Corrected exercises: chemical thermodynamics (202-SN2-RE)

Because $O_{2}(g)$ IS the reference element of oxygen in its standard state. The formation reaction would be $O_{2}(g)\rightarrow O_{2}(g)$, which changes nothing: its enthalpy is zero by definition. The same holds for $N_{2}(g)$, $H_{2}(g)$ or graphite carbon. Careful: $O_{3}(g)$ is not the standard state of oxygen, so its $\Delta H^{\circ}_{f}$ is not zero.

The equation is $CH_{4}(g)+2O_{2}(g)\rightarrow CO_{2}(g)+2H_{2}O(l)$. Then $\Delta H^{\circ}_{r}=\left[\Delta H^{\circ}_{f}\left(CO_{2}\right)+2\Delta H^{\circ}_{f}\left(H_{2}O(l)\right)\right]-\left[\Delta H^{\circ}_{f}\left(CH_{4}\right)+2\times 0\right]$, that is $\left[-393.5+2(-285.8)\right]-\left[-74.6\right]=-965.1+74.6=-890.5$ kJ/mol. The negative sign confirms an exothermic combustion.

With $2H_{2}O(g)$: $\left[-393.5+2(-241.8)\right]-\left[-74.6\right]=-877.1+74.6=-802.5$ kJ/mol.

The difference is $-890.5-(-802.5)=-88.0$ kJ/mol, which means the liquid-water version releases 88.0 kJ MORE. This difference is exactly $2\times 44.0$ kJ, twice the enthalpy of vaporization of water, since the reaction involves 2 moles of water. Physical interpretation: when the vapour condenses, it also releases its latent heat. This is the whole difference between the HIGHER heating value of a fuel, measured with condensation, and its LOWER heating value, measured without. It is also the reason condensing boilers exist: they recover precisely these 88 kJ by cooling the flue gases below the dew point.

First reaction: we start with 3 moles of gas and end with 0 moles of gas, the water being liquid. Disorder drops sharply, so $\Delta S^{\circ}_{r}<0$. Second: we start with 0 moles of gas and produce 1, so $\Delta S^{\circ}_{r}>0$. Third: 4 moles of gas give 2 moles of gas, so $\Delta S^{\circ}_{r}<0$.

$\Delta S^{\circ}_{r}=\left[S^{\circ}(CaO)+S^{\circ}\left(CO_{2}\right)\right]-S^{\circ}\left(CaCO_{3}\right)=(39.8+213.7)-92.9=253.5-92.9=+160.6$ J/(mol·K). The result is indeed POSITIVE, as predicted. Note that the dominant term is by far the $S^{\circ}$ of the gas, 213.7, larger than the two solids put together: this is what justifies the rule about moles of gas.

Because entropy is an ABSOLUTE quantity, not a change. The third law of thermodynamics fixes its origin: the entropy of a perfect crystal at absolute zero is zero. Any substance at a nonzero temperature therefore has a strictly positive entropy. The $\Delta H^{\circ}_{f}$, on the other hand, are DIFFERENCES relative to the elements, and a difference can obviously be negative. This is why we write $\Delta S^{\circ}_{r}$ from tabulated $S^{\circ}$ and $\Delta H^{\circ}_{r}$ from tabulated $\Delta H^{\circ}_{f}$: the two tables do not contain the same type of quantity.

Because spontaneity is not governed by $\Delta H$ alone, but by $\Delta G=\Delta H-T\Delta S$. Here $\Delta H>0$, which works against dissolution, but going from an ordered crystalline solid to dispersed, hydrated ions in the liquid STRONGLY increases the disorder: $\Delta S>0$, and at room temperature the term $-T\Delta S$ outweighs $\Delta H$, making $\Delta G$ negative. This is exactly the principle of the instant cold packs used for sports injuries: they get cold because the dissolution is endothermic, and they work anyway because it is favoured by entropy.

Case 1, $\Delta H<0$ and $\Delta S>0$: both terms favour the reaction, $\Delta G<0$ at EVERY temperature, the reaction is always spontaneous. Case 2, $\Delta H>0$ and $\Delta S<0$: both oppose it, $\Delta G>0$ always, NEVER spontaneous. Case 3, $\Delta H<0$ and $\Delta S<0$: enthalpy favours, entropy opposes; since the entropy term is multiplied by $T$, it wins as $T$ grows: spontaneous at LOW temperature. Case 4, $\Delta H>0$ and $\Delta S>0$: the opposite situation, spontaneous at HIGH temperature.

First convert: $\Delta S^{\circ}_{r}=160.6$ J/(mol·K) $=0.1606$ kJ/(mol·K). Then $\Delta G^{\circ}=179.2-298\times 0.1606=179.2-47.9=+131.3$ kJ/mol. The value is POSITIVE: at 25 °C, the decomposition of limestone is NOT spontaneous. Fortunately, otherwise limestone cliffs and seashells would decompose spontaneously in the open air.

$\Delta H>0$ and $\Delta S>0$: this is case 4, spontaneous at HIGH temperature only. We therefore expect a temperature above which limestone decomposes, which is the subject of the next exercise, and which corresponds to a very real industrial process, the making of lime.

The mistake is forgetting the UNIT CONVERSION: the student subtracted joules from kilojoules. By treating 160.6 J/(mol·K) as if it were in kJ, the entropy term is inflated by a factor of 1000. The order of magnitude should have raised the alarm immediately: $-47{,}679$ kJ/mol, nearly 48 megajoules per mole, exceeds by several orders of magnitude any conceivable chemical energy, a covalent bond being typically worth a few hundred kJ/mol. REFLEX: in $\Delta H-T\Delta S$, write the units next to each term BEFORE typing the calculation.

$\Delta G^{\circ}(T)=\Delta H^{\circ}-T\Delta S^{\circ}=179.2-0.1606\,T$, with $T$ in kelvins and $\Delta G^{\circ}$ in kJ/mol. The intercept is $\Delta H^{\circ}=179.2$ kJ/mol, reached for $T=0$, and the SLOPE is $-\Delta S^{\circ}=-0.1606$ kJ/(mol·K). The line therefore goes down, which is the behaviour of any reaction with $\Delta S>0$: the hotter it is, the more disorder is rewarded.

The crossover temperature makes $\Delta G^{\circ}$ vanish: $T=\dfrac{\Delta H^{\circ}}{\Delta S^{\circ}}=\dfrac{179.2}{0.1606}=1116$ K. In degrees Celsius, $1116-273=843$ °C. This is the point marked on the figure, at the intersection of the line and the horizontal axis.

$\Delta G^{\circ}(1200)=179.2-0.1606\times 1200=179.2-192.7=-13.5$ kJ/mol. The value is negative: at 1200 K, that is 927 °C, the decomposition is SPONTANEOUS. The graph confirms it, the line passing below the axis beyond 1116 K.

Yes, perfectly consistent. The kilns operate between 900 and 1000 °C, that is between 1173 and 1273 K, so clearly ABOVE the crossover temperature of 843 °C: the reaction is spontaneous there and limestone turns into quicklime. At 800 °C, that is 1073 K, we would have $\Delta G^{\circ}=179.2-0.1606\times 1073=+6.9$ kJ/mol, a value that is still positive: the decomposition would not be spontaneous and the kiln would be ineffective. The industrial margin above the theoretical 843 °C is also explained by kinetics and by the need to remove the $CO_{2}$ produced.

When $\Delta H$ and $\Delta S$ have OPPOSITE SIGNS, that is in cases 1 and 2 of the table. If $\Delta H<0$ and $\Delta S>0$, the line starts from a negative intercept and keeps going down: it stays below the axis for every $T>0$. If $\Delta H>0$ and $\Delta S<0$, it starts from a positive intercept and goes UP: it stays above. In both cases, the formula $T=\frac{\Delta H}{\Delta S}$ does give a number, but it is NEGATIVE, hence meaningless physically. This is the check to make before announcing a crossover temperature: if it comes out negative, the reaction has none.

Because we cannot stop the combustion of carbon at the monoxide stage: as soon as there is enough oxygen, part of the $CO$ keeps burning to $CO_{2}$, and if too little oxygen is supplied some carbon is left unburnt. The product obtained is always a mixture of unknown composition, so the heat measured corresponds to no precise equation. This is exactly the situation Hess's law is made to resolve.

We keep reaction (1) as it is and REVERSE reaction (2), which changes the sign of its enthalpy: $CO_{2}(g)\rightarrow CO(g)+\frac{1}{2}O_{2}(g)$, $\Delta H^{\circ}=+283.0$ kJ/mol. When we add them, the $CO_{2}$ cancels on both sides, and $\frac{1}{2}O_{2}$ is left on the left: $C(s)+\frac{1}{2}O_{2}(g)\rightarrow CO(g)$.

$\Delta H^{\circ}_{f}(CO)=(-393.5)+(+283.0)=-110.5$ kJ/mol.

On the fact that enthalpy is a STATE FUNCTION: its change depends only on the initial and final states, never on the path followed. Going from carbon to monoxide directly, or via the dioxide and then back, costs exactly the same. If enthalpy were not a state function, we could choose a cheaper outbound path than the return path and loop indefinitely, producing energy from nothing: a perpetual motion machine. Hess's law is therefore not a calculation trick, it is the first law expressing itself.

Burning one mole of carbon to $CO_{2}$ releases 393.5 kJ; stopping at $CO$ releases only 110.5. The loss is $393.5-110.5=283.0$ kJ per mole, that is $\frac{283.0}{393.5}=71.9$% of the available energy. A poorly adjusted appliance therefore wastes nearly three quarters of its fuel, but the real danger is not there: this unreleased energy stays stored in an ODOURLESS, COLOURLESS gas that binds to hemoglobin about 200 times better than oxygen does. Carbon monoxide kills precisely because it is a fuel that has not finished burning, and nothing in the air signals its presence.

$\Delta G^{\circ}(298)=+131.3$ kJ/mol, calculated in exercise 3. From $\Delta G^{\circ}=-RT\ln K$ we get $\ln K=-\dfrac{\Delta G^{\circ}}{RT}=-\dfrac{131{,}300}{8.314\times 298}=-53.0$, so $\log K=-23.0$ and $K=9.5\times 10^{-24}$. A constant this tiny means the reaction lies entirely on the limestone side: at room temperature, it does not decompose in any measurable way, which agrees with the qualitative conclusion of exercise 3. Rule of thumb: at 298 K, each slice of about 6 kJ/mol of $\Delta G^{\circ}$ multiplies or divides $K$ by ten, since $RT\ln 10=5.7$ kJ/mol.

$\Delta G^{\circ}(1200)=-13.5$ kJ/mol. Then $\ln K=-\dfrac{-13{,}520}{8.314\times 1200}=+1.36$, so $K=3.9$. We went from $10^{-23}$ to about 4, more than twenty-three orders of magnitude, for a temperature change of a factor of only 4. It is the most striking illustration of the exponential sensitivity of $K$ to temperature, through the factor $e^{-\Delta G^{\circ}/RT}$.

At the crossover temperature, $\Delta G^{\circ}=0$ by definition. So $-RT\ln K=0$, and since $R$ and $T$ are nonzero, $\ln K=0$, that is $K=1$. The crossover temperature is therefore exactly the one at which the equilibrium constant equals 1: reactants and products are, in terms of the constant, on an equal footing.

The solids $CaCO_{3}$ and $CaO$ have an activity equal to 1 and do not appear. What remains is $K=P_{CO_{2}}$, the partial pressure of carbon dioxide, expressed relative to the standard pressure. The equilibrium constant of this reaction can therefore be read DIRECTLY on a pressure gauge: it is the pressure of $CO_{2}$ above limestone heated in a closed vessel, called the decomposition pressure. At 1200 K it is about 3.9 times the standard pressure, nearly 4 atm.

$\Delta G^{\circ}$ is a CONSTANT for a given reaction and temperature: it is the change in free energy when the reaction goes to completion entirely under standard conditions. $\Delta G$, on the other hand, depends on the instantaneous composition of the mixture, according to $\Delta G=\Delta G^{\circ}+RT\ln Q$. At equilibrium, it is $\Delta G$ that vanishes, never $\Delta G^{\circ}$: setting $\Delta G=0$ gives $\Delta G^{\circ}=-RT\ln Q_{\text{eq}}=-RT\ln K$, which proves the starting relation again. Confusing the two is the most frequent conceptual mistake of the chapter: a reaction whose $\Delta G^{\circ}$ is positive can perfectly well proceed a little, until $Q$ reaches $K$.

$\Delta H^{\circ}_{r}=2\Delta H^{\circ}_{f}\left(NH_{3}\right)-\left[0+3\times 0\right]=2(-45.9)=-91.8$ kJ/mol: the synthesis is EXOTHERMIC. $\Delta S^{\circ}_{r}=2(192.8)-\left[191.6+3(130.7)\right]=385.6-583.7=-198.1$ J/(mol·K): negative, as expected since we go from 4 moles of gas to 2.

$\Delta G^{\circ}(298)=-91.8-298\times(-0.1981)=-91.8+59.0=-32.8$ kJ/mol. The value is negative: at room temperature, the synthesis IS thermodynamically spontaneous. This is case 3 of the table, $\Delta H<0$ and $\Delta S<0$, so spontaneous at low temperature.

$T=\dfrac{\Delta H^{\circ}}{\Delta S^{\circ}}=\dfrac{-91.8}{-0.1981}=463$ K, that is 190 °C. BEYOND this temperature, $\Delta G^{\circ}$ becomes positive and the synthesis stops being spontaneous: ammonia tends on the contrary to decompose.

At 450 °C, that is 723 K: $\Delta G^{\circ}=-91.8+723\times 0.1981=-91.8+143.2=+51.4$ kJ/mol, clearly positive. The constant is $\ln K=-\dfrac{51{,}400}{8.314\times 723}=-8.56$, so $K=1.9\times 10^{-4}$: the yield at equilibrium is very low. The choice is explained by KINETICS, which thermodynamics ignores completely. At 25 °C, the $N\equiv N$ triple bond is so strong that the activation energy is enormous and the rate effectively zero: the reaction is spontaneous but infinitely slow, hence industrially nonexistent. Heating to 450 °C sacrifices yield to gain speed. This is the central compromise of chemical engineering: a reaction that is thermodynamically favourable but kinetically blocked is worth nothing.

They adjust the PRESSURE, raised to 150 to 300 bar. The reaction converts 4 moles of gas into 2: according to Le Chatelier's principle, an increase in pressure shifts the equilibrium toward the side that occupies the least volume, that is toward ammonia. It is precisely because $\Delta n_{\text{gas}}=-2$ is nonzero that the pressure lever works; on a reaction where the number of moles of gas is conserved, it would have no effect. An iron catalyst is added as a complement: it increases the rate without shifting the equilibrium, which makes it possible not to raise the temperature further. Yield, rate and cost are thus balanced.

Freezing is the reverse of melting: $\Delta H_{sys}=-6.01$ kJ/mol and $\Delta S_{sys}=-22.0$ J/(mol·K). The system loses entropy, which seems forbidden by the second law if we forget that it does NOT speak of the system alone. A system can perfectly well become more ordered, provided it disorders its surroundings even more.

The heat given up by the system is received by the surroundings, at temperature $T$: $\Delta S_{surr}=\frac{-\Delta H_{sys}}{T}$. At $-10$ °C, that is 263 K: $\Delta S_{surr}=\frac{+6010}{263}=+22.85$ J/(mol·K). So $\Delta S_{univ}=-22.0+22.85=+0.85$ J/(mol·K) $>0$: freezing is spontaneous, in agreement with experiment.

At $+10$ °C (283 K): $\Delta S_{surr}=\frac{6010}{283}=+21.24$, so $\Delta S_{univ}=-22.0+21.24=-0.76$ J/(mol·K) $<0$: freezing is impossible, it is melting that is spontaneous. At 0 °C (273 K): $\Delta S_{surr}=\frac{6010}{273}=+22.0$, so $\Delta S_{univ}=0$: neither one nor the other, the two phases coexist at EQUILIBRIUM. We recover exactly the melting point, not as a tabulated value but as the temperature at which the two terms cancel.

$\Delta G=\Delta H-T\Delta S=-6010-(263)(-22.0)=-6010+5786=-224$ J/mol, negative, so spontaneous. And $-T\Delta S_{univ}=-(263)(0.85)=-224$ J/mol: both calculations give the same number, as the identity $\Delta G=-T\Delta S_{univ}$ requires, which is obtained by multiplying $\Delta S_{univ}=\Delta S_{sys}-\frac{\Delta H_{sys}}{T}$ by $-T$.

Because $\Delta G$ is nothing other than $-T\Delta S_{univ}$ in disguise: saying $\Delta G<0$ is EXACTLY saying $\Delta S_{univ}>0$, since $T$ is always positive and so does not change the direction of the inequality. Free energy adds no physics to the second law, it rewrites it. Its advantage is purely practical and considerable: $\Delta G$ involves only quantities of the SYSTEM, measurable and tabulated, whereas $\Delta S_{univ}$ would force us to describe the surroundings, that is the rest of the world. We traded a universal bookkeeping for a local one, losing nothing.

$\Delta G^{\circ}=179{,}200-(1000)(160.6)=179{,}200-160{,}600=+18{,}600$ J/mol, positive: under standard conditions, that is under 1 atm of $CO_{2}$, the decomposition is NOT spontaneous at 1000 K. The constant is $K=e^{-\Delta G^{\circ}/RT}=e^{-18{,}600/((8.314)(1000))}=e^{-2.237}=0.107$ atm: at equilibrium, limestone imposes only a tenth of an atmosphere of $CO_{2}$.

$\Delta G=\Delta G^{\circ}+RT\ln Q$. The $\Delta G^{\circ}$ describes the reaction carried out under 1 atm of each gas; the term $RT\ln Q$ corrects for the actual conditions. Here the two solids, $CaCO_{3}$ and $CaO$, do not appear and $Q=P_{CO_{2}}$: the quotient reduces to the pressure of carbon dioxide actually present above the solid.

$\Delta G=18{,}600+(8.314)(1000)\ln(0.010)=18{,}600+(8314)(-4.605)=18{,}600-38{,}290=-19{,}690$ J/mol, clearly negative: the decomposition is spontaneous, and it remains so as long as the sweeping keeps the pressure of $CO_{2}$ below the equilibrium value of 0.107 atm.

$T_{cross}=\frac{\Delta H^{\circ}}{\Delta S^{\circ}}=\frac{179{,}200}{160.6}=1116$ K, that is 843 °C. The paradox is only apparent: the crossover temperature is the one where $\Delta G^{\circ}$ changes sign, so it concerns ONLY standard conditions, under 1 atm of $CO_{2}$. It answers the question “from what temperature does limestone decompose under an atmosphere of its own gas?”. That is not the question an open kiln poses, where the $CO_{2}$ is removed continuously: there, 1000 K is amply sufficient, and even less.

Rule: any gaseous product removed from the medium makes $Q$ drop, so $RT\ln Q$ becomes very negative and pulls $\Delta G$ down. A reaction that is thermodynamically unfavourable under standard conditions can thus become spontaneous, without heating further, simply because it is prevented from reaching its equilibrium. This is Le Chatelier's principle read in energy language. Industrially, it is the principle of the air-swept lime kiln and, more generally, of any reactor with continuous removal of the product: reactive distillation, evaporation of a condensate, sweeping with an inert gas. We do not buy spontaneity with heat, we buy it by emptying the reactor.

$+742.2$ kJ/mol is an enormous value: hematite is extraordinarily stable, which explains why it is an ore and not a native metal. Heating would help, since the decomposition produces gas and so has a positive $\Delta S^{\circ}$, but the order of magnitude makes the operation illusory: we would need to exceed several thousand kelvins, far beyond what a furnace can withstand and beyond the temperature at which iron itself boils. The direct route is therefore doomed, not in principle but by the sheer size of the temperature required.

$\Delta G^{\circ}=492{,}700-(298)(543.2)=492{,}700-161{,}874=+330{,}800$ J/mol, that is $+331$ kJ/mol. At room temperature, carbon therefore does not reduce hematite: this is fortunate, otherwise the ore would not coexist with coal in the ground. But the value is already less than half that of the direct route, and above all $\Delta S^{\circ}$ is very strongly positive.

$T_{cross}=\frac{\Delta H^{\circ}}{\Delta S^{\circ}}=\frac{492{,}700}{543.2}=907$ K, that is 634 °C: a temperature perfectly accessible to a coal-fired furnace, and known empirically long before it was explained.