The heat of a reaction

The heat given out by a reaction can be computed from tables for reactions that have never been carried out, and the reason that works is that heat measured in the right way is a property of the change rather than of the apparatus.

The previous lesson counted what a reaction consumes and produces, and ended with the observation that conservation says nothing about direction. Energy is the obvious candidate for what does. This lesson makes the energy of a reaction computable; the next one shows that computing it is not enough.

Heat measured in an open flask is a property of the reaction

Heat is not a property of a system. How much of an energy change arrives as heat rather than as work depends on the path taken, so in general asking for the heat of a process has no answer until the constraints are stated. Chemistry gets lucky, because almost every reaction of interest happens in a vessel open to the atmosphere, which fixes the pressure.

At constant pressure the only work a reaction usually does is pushing the atmosphere back, $p Δ V$ . Conservation of energy then gives $Δ U = Q_{p} - p Δ V$ , and rearranging,

Q_{p} = Δ U + p Δ V = (U_{2} + p V_{2}) - (U_{1} + p V_{1})

The heat is the change in the combination $U + p V$ , evaluated at the start and the end and nowhere in between. That combination is the enthalpy $H$ , and since $U$ , $p$ and $V$ are all properties of the state, so is $H$ . The result is $Δ H = Q_{p}$ : at constant pressure, with only expansion work, the heat absorbed is the change in a state function.

This is why every table in chemistry is a table of enthalpies. A reaction run in a flask, in a bucket or in a chemical plant gives the same $Δ H$ , because the $p Δ V$ term the atmosphere absorbs has already been folded into the bookkeeping. Nothing is stored in the $p V$ term in any physical sense; it is the correct accounting for the commonest constraint. A reaction with $Δ H < 0$ releases heat and is exothermic, one with $Δ H > 0$ absorbs it and is endothermic.

Formation enthalpies, and Hess's law as a consequence

Enthalpy has no absolute zero, so only differences are tabulated, and the convention picks one reference. The standard enthalpy of formation $Δ_{f} H^{ominus}$ is the enthalpy change on forming one mole of a substance from its elements in their standard states at $10^{5}$ Pa, and an element in its standard state is assigned zero. Graphite is zero and diamond is $+ 1.9$ kJ mol⁻¹, because graphite is the stable form at ordinary conditions.

Because $H$ is a state function, the enthalpy change around any closed path is zero, so a reaction enthalpy does not depend on the route. That is Hess's law, published by Germain Hess in 1840, before the First Law of thermodynamics was settled. It is not an extra assumption but a restatement of what a state function is, and it licenses routing every reaction through the elements:

Δ_{r} H^{ominus} = \sum ν_{i} Δ_{f} H^{ominus} (i)

with the stoichiometric numbers of the previous lesson, negative for reactants and positive for products. The practical payoff is that a table of a few thousand formation enthalpies gives the enthalpy of any reaction that can be written between them, including reactions too slow, too violent or too impure to measure.

Example. The thermite reaction is $F(s) + 2 Al(s) \to A(s) + 2 Fe(s)$ . Given $Δ_{f} H^{ominus} = - 824.2$ kJ mol⁻¹ for $F$ and $- 1675.7$ for $A$ , find $Δ_{r} H^{ominus}$ .

Aluminium and iron are elements in their standard states, so both are zero. The products sum to $- 1675.7$ and the reactants to $- 824.2$ , giving $Δ_{r} H^{ominus} = - 1675.7 - (- 824.2) = - 851.5$ kJ mol⁻¹. Released into a small mass with no cooling, that is enough to raise the iron produced past its melting point of $1538$ °C, which is exactly what the reaction is used for when welding rail.

Now you. Find $Δ_{r} H^{ominus}$ for $2 S(g) + 3 (g) \to 2 S(g) + 2 O(l)$ , given $Δ_{f} H^{ominus} = - 20.6$ kJ mol⁻¹ for $S$ , $- 296.8$ for $S$ and $- 285.8$ for liquid water.

Answer

Products: $2 (- 296.8) + 2 (- 285.8) = - 1165.2$ kJ mol⁻¹. Reactants: $2 (- 20.6) = - 41.2$ , oxygen being zero. So $Δ_{r} H^{ominus} = - 1165.2 + 41.2 = - 1124.0$ kJ mol⁻¹.

The state symbols are load-bearing. Taking the water as vapour instead would change the answer by $2 \times 44 = 88$ kJ, the enthalpy of vaporisation of two moles, and that difference is precisely why natural gas has two quoted heating values that differ by about ten per cent.

Bond enthalpies: a cruder estimate that says why

Formation enthalpies give the number without explaining it. Bonds do the reverse. A reaction breaks some bonds and makes others, breaking costs energy and making releases it, so to a first approximation

Δ_{r} H \approx \sum (bonds broken) - \sum (bonds formed)

The values used are mean bond enthalpies, averaged over many molecules, since the strength of a C to H bond is not quite the same in methane as in ethanol. That averaging is the source of the method's error and also of its reach: one table of about thirty numbers covers organic chemistry.

Example. Estimate the enthalpy of combustion of methane to $C$ and water vapour, using mean bond enthalpies of $413$ kJ mol⁻¹ for C to H, $498$ for O to O double, $799$ for C to O double in carbon dioxide and $463$ for O to H.

Breaking four C to H bonds and two oxygen double bonds costs $4 (413) + 2 (498) = 2648$ kJ. Forming two carbon to oxygen double bonds and four O to H bonds releases $2 (799) + 4 (463) = 3450$ kJ. The estimate is $2648 - 3450 = - 802$ kJ mol⁻¹, against $- 802.5$ from formation enthalpies.

Now you. Estimate $Δ_{r} H$ for the hydrogenation of ethene, $+ \to$ , given C to C double $614$ , H to H $436$ , C to C single $347$ and C to H $413$ kJ mol⁻¹. Compare with the value from $Δ_{f} H^{ominus} = + 52.4$ kJ mol⁻¹ for ethene and $- 84.0$ for ethane.

Answer

Broken: the carbon double bond and the hydrogen bond, $614 + 436 = 1050$ kJ. Formed: one carbon single bond and two new C to H bonds, $347 + 2 (413) = 1173$ kJ. The estimate is $- 123$ kJ mol⁻¹; the formation data give $- 84.0 - 52.4 = - 136.4$ . The estimate is out by $13$ kJ mol⁻¹, about ten per cent.

The agreement in the methane example is better than the method deserves, and treating it as typical would be a mistake. Errors of $10$ to $30$ kJ mol⁻¹ are normal, and they become much worse wherever a real molecule is not a collection of independent bonds. Benzene is the standard casualty: bond enthalpies predict its hydrogenation enthalpy about $150$ kJ mol⁻¹ too negative, because the delocalised ring is more stable than three isolated double bonds. Bond enthalpies also apply only to gases, since they say nothing about the energy of holding a liquid or a lattice together. Use them to see why a reaction is exothermic, and formation enthalpies when the number has to be right.

Measuring it: two calorimeters

Both tables above are ultimately built from measurements, and the measurement means choosing a constraint. A coffee cup calorimeter, an insulated vessel open to the air, holds the pressure fixed and so measures $Δ H$ directly. It suits reactions in solution: neutralisations, dissolutions, precipitations. The heat released warms the solution, and if the solution is dilute its specific heat capacity is close to that of water, $4.18$ J g⁻¹ K⁻¹.

Example. Mixing $50.0$ mL of $1.00$ mol dm⁻³ hydrochloric acid with $50.0$ mL of $1.00$ mol dm⁻³ sodium hydroxide, both at $21.0$ °C, raises the temperature to $27.8$ °C. Taking the mixture as $100.0$ g of water, find the enthalpy of neutralisation per mole of water formed.

The heat gained by the solution is $q = m c Δ T = 100.0 \times 4.18 \times 6.8 = 2842$ J. The acid supplies $0.0500$ mol of protons and the base $0.0500$ mol of hydroxide, so $0.0500$ mol of water forms. Since the solution gained that heat, the reaction lost it: $Δ H = - 2842 / 0.0500 = - 56.8$ kJ mol⁻¹, against a tabulated $- 57.3$ . The shortfall is heat that went into the polystyrene cup and the thermometer rather than the liquid.

Now you. The same experiment is done with $50.0$ mL of each solution at $2.00$ mol dm⁻³, again treating the mixture as $100.0$ g of water with $c = 4.18$ J g⁻¹ K⁻¹, and the true value of $- 57.3$ kJ mol⁻¹. What temperature rise should a perfect calorimeter show?

Answer

Now $0.100$ mol of water forms, releasing $0.100 \times 57300 = 5730$ J. So $Δ T = 5730 / (100.0 \times 4.18) = 13.7$ K. Doubling the concentration doubles the rise, because both the heat and the mass of solution scale, but only the heat depends on concentration.

A bomb calorimeter seals the sample in a rigid steel vessel under excess oxygen, so the volume rather than the pressure is fixed. No expansion work is possible, and the heat measured is $Δ U$ , not $Δ H$ . The two are related by $Δ H = Δ U + Δ n_{gas} R T$ , where only the change in moles of gas matters, since solids and liquids occupy a negligible volume by comparison. For methane, three moles of gas become one, so $Δ n_{gas} = - 2$ and the correction is $2 \times 2.48 = 5.0$ kJ mol⁻¹, half a per cent and far larger than the instrument's error.

The same reaction at a different temperature

Tables are at $298.15$ K and reactions are run wherever they are run. The correction follows from Hess's law again: instead of reacting at $T_{2}$ , cool the reactants to $298$ K, react there, and warm the products back. The two temperature changes cost the heat capacities of the reactants and products respectively, so

Δ_{r} H (T_{2}) = Δ_{r} H (T_{1}) + Δ_{r} C_{p} (T_{2} - T_{1})

with $Δ_{r} C_{p}$ the same stoichiometric sum over molar heat capacities. This is Kirchhoff's law, and treating $Δ_{r} C_{p}$ as constant is an approximation good over a hundred kelvin or so.

Apply it to ammonia synthesis, $+ 3 \to 2 N$ , with $Δ_{r} H^{ominus} (298) = - 91.8$ kJ mol⁻¹ and molar heat capacities of $29.12$ , $28.82$ and $35.06$ J mol⁻¹ K⁻¹. Then $Δ_{r} C_{p} = 2 (35.06) - 29.12 - 3 (28.82) = - 45.5$ J K⁻¹, and at $500$ K, a typical plant temperature, $Δ_{r} H = - 91.8 - 0.0455 (202) = - 101.0$ kJ mol⁻¹. The reaction is nine per cent more exothermic hot than the table suggests, which matters when the reactor has to be cooled continuously.

The sign of $Δ_{r} C_{p}$ has a plain reading. It is negative here because four moles of gas become two, and fewer molecules store less thermal energy, so the products carry less of the temperature change than the reactants did.

Dissolving: two large numbers that nearly cancel

Dissolving a salt is a good test of the machinery, because it splits cleanly into two steps that can be looked at separately. Pull the lattice apart into gaseous ions, which costs the lattice enthalpy, then let water surround each ion, which releases the hydration enthalpy. For sodium chloride the lattice enthalpy is $+ 787$ kJ mol⁻¹ and the hydration enthalpies of the two ions sum to about $- 783$ , so the enthalpy of solution is the difference of two numbers near $800$ : about $+ 4$ kJ mol⁻¹, against a measured $+ 3.9$ .

That near-cancellation is the whole story of solubility enthalpies. Both terms scale with charge and with the inverse of ionic radius, so they track each other, and what survives is a small residue whose sign is genuinely hard to predict. Sodium chloride absorbs a little heat as it dissolves, lithium chloride releases $37$ kJ mol⁻¹, and ammonium nitrate absorbs $25.7$ . That last one is sold in instant cold packs, and it is about to cause trouble.

Exothermic is not the same as spontaneous

The intuition that reactions happen because they release energy is old, respectable and false. Marcellin Berthelot stated it as a principle in 1867: every spontaneous change is the one that releases the most heat. It accounts for a great many reactions, which is why it survived as long as it did.

The counterexamples are on any bench. Ammonium nitrate dissolves eagerly in water while chilling the beaker, with $Δ_{sol} H^{ominus} = + 25.7$ kJ mol⁻¹. Ice in a room at $10$ °C melts, absorbing $6.01$ kJ mol⁻¹. Mix solid barium hydroxide octahydrate with solid ammonium thiocyanate and the two powders liquefy while the beaker drops to around $- 20$ °C, cold enough to freeze it to a wooden block, absorbing some $80$ kJ per mole. All three go uphill in enthalpy and all three go anyway.

Notice what the three have in common: in every case the products are more spread out, more mobile, more numerous than the reactants. A crystal becomes free ions in solution; a solid becomes a liquid; two solids become a liquid and a gas. Something is being gained that is not energy, and gained enough to pay an energy penalty of tens of kilojoules per mole.

Naming that quantity, tabulating it as reliably as formation enthalpies are tabulated, and combining it with $Δ H$ into a single criterion that decides the direction of any reaction, is the next lesson.