Cambridge International AS & A Level Chemistry 9701 · Topic 5

Chemical energetics

Every chemical reaction is a trade. Bonds in the reactants have to be pulled apart, which costs energy, and bonds in the products snap together, which pays it back. The enthalpy change of a reaction is simply the balance of that trade — and almost everything in this topic is a different way of getting at the same number: measure it with a thermometer, estimate it from bond energies, or route around it with a Hess cycle when the reaction refuses to be measured at all.

Enthalpy and enthalpy change5.1.1

Chemical substances store energy. Not in any mystical sense — the energy is in the electrostatic arrangement of nuclei and electrons, and a different arrangement holds a different amount. When a reaction rearranges those atoms, the stored energy changes, and the difference has to go somewhere: it is released to the surroundings as heat, or absorbed from them.

Enthalpy, given the symbol H, is the name for that stored heat energy measured at constant pressure. You never measure H itself and you will never be asked to — there is no zero to measure it from. What you can always measure is the change:

Two details in that definition carry marks. The first is the direction of the subtraction: products minus reactants, always, which is what makes the sign come out right without having to think about it. The second is constant pressure. A reaction in an open beaker or a polystyrene cup is at constant pressure — the atmosphere presses down with the same force throughout, and the mixture can expand or contract freely against it. That is the condition enthalpy is defined for, and it is why nearly every enthalpy measurement in this course is done in an open container.

Units and what they mean

Enthalpy changes are quoted in kJ mol⁻¹. The "per mole" is not decoration — it is per mole of the reaction as you wrote it, and changing the equation changes the number:

Those two rules — double the equation, double ΔH; reverse the equation, reverse the sign — are the whole basis of Hess's law arithmetic later on. They are worth fixing now.

Exothermic and endothermic reactions5.1.1

Every reaction falls into one of two camps, and the test is whether the products sit below the reactants on the enthalpy scale or above them.

The row that catches people out is the fourth one, because the sign and the thermometer point opposite ways. An exothermic reaction has a negative ΔH but makes the thermometer go up. There is no contradiction: the reacting chemicals are the system and they have lost enthalpy, while the water around them is the surroundings and it has gained it. The minus sign is bookkeeping from the system's point of view.

Why exothermic reactions are the common ones

Most spontaneous reactions you meet at AS are exothermic, and there is a reason worth carrying: a system tends towards the arrangement of lower energy, in the same way a ball rolls downhill. That is a rule of thumb rather than a law, and endothermic reactions certainly happen — ammonium nitrate dissolving in water chills the beaker, and thermal decompositions need a Bunsen precisely because they are uphill. The complete account of why some uphill processes still go is entropy, which is A Level material. For topic 5, "exothermic usually, but not always" is exactly the right level of commitment.

Reaction pathway diagrams5.1.2

A reaction pathway diagram plots enthalpy up the page against the progress of the reaction along it. It is the single most efficient way to show three quantities at once — how far the reaction falls or climbs, how big the barrier is on the way, and whether there is an intermediate in the middle. The syllabus calls it a reaction pathway diagram; you will also see it called an energy profile or an enthalpy level diagram, and they are the same picture.

Four things have to be on it before it earns full marks:

• a labelled vertical axis — enthalpy, H, or "energy";
• a labelled horizontal axis — progress of reaction, or reaction pathway;
• horizontal lines for reactants and products, each labelled with the species, not just the word;
• arrows for ΔH (between the two levels) and E_a (from the reactant level up to the top of the hump), each with its head pointing the right way.

The direction of the ΔH arrow is the answer to "is it exothermic?", so draw it from reactants to products, never the other way round. Pointing down means negative.

The hump, and what it is

Reactants do not slide straight into products. Bonds have to be stretched and partly broken before the new ones can form, and that intermediate arrangement — the transition state at the top of the hump — is higher in energy than either end. The climb from the reactants to that peak is the activation energy, E_a: the minimum energy a colliding pair must bring with them if the collision is to produce anything.

Notice what that means for the reverse reaction. Going backwards, the climb starts from the product level instead, so:

One step or two

A reaction that happens in a single step has a single hump. A reaction that happens in two steps has two humps with a dip between them, and the species sitting in that dip is an intermediate — a real substance, with real bonds, that exists for a measurable time, unlike a transition state.

Two features of the two-step picture are examinable. First, the higher of the two humps is the rate-determining step: the overall reaction can go no faster than its slowest stage, and the slowest stage is the one with the biggest barrier. Second, ΔH is measured from the first reactant level to the final product level and ignores the intermediate entirely — which is Hess's law showing up early, in picture form. The route does not matter; only the ends do.

Standard conditions and standard states5.1.3(a)

An enthalpy change depends on the conditions. Burn methane and collect the water as liquid and you measure −890 kJ mol⁻¹; let the water leave as vapour and you measure −802, because vaporising it costs energy that never reaches your thermometer. Warm the reaction up and the number shifts again. So a quoted value is meaningless unless everybody agrees on the conditions, and that agreement is what standard conditions are.

Note what is not in the definition: 298 K is not "room temperature" in any loose sense, it is a precise reference point, and it is emphatically not 273 K. Confusing standard conditions with the s.t.p. of gas calculations (273 K, 100 kPa) is a routine and expensive error, because the two sets of conditions belong to different topics and serve different purposes.

Standard states

Fixing the temperature and pressure is only half of it. You also have to fix which form of each substance you mean.

Carbon is the one to remember, because it appears in almost every Hess cycle you will ever build. Written correctly it is C(s, graphite). Diamond has a standard enthalpy of formation of +1.9 kJ mol⁻¹ — small, but not zero, and it is not zero precisely because graphite, not diamond, is the standard state.

The standard enthalpy changes5.1.3(b)

ΔH^⊖ on its own means "the enthalpy change of the reaction as written". The subscripts name particular kinds of reaction, each defined so tightly that the equation writes itself once you know which one is being asked for. Four are named in 9701, and each definition contains a quantity — "one mole of what" — that is the key to writing the equation.

Standard enthalpy change of reaction, ΔH_r^⊖

This is the general case, and the phrase "as written" is doing the work: ΔH_r for 2H₂ + O₂ → 2H₂O(l) is −572 kJ mol⁻¹, not −286, because the equation as written makes two moles of water. Quote the equation whenever you quote a ΔH_r.

Standard enthalpy change of formation, ΔH_f^⊖

"One mole of the compound" fixes the right-hand side at exactly 1, and everything else on the left has to bend to fit — which is where the fractions come from:

One consequence of the definition does an enormous amount of work later:

So ΔH_f^⊖[O₂(g)] = 0 and ΔH_f^⊖[C(s, graphite)] = 0 — but ΔH_f^⊖[C(s, diamond)] = +1.9 kJ mol⁻¹ and ΔH_f^⊖[O(g)] = +249 kJ mol⁻¹, because neither of those is the standard state. The zero belongs to the standard state, not to the element.

Standard enthalpy change of combustion, ΔH_c^⊖

Three words are load-bearing. One mole of the substance being burned — so the fractions go on the oxygen this time. Completely — carbon to CO₂, never CO or soot; hydrogen to H₂O; sulfur to SO₂. And in oxygen, in excess.

Standard enthalpy change of neutralisation, ΔH_neut^⊖

Per mole of water — not per mole of acid, which is why H₂SO₄ releases roughly twice as much heat per mole as HCl and yet has the same enthalpy of neutralisation.

For a strong acid with a strong alkali the value is almost constant at about −57 kJ mol⁻¹, whichever acid and alkali you choose, and the reason is the ionic equation:

H⁺(aq) + OH⁻(aq) → H₂O(l)    ΔH^⊖ = −57.1 kJ mol⁻¹

Strong acids and strong alkalis are fully dissociated already, so the spectator ions take no part and the only chemistry happening is that one reaction. With a weak acid such as ethanoic acid the value comes out a little less exothermic — around −56 kJ mol⁻¹ — because some of the heat released is spent dissociating the remaining undissociated acid molecules. That difference is a favourite short-answer question.

Measuring an enthalpy change5.1.7

You cannot put a thermometer into a chemical bond. What you can do is let the reaction dump its heat into something whose temperature you can measure — almost always water — and work backwards. That is calorimetry, and the whole of it rests on one equation.

Three of those rows cause almost every lost mark in this part of the syllabus, so take them one at a time.

m is the water, not the chemical

The specific heat capacity you are given is water's, so the mass in the equation must be the mass of water being warmed. In a combustion experiment that is the water in the calorimeter, not the fuel that burned. In a neutralisation or dissolving experiment the reaction happens in the water, so m is the total mass of the final solution — both volumes added together.

Volumes convert to masses using the assumption that dilute aqueous solutions have a density of 1.00 g cm⁻³: 50.0 cm³ of solution weighs 50.0 g. Stating that assumption when you use it is worth a mark in "suggest the assumptions you have made" questions.

ΔT in kelvin — which is the easy part

A temperature change of 1 K is a change of 1 °C: the two scales are offset, but their degrees are the same size. A rise from 19.4 °C to 26.2 °C is a ΔT of 6.8 °C and of 6.8 K, identically. So you never convert a ΔT — only an absolute temperature, and this equation never contains one.

The minus sign, and where n goes

The minus sign in ΔH = −mcΔT/n is the conversion between the surroundings' point of view and the system's. The water warmed up, so the water gained heat, so the reaction lost it:

And n is the number of moles of whatever the answer is "per mole" of, which is set by the definition you are quoting — the fuel for ΔH_c, the water formed for ΔH_neut, the solid dissolved for an enthalpy of solution. If two reagents are present in different amounts, n is fixed by the limiting one; the excess contributes heat capacity but no chemistry.

Enthalpy of combustion by experiment5.1.7

The school version is a spirit burner of liquid fuel under a metal calorimeter of water. You weigh the burner, light it, stir the water, and weigh the burner again when the temperature has risen by a useful amount. The mass lost is the fuel burned.

Why the answer always comes out too small

A spirit-burner measurement of ethanol typically returns −1000 to −1100 kJ mol⁻¹ against a true value of −1367. The error is large — of the order of 25% — and it is always in the same direction: the experiment under-estimates how exothermic the reaction is. Four causes, in roughly descending order of size:

• Heat lost to the surroundings. Most of it. The flame heats the air, the calorimeter, the tripod and the room as well as the water, and a metal can radiates freely.
• Incomplete combustion. A yellow, sooty flame is producing carbon and carbon monoxide rather than carbon dioxide, and that releases less energy per mole. The soot on the bottom of the can is the evidence.
• Evaporation of the fuel. Volatile alcohols evaporate from the wick between weighings, so the recorded mass loss is larger than the mass actually burned — which makes n too big and ΔH too small.
• Non-standard conditions. The experiment is not at 298 K and 101 kPa, and some water leaves as vapour rather than condensing to the liquid the definition requires.

Improving it

"Suggest two improvements" is a standard three-mark question, and the good answers attack the largest error first:

• shield the flame from draughts and reduce the distance between the flame and the calorimeter;
• insulate the calorimeter — a lid, and lagging round the sides;
• use a copper calorimeter, which conducts the heat into the water quickly instead of storing it;
• supply extra oxygen, or burn in pure oxygen, to make combustion complete;
• stir continuously, so the measured temperature is the whole body of water rather than the top of it.

"Repeat the experiment and average" is not one of them. Repeating reduces random error; every cause listed above is systematic, and an average of ten runs that all lose heat to the room is just a more precise wrong answer. That distinction — random versus systematic — is what the question is really testing.

The bomb calorimeter

The research answer to the same problem is a bomb calorimeter: a sealed, thick-walled steel vessel charged with the sample and with pure oxygen at high pressure, immersed in a known mass of water inside an insulated jacket, and ignited electrically.

Neutralisation and solution by experiment5.1.7

Reactions in solution are far easier to measure accurately than combustions, for one reason: the reaction happens inside the water it is heating, so almost none of the heat has anywhere else to go. An expanded-polystyrene cup with a lid is a perfectly respectable calorimeter, and school results for enthalpy of neutralisation routinely land within a few per cent of the accepted value.

The method

1. Measure a known volume of the acid into the cup and record its temperature.
2. Measure a known volume of the alkali separately and record its temperature; if they differ, average them for the start temperature.
3. Add the alkali quickly, put the lid on, stir, and record the highest temperature reached.
4. Take m as the total mass of the mixed solution, and n from the limiting reagent.

Where the excess reagent goes

If the two solutions are not equimolar, one is in excess, and the excess does two things at once: it takes no chemical part, so it does not appear in n, but it is still water being warmed, so it does appear in m. Mix 50.0 cm³ of 1.00 mol dm⁻³ HCl with 50.0 cm³ of 2.00 mol dm⁻³ NaOH and only 0.0500 mol of water can form; m is still 100 g. Getting that pair the right way round is the whole difficulty of the harder version of this question.

Why the solution methods are better

Polystyrene wins because it is a poor conductor, has a very low heat capacity of its own, and — the decisive point — the reaction is not separated from the thermometer by a flame, a gap of air and a sheet of metal.

Cooling curves and the assumptions behind the number5.1.7

There is a problem with "record the highest temperature reached". The mixture starts losing heat to the room the instant it starts warming up, so the peak the thermometer shows is already lower than the temperature the mixture would have reached if the reaction had been instantaneous and the cup perfect. The slower the reaction, the worse the shortfall — and for a reaction such as a metal displacing another from solution, which takes minutes, the shortfall is substantial.

The standard fix is to extrapolate a cooling curve.

The method

1. Record the temperature every minute for about four minutes before adding the second reagent, to establish the starting line.
2. Add the reagent at a recorded time — conventionally the fifth minute — and do not take a reading at the moment of mixing.
3. Carry on recording every minute for another ten minutes or so, through the peak and well into the decline.
4. Plot temperature against time. Draw a straight line of best fit through the cooling points after the peak, and extend it back to the moment of mixing.
5. The temperature where that extrapolated line crosses the time of mixing is the corrected maximum. Use it, minus the starting temperature, as ΔT.

Two features of the graph are commonly asked about directly. The flat section before mixing establishes the initial temperature and shows the room is not already heating the cup. The straight cooling line afterwards works because Newtonian cooling over a small temperature range is close enough to linear to extrapolate over one or two minutes — which is also why you extrapolate back a short way and not a long one.

The assumptions you are always making

Every calorimetry answer in this topic rests on a short list of assumptions, and "state the assumptions you have made" is a recurring question. They are:

• the solution has the same specific heat capacity as water, 4.18 J g⁻¹ K⁻¹;
• the solution has the density of water, 1.00 g cm⁻³, so cm³ and g are interchangeable;
• no heat is lost to the surroundings, or to the cup, thermometer and stirrer;
• the apparatus itself absorbs no heat — the polystyrene and the thermometer have negligible heat capacity;
• the reaction is complete.

Each one is slightly false, and each one is false in a knowable direction, which is what makes them worth stating. The heat that went into the cup and the thermometer never reached the water, so it never reached ΔT; a real salt solution has a specific heat capacity a little below water's, so the true q is a little smaller than calculated. Most of these push the same way, which is why calorimetry in a school lab reliably under-estimates rather than scattering randomly about the true value.

Where the energy actually comes from5.1.4

Everything so far has treated ΔH as a number to be measured. This part of the topic asks why it has the value it has, and the answer is entirely about bonds.

It is worth being clear why breaking costs energy, because stated as a rule it sounds arbitrary. A covalent bond exists because the shared electrons are attracted to both nuclei; that attraction is what holds the atoms together, and pulling the atoms apart means working against it. Work against an attraction and you put energy in. Letting two atoms come together and settle into the bonded arrangement is the reverse, so it gives the energy back out.

So why is any reaction exothermic?

Because the two sums are not equal. Every reaction breaks some bonds and makes others, and the enthalpy change is the net result:

Combustion is strongly exothermic because C=O bonds in carbon dioxide and O–H bonds in water are exceptionally strong — far stronger than the C–H, C–C and O=O bonds that had to be broken to reach them. That is the entire explanation of why fuels are fuels.

Exact and mean bond energies5.1.6

A bond energy is a measured quantity, and the subtlety 5.1.6 is after is that some of these measurements are exact and others are averages — which matters enormously for how much you should trust an answer calculated from them.

The exact ones

For a diatomic molecule there is only one bond, so there is nothing to average. H–H in H₂ is +436 kJ mol⁻¹ and that is the whole story:

H₂(g) → 2H(g)    ΔH = +436 kJ mol⁻¹

The same is true of Cl–Cl (+244), O=O (+496), N≡N (+994) and every H–halogen bond. These values are exact in the sense 5.1.6 means: they describe one real bond in one real molecule, not a family of them.

The averaged ones

A C–H bond, by contrast, exists in methane, in ethane, in ethanol, in benzene and in ten million other compounds, and it is slightly different in every one, because the rest of the molecule is different. Methane alone makes the point, because even its four identical-looking C–H bonds come off at four different energies:

The four successive dissociations of methane total 1646 kJ mol⁻¹, so the mean C–H bond energy in methane is about 412 kJ mol⁻¹ — while no individual step is anywhere near that value. Widen the average across many compounds and you get the figure quoted in a data table, around 410. It is a genuinely useful number and it is not the energy of any actual bond.

Calculating ΔH from bond energies5.1.5

The calculation is short and the marks are almost entirely in the bookkeeping. Four steps, and the third is the one people skip.

1. Write the equation out in full displayed form, or at least list every bond in every molecule. H–H is one bond; CH₄ is four C–H bonds; CO₂ is two C=O bonds.
2. Multiply by the coefficients. 2H₂O is four O–H bonds, not two.
3. Add up each side separately, keeping breaking and making apart.
4. Subtract: broken − formed, and give the answer a sign and a unit.

Only the bonds that change

There is a legitimate shortcut. Bonds that appear unchanged on both sides of the equation contribute the same amount to each sum and cancel exactly, so you may leave them out — provided you leave out all of them and are certain they really are unchanged:

Hess's law5.2.1

Some enthalpy changes cannot be measured. You cannot put carbon and hydrogen in a flask and get methane, so ΔH_f[CH₄] has never been measured directly by anybody. You cannot heat calcium carbonate in a calorimeter and get a clean figure, because the decomposition needs a temperature far outside anything a polystyrene cup will survive. Reactions that are too slow, too incomplete, too violent, or that simply produce something else instead, are all out of reach of a thermometer.

Hess's law is the way round every one of those obstacles.

So if you cannot walk from A to B directly, walk from A to C to B instead and add up the two legs. The answer is identical, and that is not an approximation — it follows from the fact that enthalpy is a property of a state, not of a journey. If the two routes gave different answers you could go round the loop repeatedly and create energy out of nothing, which the first law of thermodynamics forbids.

Reading a cycle: follow the arrows

The one skill that makes cycles reliable is a mechanical rule, applied without thinking about whether the answer "looks exothermic":

That is the whole technique, and it handles the signs automatically. You never have to decide whether a step "should" be positive — the arrow on the diagram has already decided, and going backwards along it simply reverses what it says.

Two standard shapes

Every Hess question you will meet at AS is one of two cycles, distinguished entirely by which kind of data you were handed:

The two formulae are mirror images of each other, and mixing them up is the commonest single error in the topic. There is a way to remember which is which that does not rely on memory at all, and it is in the next section.

Cycles built from enthalpies of formation5.2.2

When the data you are given is a list of ΔH_f values, the third corner of the cycle is the elements in their standard states — because that is what a formation value is measured from. Every arrow runs from the elements to a compound, which is to say upwards on the page.

Apply the route rule. Starting at the reactants, the direct route is ΔH_r. The indirect route goes down to the elements — against the left arrow, so subtract ΣΔH_f(reactants) — and back up to the products, with the right arrow, so add ΣΔH_f(products). Equate them:

Cycles built from enthalpies of combustion5.2.2

When the data is a list of ΔH_c values, the third corner is the combustion products — carbon dioxide and water — and every arrow points down towards them, because that is the direction a combustion goes.

Choosing the cycle, and checking the answer5.2.1, 5.2.2

The two formulae are easy to confuse and there is no need to memorise either, because both fall out of one question: which way do the arrows point?

Sanity checks worth thirty seconds

• Does the sign make sense? A combustion is exothermic; a thermal decomposition is endothermic; a neutralisation is exothermic. If your arithmetic says otherwise, it is your arithmetic.
• Is the magnitude plausible? Reaction enthalpies for the reactions in this course run from tens to a few thousand kJ mol⁻¹. An answer of 0.07 or of 90 000 means a factor of a thousand went astray — almost always J against kJ.
• Did every element get a zero? In a formation cycle, O₂, N₂, H₂, Cl₂, C(s, graphite) and every other element in its standard state contributes nothing. In a combustion cycle they emphatically do not — carbon and hydrogen both burn, and both values are needed.
• Are the coefficients in? The most frequent single error in this whole topic is multiplying by 1 where the equation says 2 or 6.

Self-test

Thirty questions across the whole of topic 5, in random order, each with an explanation of why the right answer is right and — where there is a classic wrong answer — why the tempting one is wrong. Work the number out before you pick.

Definitions to learn

These are the statements that are marked word by word. Each one has a phrase that carries the mark, shown in bold; dropping it loses the mark however good the rest of the sentence is.

Data used on this page

Physical constants

Bond energies / kJ mol⁻¹

Values for diatomic molecules are exact bond dissociation energies; the rest are means averaged over a range of compounds.

Standard enthalpies of formation / kJ mol⁻¹

Standard enthalpies of combustion / kJ mol⁻¹

Note the two rows that do double duty: the combustion of C(s, graphite) is the formation of CO₂(g), and the combustion of H₂(g) is the formation of H₂O(l). The same value appears in both tables above, which is a consequence of the definitions rather than a coincidence.