hydrogen burning in chlorine, $\mathrm{H_2 + Cl_2 \rightarrow 2HCl}$, the bonds broken are one hydrogen-hydrogen bond and one chlorine-chlorine bond and the bonds made are two hydrogen-chlorine bonds. Breaking every one of them together takes $679$ kJ and making every one of them together gives out $864$ kJ. Complete the ledger, and say which kind of change it is.
| energy in kJ | exothermic or endothermic? | |
|---|---|---|
| energy taken in to break every bond on the left, in kJ | — | |
| energy given out as every bond on the right forms, in kJ | — | |
| the enthalpy change, in kJ |
You can estimate an enthalpy change from a table of bond enthalpies, and you know which way round the subtraction goes. Say what makes a reaction exothermic in terms of bonds, and give one reason an estimate from this method misses the measured value. Next unit: the nucleus, and why some atoms are unstable enough to change into others.
15. Your turn: estimate the enthalpy change of $\mathrm{H_2 + Br_2 \rightarrow 2HBr}$, given $\mathrm{H-H}$ at $436$, $\mathrm{Br-Br}$ at $193$ and $\mathrm{H-Br}$ at $366$ kJ per mole., step 3
$629 - 732 = -103 \text{ kJ}$
Exothermic, but far less than with fluorine.