Corrected exercises: thermochemistry and the first law (202-SN1-RE)

We take the point of view of the SYSTEM: $q>0$ if the system receives heat, $q<0$ if it gives some off; $w>0$ if the surroundings do work on the system, that is, if it is compressed, and $w<0$ if the system does work, that is, if it expands. The first law is then written $\Delta U=q+w$, and it states that $\Delta U$ depends only on the initial and final states, whereas $q$ and $w$ depend separately on the path followed.

$\Delta V=5.00-2.00=+3.00$ L. The work is $w=-P\Delta V=-101.3\times 3.00=-304$ kPa·L, that is $-304$ J. The sign is negative because the gas PUSHES the atmosphere: it does work. Then $\Delta U=q+w=500+(-304)=+196$ J.

The system GAINED 196 J of internal energy. It received 500 J as heat, but it immediately gave 304 J back to the surroundings as mechanical work, by pushing back the atmosphere. Only the balance stays stored in the system, as molecular motion, which shows up as a rise in temperature. This is exactly the principle of an engine: converting the heat received into work delivered.

In a vacuum the external pressure is zero, so $w=-P_{ext}\Delta V=0$ whatever the change in volume: the gas pushes against NOTHING, so it does no work. We would then have $\Delta U=q=500$ J. This expansion, called free expansion (the Joule expansion), shows that work is not a property of the system alone: it depends on what the system works against. Two expansions leading to the same final state can exchange completely different amounts of work, which illustrates that $w$ is NOT a state function, unlike $\Delta U$.

The water warms by $24.1-20.0=4.1$ °C, so it receives $q_{\text{water}}=250\times 4.18\times 4.1=4285$ J. The metal cools by $24.1-100.0=-75.9$ °C and gives up exactly this heat: $q_{\text{metal}}=-4285$ J $=125\times c\times(-75.9)$, so $c=\dfrac{4285}{125\times 75.9}=0.452$ J/(g·°C). This value is very close to that of iron, 0.449 J/(g·°C), and far from copper and from aluminum: the metal is iron.

The calorimeter also warms by 4.1 °C and absorbs $32.0\times 4.1=131$ J, which also come from the metal. The heat given up is therefore $4285+131=4416$ J, so $c=\dfrac{4416}{125\times 75.9}=0.465$ J/(g·°C), which is 3.1% higher. The omission UNDERESTIMATED the specific heat, and this direction was predictable without any calculation: neglecting the calorimeter amounts to forgetting part of the heat received, hence underestimating the heat given up by the metal, hence its specific heat. A line of reasoning like this should always accompany a correction: you should know which way it goes before putting a number on it.

The mass of 1.50 L of water is 1500 g, and the required rise is $100.0-15.0=85.0$ °C. Then $q=1500\times 4.18\times 85.0=5.33\times 10^{5}$ J, that is 533 kJ. In 4.00 min $=240$ s, the power required is $P=\dfrac{532{,}950}{240}=2221$ W. This is exactly the power of household kettles, around 2000 to 2200 W, which confirms the order of magnitude.

$\dfrac{4.18}{0.449}=9.31$: it takes well over nine times more energy to raise a gram of water by one degree than a gram of iron, and conversely water gives back as much when it cools. Climatic consequence: a large body of water warms and cools very slowly, which moderates the climate of coastal and lakeside regions. Montreal, surrounded by water but in a continental climate, has much larger temperature swings than a coastal city at the same latitude. Technical consequence: water is the ideal coolant, in an engine as in a power plant, because it carries away a lot of heat while warming very little. This same property stabilizes the temperature of the human body, which is mostly water.

$HCl+NaOH\rightarrow NaCl+H_{2}O$, or in net ionic form $H^{+}+OH^{-}\rightarrow H_{2}O$. Each solution supplies $n=0.0500\times 1.00=0.0500$ mol: the reactants are in exact proportions and 0.0500 mol of water is formed.

The total mass of the mixture is $50.0+50.0=100.0$ mL, that is 100.0 g. The rise is $\Delta T=27.8-21.0=6.8$ °C. Then $q_{\text{mix}}=mc\Delta T=100.0\times 4.18\times 6.8=2.84\times 10^{3}$ J. The mixture warms up, so the reaction GAVE OFF this heat.

The reaction gives off 2842 J for 0.0500 mol, so $\Delta H=-\dfrac{2842}{0.0500}=-5.68\times 10^{4}$ J/mol, that is $-56.8$ kJ/mol. The difference from the accepted value of $-57.3$ kJ/mol is 0.9%: excellent for such a rudimentary setup.

The calorimeter also warms up, and this heat also comes from the reaction: $q_{\text{total}}=\left(mc+C_{cal}\right)\Delta T=\left(100.0\times 4.18+15.0\right)\times 6.8=433\times 6.8=2944$ J. Then $\Delta H=-\dfrac{2944}{0.0500}=-58.9$ kJ/mol. By neglecting the calorimeter, we therefore UNDERESTIMATED the heat released: part of the energy went into heating the cup and the thermometer without being counted. The omission always leads to a value that is too small in absolute value.

Because the reaction actually taking place is ALWAYS the same. A strong acid is completely dissociated, and so is a strong base: before reaction, the solution contains separate $H^{+}$, $Cl^{-}$, $Na^{+}$ and $OH^{-}$ ions. The $Na^{+}$ and $Cl^{-}$ ions are mere SPECTATORS, and they are unchanged at the end. The only chemical change is $H^{+}+OH^{-}\rightarrow H_{2}O$, hence an enthalpy close to $-57$ kJ/mol whichever pair is chosen. The value would depart noticeably only for a WEAK acid or base, because you would then have to pay the dissociation energy as well.

Forming an element from itself is not a reaction: the corresponding enthalpy is zero by convention, and this convention is essential to give the whole scale a common zero. It applies only to the standard state of the element, that is, its most stable form at 25 °C under 100 kPa: graphite carbon (not diamond), dioxygen $O_{2}$ (not ozone), liquid bromine (not gaseous). For water, $H_{2}O(l)$ and $H_{2}O(g)$ are not the same species: the difference $-241.8-(-285.8)=+44.0$ kJ/mol is exactly the enthalpy of vaporization of water at 25 °C. Specifying the physical state in a table of enthalpies is therefore never a detail.

$\Delta H^{\circ}=\left[3(-393.5)+4(-285.8)\right]-\left[(-103.8)+5(0)\right]=(-1180.5-1143.2)+103.8=-2219.9$ kJ/mol. The combustion of propane therefore releases nearly 2220 kJ per mole, which makes it an excellent fuel.

With $H_{2}O(g)$: $\Delta H^{\circ}=\left[3(-393.5)+4(-241.8)\right]+103.8=-2043.9$ kJ/mol. The difference is $176.0$ kJ/mol, exactly $4\times 44.0$: the four moles of water formed stay as vapour instead of condensing, and the heat of condensation is not recovered. The value with liquid water, 2219.9 kJ/mol, is the higher heating value; the one with vapour, 2043.9 kJ/mol, is the lower heating value. An ordinary burner vents the hot flue gases, so the water leaves as vapour and it recovers only the lower heating value. Condensing boilers are designed precisely to cool the flue gases below the dew point and recover that extra $176.0$ kJ/mol, that is $7.9$% of the higher heating value, which explains why their efficiency is advertised as above 100% when it is based on the lower heating value.

The equation is $C_{2}H_{2}(g)+\dfrac{5}{2}O_{2}(g)\rightarrow 2\,CO_{2}(g)+H_{2}O(l)$, with $\Delta H^{\circ}=-1299.6$ kJ/mol. We write $-1299.6=\left[2(-393.5)+(-285.8)\right]-\Delta H_{f}^{\circ}(C_{2}H_{2})$, that is $-1299.6=-1072.8-\Delta H_{f}^{\circ}(C_{2}H_{2})$, so $\Delta H_{f}^{\circ}(C_{2}H_{2})=-1072.8+1299.6=+226.8$ kJ/mol. The positive sign is surprising at first, because most common compounds have a negative enthalpy of formation. It means that acetylene sits higher in energy than the graphite carbon and the dihydrogen it comes from: it is an endothermic compound, thermodynamically unstable with respect to its elements. This energy stored at formation is released at combustion, which explains the oxyacetylene flame and its 3000 °C, and also why pure acetylene decomposes explosively under pressure, without even needing oxygen.

$\Delta H\approx\sum E(\text{broken})-\sum E(\text{formed})$. Breaking a bond always requires supplying energy, which contributes POSITIVELY to $\Delta H$; forming a bond releases energy, which contributes NEGATIVELY. A reaction is therefore exothermic when the bonds formed are, overall, stronger than those that were broken.

You must count only the bonds that CHANGE. Broken: one $C-H$ bond of methane and the $Cl-Cl$ bond, that is $414+243=657$ kJ/mol. Formed: one $C-Cl$ bond and one $H-Cl$ bond, that is $339+431=770$ kJ/mol. Then $\Delta H\approx 657-770=-113$ kJ/mol. The other three $C-H$ bonds of methane are intact on both sides: listing them on both sides would change nothing, but counting them on one side only would be a mistake.

Broken: $H-H$ and $Cl-Cl$, that is $436+243=679$ kJ/mol. Formed: two $H-Cl$ bonds, that is $2\times 431=862$ kJ/mol. Hence $\Delta H\approx 679-862=-183$ kJ/mol. From the enthalpies of formation, $\Delta H^{\circ}_{r}=2\times(-92.3)-0-0=-184.6$ kJ/mol. The difference is 1.6 kJ/mol, less than 1%: the estimate is excellent here.

First limitation: the tabulated energies are AVERAGES over many molecules. The $C-H$ bond is not worth exactly the same in methane, in ethanol or in benzene, yet it is assigned a single value of 414 kJ/mol. Second limitation: the method is valid only for GASEOUS species. As soon as a reactant or a product is a liquid or a solid, you would have to add the enthalpies of change of state, which a bond count ignores. That is why the result in b), where every species is a gas, is reliable, and why the same method would be poor for a combustion producing liquid water.

When the compounds involved appear in NO table of enthalpies of formation, which is the usual situation in organic chemistry: there are tens of millions of organic compounds and tables for only a few thousand. A chemist who wants to know whether a planned reaction will be exothermic then counts the bonds, gets an order of magnitude in a few minutes, and decides whether the experiment is worth doing. It is a tool for DECIDING, not for measuring.

Because the volume is CONSTANT: $\Delta V=0$, so $w=-P\Delta V=0$. The first law then reduces to $\Delta U=q_{V}$: all the heat exchanged directly measures the change in internal energy. In an open calorimeter, by contrast, the system can expand and do work on the atmosphere, and the heat measured is $q_{P}=\Delta H$.

$q=C_{cal}\Delta T=9.50\times 1.65=15.68$ kJ released by 1.00 g. The amount burned is $n=\dfrac{1.00}{180.16}=5.55\times 10^{-3}$ mol. Per mole: $\dfrac{15.68}{5.55\times 10^{-3}}=2824$ kJ, so $\Delta U=-2824$ kJ/mol.

Count ONLY the gases: on the left 6 moles of $O_{2}$, on the right 6 moles of $CO_{2}$, the water being liquid and the glucose solid. So $\Delta n_{\text{gas}}=6-6=0$, and $\Delta H=\Delta U+\Delta n_{\text{gas}}RT=\Delta U=-2824$ kJ/mol. The two quantities coincide exactly. This is a happy accident of stoichiometry, and it is why the food energy of glucose, about 15.7 kJ per gram, can be read directly from a bomb measurement.

Here $\Delta n_{\text{gas}}=1-\left(1+2\right)=-2$, because $CH_{4}$ and $O_{2}$ are gases on the left and only $CO_{2}$ is on the right. Then $\Delta H-\Delta U=\Delta n_{\text{gas}}RT=(-2)\times 8.314\times 10^{-3}\times 298=-4.96$ kJ/mol. The enthalpy is therefore more negative than the internal energy by nearly 5 kJ/mol: when methane burns at constant pressure, the system CONTRACTS, the atmosphere does work on the system, and that much more heat has to be given off.

When $\Delta n_{\text{gas}}=0$, as in c), but also, and above all, for every reaction taking place entirely in the CONDENSED phase, solids and liquids only: volume changes are then thousands of times smaller, and the term $\Delta n_{\text{gas}}RT$ is zero by construction. In practice, the correction reaches a few kJ/mol only when gases appear or disappear, which remains small next to combustion enthalpies of several thousand kJ/mol: less than 0.2% here.

You may REVERSE an equation, which changes the sign of $\Delta H$; you may MULTIPLY it by a coefficient, which multiplies $\Delta H$ by that same coefficient; and you may ADD two equations, which adds their $\Delta H$ values. These three operations are legitimate because enthalpy is a state function: its change depends only on the starting and ending states, never on the path.

We want $C(s)+\frac{1}{2}O_{2}\rightarrow CO$. It is enough to take (1) and SUBTRACT (2), that is, to add (2) reversed: $C+O_{2}\rightarrow CO_{2}$ then $CO_{2}\rightarrow CO+\frac{1}{2}O_{2}$. When we add, $CO_{2}$ cancels and we are indeed left with $C+\frac{1}{2}O_{2}\rightarrow CO$. So $\Delta H=-393.5-(-283.0)=-110.5$ kJ/mol. Direct measurement is impossible because the combustion of carbon cannot be stopped at the monoxide: a mixture of $CO$ and $CO_{2}$ always forms, in uncontrolled proportions. This is precisely the kind of situation Hess's law was designed for.

We want $2C(s)+H_{2}(g)\rightarrow C_{2}H_{2}(g)$. The combination is $2\times(1)+(4)-(3)$: we form two $CO_{2}$ from the carbon, one $H_{2}O$ from the dihydrogen, then run the combustion of acetylene backwards by reversing it. Then $\Delta H=2(-393.5)+(-285.8)-(-1299.6)=-787.0-285.8+1299.6=+226.8$ kJ/mol. Check: the $O_{2}$ cancels, $2+\frac{1}{2}-\frac{5}{2}=0$, and both $CO_{2}$ and $H_{2}O$ disappear.

By its very definition, this value is the standard enthalpy of FORMATION of acetylene, since the reaction starts from the elements in their standard states. Its positive sign means that acetylene is less stable than the elements it comes from: it is an ENDOTHERMIC compound, whose formation requires energy. This is not surprising when you consider that an electric arc furnace is needed to produce it, and it also explains its energetic character in combustion, which makes it the gas of the oxyacetylene torch, as well as its instability under compression.

By adding (1) and (2) as they stand, the student obtains $C+O_{2}+CO+\frac{1}{2}O_{2}\rightarrow 2CO_{2}$, that is, after simplification, $C(s)+CO(g)+\frac{3}{2}O_{2}(g)\rightarrow 2CO_{2}(g)$, whose enthalpy is indeed $-676.5$ kJ/mol. The number is therefore not wrong: it simply describes a DIFFERENT reaction from the one requested. This is the characteristic trap of Hess's law, where an arithmetically correct calculation answers the wrong question. The essential check is to rewrite the summed equation and verify that it coincides exactly with the target.

The cycle links the initial state $Na(s)+\dfrac{1}{2}Cl_{2}(g)$ to the final state $NaCl(s)$ by a detour in five steps: sublimation of sodium, $Na(s)\rightarrow Na(g)$, $+107.3$; ionization, $Na(g)\rightarrow Na^{+}(g)+e^{-}$, $+495.8$; dissociation of dichlorine, $\dfrac{1}{2}Cl_{2}(g)\rightarrow Cl(g)$, $+121.3$; attachment of the electron, $Cl(g)+e^{-}\rightarrow Cl^{-}(g)$, $-349.0$; and finally condensation of the gaseous ions into a lattice, $Na^{+}(g)+Cl^{-}(g)\rightarrow NaCl(s)$. The dissociation term counts for only half because the formation equation consumes only half a $Cl_{2}$: the tabulated value, 242.6 kJ/mol, corresponds to breaking a whole mole of $Cl-Cl$ bonds to give two moles of atoms, and we use only one mole of them.

Hess's law says that the sum of the steps along the detour equals the enthalpy of the direct path: $\Delta H_{f}^{\circ}=107.3+495.8+121.3-349.0+\Delta H_{\text{lattice}}$. Since $\Delta H_{f}^{\circ}=-411.2$ and the sum of the first four terms is $+375.4$, we get $\Delta H_{\text{lattice}}=-411.2-375.4=-786.6$ kJ/mol. The lattice energy, defined as the energy needed to separate the crystal into gaseous ions, is the opposite: $+786.6$ kJ/mol. The signs are consistent: the first three steps cost energy, only the attachment of the electron to chlorine releases a little, and it is the formation of the lattice that releases most of the energy and makes the whole process exothermic. Without it, the overall balance would be strongly positive and $NaCl$ would not exist.

Dissolving the salt first requires taking the lattice apart into gaseous ions, which costs the lattice energy, $+786.6$ kJ/mol, then hydrating these ions, which releases energy: $-406-377=-783$ kJ/mol. The balance is $\Delta H_{\text{solution}}=786.6-783=+3.6$ kJ/mol, to be compared with the $+3.9$ kJ/mol measured. The agreement is excellent, but you need to see how fragile it is: the result, a few kilojoules, is the difference of two numbers close to 800. An error of 1% on either one, that is 8 kJ, would be enough to change the sign of the result and therefore the physical conclusion. This is a typical example of a calculation in which the relative precision of the result is far worse than that of the data, and in which you must avoid rounding along the way.

Everything comes down to the competition between two large terms of opposite signs: the lattice energy, which opposes dissolution, and the hydration, which favours it. For ammonium nitrate, the lattice clearly wins: dissolution absorbs 25.7 kJ/mol and cools the solution, which is exactly the principle of instant cold packs. For calcium chloride, the $Ca^{2+}$ ion carries a double charge and has a small radius, so its hydration releases an enormous amount of energy and wins over the lattice: dissolution gives off 81.3 kJ/mol and warms the solution, hence its use in hot packs and as a de-icer that works even in severe cold. As for table salt, its endothermic character does not prevent it from dissolving because enthalpy does not decide alone: going from a perfectly ordered crystal to ions dispersed and free in water greatly increases the disorder. This gain in entropy more than pays for an enthalpy cost of only a few kilojoules, and it is the combined balance, the free energy, that really governs spontaneity. That is the subject of the 202-SN2-RE course.

In both cases, the mixture weighs $100+100=200$ g and contains $0.100\;\text{L}\times 1.00\;\text{mol/L}=0.100$ mol of each reactant, so 0.100 mol of water is formed. First experiment: $\Delta T=27.9-21.0=6.9$ °C, so $q=200\times 4.18\times 6.9=5768$ J absorbed by the mixture. The reaction therefore gave off this heat: $\Delta H=-\dfrac{5768}{0.100}=-57.7$ kJ/mol, very close to the accepted value of $-57.3$ kJ/mol for a strong acid and a strong base. Second experiment: $\Delta T=6.7$ °C, $q=200\times 4.18\times 6.7=5601$ J, so $\Delta H=-56.0$ kJ/mol.

With hydrochloric acid, both reactants are completely dissociated in solution and the only real change is $H^{+}(aq)+OH^{-}(aq)\rightarrow H_{2}O(l)$: this is why the value depends on neither the strong acid nor the strong base chosen. With acetic acid, the equation is $CH_{3}COOH(aq)+OH^{-}(aq)\rightarrow CH_{3}COO^{-}(aq)+H_{2}O(l)$, and by Hess's law it breaks into two steps: first ionize the acid, $CH_{3}COOH\rightarrow CH_{3}COO^{-}+H^{+}$, then neutralize, $H^{+}+OH^{-}\rightarrow H_{2}O$. The second step is the same as in the first experiment. The whole difference therefore comes from the first step, which consumes energy to break the $O-H$ bond of the undissociated acid: part of the heat released by the neutralization goes to pay for this ionization instead of warming the mixture, and the measured balance is less exothermic.

Hess's law gives directly $\Delta H_{\text{ionization}}=\Delta H_{\text{weak}}-\Delta H_{\text{strong}}=(-56.0)-(-57.7)=+1.7$ kJ/mol. The positive sign is consistent with the interpretation in b): ionizing a weak acid costs energy. The value is small, of the order of a kilojoule, which also explains why the acidity of a weak acid depends rather little on temperature: the decisive factor in its acid dissociation constant is not the enthalpy but the entropy of solvation of the ions formed.

Each temperature is known to about $\pm 0.1$ °C, so each $\Delta T$ to $\pm 0.2$ °C, and the difference of the two rises to $\pm 0.4$ °C in the worst case. Now 0.2 °C already corresponds to $200\times 4.18\times 0.2/0.100=1.7$ kJ/mol, exactly the value we claim to measure. The uncertainty is therefore of the same order as the result, or even larger: the method gives the right sign and the right order of magnitude, but no reliable significant figure. This is the structural weakness of any measurement obtained as the difference of two nearby quantities, already met with the Born-Haber cycle. To get a trustworthy value, you need a precision calorimeter and thermistors reading to hundredths of a degree. With ammonia, a weak base, the same phenomenon would be seen on the base side: part of the energy would go into ionizing $NH_{3}$, and the temperature rise would be even smaller, of the order of 6.4 °C. Neutralizing a weak acid with a weak base would add up both costs and give the least exothermic value of the four combinations.

We divide the molar enthalpy by the molar mass. Dihydrogen: $\dfrac{285.8}{2.016}=141.8$ kJ/g. Octane: $\dfrac{5470}{114.23}=47.9$ kJ/g. Ethanol: $\dfrac{1367}{46.07}=29.7$ kJ/g.

Ranking by energy per mass: dihydrogen 141.8, then octane 47.9, then ethanol 29.7 kJ/g. Dihydrogen crushes the competition, with nearly THREE times the energy per mass of gasoline. The reason is its tiny molar mass: there is no heavy atom to carry around, only the hydrogen that holds the energy. This is why it is the fuel of space launch vehicles, where every kilogram carried costs a great deal.

We multiply the energy per mass by the density, noting that kJ/g multiplied by g/mL gives kJ/mL, that is MJ/L. Octane: $47.9\times 0.703=33.7$ MJ/L. Ethanol: $29.7\times 0.789=23.4$ MJ/L. Dihydrogen at 700 bar: $141.8\times 0.040=5.7$ MJ/L. New ranking: octane, then ethanol, then dihydrogen DEAD LAST.

The reversal comes entirely from the density. Dihydrogen, even compressed to 700 bar, remains seventeen times less dense than gasoline: its advantage of 3 in mass is more than offset by a disadvantage of 17 in volume, giving a final ratio of about 6 in favour of gasoline. Practical consequence for the automobile, where the available volume is constrained: a hydrogen tank must be about six times larger, on top of being a 700 bar tank, hence thick, heavy and expensive. This is the main technical obstacle for the hydrogen car, and it is as much thermochemical as mechanical in nature.

For reasons that have nothing to do with energy content. Ethanol is made from biomass, so it is partly renewable, and its carbon balance is better since the carbon released was taken from the atmosphere the year before. It also has a high octane rating, which allows the engine's compression ratio to be raised and part of the lost efficiency to be recovered. Finally, its molecule already contains an oxygen atom, which improves combustion and reduces emissions of carbon monoxide and unburned hydrocarbons. The price to pay is a higher consumption in litres, in exactly the ratio calculated in c).