Chemical Thermodynamics

Internal energy (\(U\) or \(E\)) is the sum of all forms of energy of a system: the kinetic energy (translational, rotational and vibrational motion) and the potential energy (attractions between the particles) of all its particles. It is a state function and it depends on the amount of substance (an extensive property).

Unit: joule (J) or kilojoule (kJ); for one mole, kJ mol^-1.

Enthalpy (\(H\)) is the total heat content of a system, defined as the internal energy plus the product of pressure and volume:

The change in enthalpy is equal to the heat gained or lost at constant pressure, \(\Delta H=q_p\). Unit: J or kJ (kJ mol^-1 for molar enthalpy change).

Entropy (\(S\)) is the measure of the disorder or randomness of a system. For a reversible isothermal process \(\Delta S=\dfrac{q_{rev}}{T}\). A highly ordered system has low entropy and a disordered system has high entropy.

Unit: J K^-1 (J K^-1 mol^-1 for one mole), or cal K^-1.

Gibbs free energy (\(G\)) is the part of the energy of a system that is available to do useful work. It combines enthalpy and entropy:

The decrease in free energy in a spontaneous process at constant temperature and pressure is equal to the maximum useful work, \(\Delta G=-W_{max}\). Unit: J or kJ (kJ mol^-1).

Statement: Energy can neither be created nor destroyed, but it can be changed from one form to another (the energy of the universe is constant). When heat is added to a system, it appears as some other form of energy in the same amount.

where \(\Delta U\) is the change in internal energy, \(q\) is the heat absorbed by the system and \(w\) is the work done on the system. Sign convention: \(q\) is positive when heat is absorbed and negative when heat is released; \(w\) is positive when work is done on the system and negative when the system does work. (If the work done by the system is written as \(W\), the law reads \(\Delta E=Q-W\), which is the same statement.) For a gas at constant pressure \(w=-P\Delta V\).

Statement: The total entropy of the universe always increases in a spontaneous (natural) process.

For a reversible process \(\Delta S_{univ}=0\) (equilibrium), and for a non-spontaneous process \(\Delta S_{univ}<0\). It explains why heat flows from a hot body to a cold body, why no engine is 100% efficient, and why a gas spreads into an empty flask but never collects back by itself.

Statement: The heat required to decompose a compound into its elements is equal in magnitude but opposite in sign to the heat evolved when the compound is formed from its elements.

Example: \(H_{2(g)}+\frac12O_{2(g)}\rightarrow H_2O_{(l)}\), \(\Delta H=-285.8\) kJ/mol, and \(H_2O_{(l)}\rightarrow H_{2(g)}+\frac12O_{2(g)}\), \(\Delta H=+285.8\) kJ/mol. This is the first law of thermochemistry.

Statement: The total enthalpy change of a chemical reaction is the same whether the reaction takes place in one step or in several steps, provided the initial and the final conditions are the same.

Example: \(C_{(s)}+\frac12O_2\rightarrow CO\), \(\Delta H_1=-110.5\) kJ and \(CO+\frac12O_2\rightarrow CO_2\), \(\Delta H_2=-283.0\) kJ. Then for \(C_{(s)}+O_2\rightarrow CO_2\), \(\Delta H=-110.5-283.0=-393.5\) kJ, the same as the direct burning of carbon. This is the second law of thermochemistry and it follows from the fact that enthalpy is a state function.

A state function is a property of a system whose value depends only on the present state of the system (given by \(P, V, T\) and the amount), and not on the path by which that state was reached. Its change is \(\Delta X=X_{final}-X_{initial}\).

Enthalpy is a state function because \(H=E+PV\), and \(E\), \(P\) and \(V\) are all state functions, so \(H\) is fixed by the state of the system. The change \(\Delta H=H_{products}-H_{reactants}\) is the same by whatever route the reaction takes. Hess’s law is the direct proof: the enthalpy change of \(C_{(s)}\rightarrow CO_2\) is \(-393.5\) kJ whether it takes place in one step or through \(CO\).

Entropy is a state function because it measures the disorder of the system, which depends on its present condition (temperature, pressure, volume and amount of substance). The change \(\Delta S=S_{final}-S_{initial}\) depends only on the two end states, for example \(\Delta S_{fusion}\) of ice at 273 K is the same, \(S_{liquid}-S_{solid}\), however the melting is carried out. Heat and work, in contrast, are path functions.

Internal energy is the sum of all the energies of the particles of a system (kinetic energy of translation, rotation and vibration, and potential energy due to intermolecular forces).

Relationship: \(H=E+PV\), so for a change at constant pressure

For gases, \(P\Delta V=\Delta n_gRT\) (ideal gas, \(PV=nRT\)), so

where \(\Delta n_g\) is the moles of gaseous products minus the moles of gaseous reactants.

Formation of HCl:

\(\Delta n_g=2-(1+1)=0\). Therefore \(\Delta n_gRT=0\) and

So the change in enthalpy equals the change in internal energy (both are \(-184.6\) kJ for 2 mol of HCl, as \(\Delta H_f\) of \(HCl\) is \(-92.3\) kJ/mol). The same is true for every reaction in which the number of moles of gas does not change.

The enthalpy of reaction is the heat absorbed or released when the number of moles of reactants shown in the balanced equation react completely to form products at constant pressure. \(\Delta H_{rxn}=H_{products}-H_{reactants}\), and under standard conditions (298 K and 1 atm) it is the standard enthalpy of reaction \(\Delta H^\circ_{rxn}=\sum\Delta H_f^\circ(\text{products})-\sum\Delta H_f^\circ(\text{reactants})\).

The sign is positive, so the reaction is endothermic: 180 kJ of heat is absorbed from the surroundings when 1 mole of \(N_2\) reacts with 1 mole of \(O_2\) to give 2 moles of \(NO\). The enthalpy of the product is 180 kJ higher than that of the reactants.

This is the reverse reaction. By the Laplace and Lavoisier law the sign changes:

The decomposition of 2 moles of \(NO\) is exothermic and releases 180 kJ.

The enthalpy of formation refers to 1 mole of the compound. The equation gives 2 moles of \(NO\) for 180 kJ:

(The tabulated value is \(+90.3\) kJ/mol.) \(NO\) is formed with absorption of energy, so it is an endothermic and relatively unstable compound.

Laplace and Lavoisier law (first law of thermochemistry, 1780): The heat required to decompose a compound into its elements is equal in magnitude but opposite in sign to the heat evolved when the compound is formed from its elements. In general, \(\Delta H_{forward}=-\Delta H_{reverse}\).

Applications:

1. The enthalpy of decomposition of a compound is found from its enthalpy of formation: \(S_{(s)}+O_{2(g)}\rightarrow SO_{2(g)}\), \(\Delta H=-296.9\) kJ, so \(SO_{2(g)}\rightarrow S_{(s)}+O_{2(g)}\), \(\Delta H=+296.9\) kJ.

2. The enthalpy of a reverse reaction (which may be difficult to measure) is known when the forward value is known, e.g. \(H_2O_{(l)}\rightarrow H_2+\frac12O_2\), \(+285.8\) kJ/mol.

3. When reactions are added in Hess’s law calculations, an equation may be reversed and the sign of its \(\Delta H\) changed.

4. It follows from the law of conservation of energy: if the forward and reverse values were not equal and opposite, a cycle could create or destroy energy.

A process in which the particles move apart absorbs energy, and a process in which they come closer releases energy.

Melting, evaporation and sublimation proceed with an increase in entropy (\(\Delta S>0\)), since the particles become more free and more disordered. Sublimation has the largest increase because the solid changes directly to a gas. Freezing, condensation and deposition decrease the entropy.

In \(\Delta G=\Delta H-T\Delta S\), a process is spontaneous at low temperature when \(\Delta H\) is negative and \(\Delta S\) is also negative. At low temperature the term \(T\Delta S\) is small, so \(\Delta H\) controls the sign and \(\Delta G\) is negative. At high temperature the term \(-T\Delta S\) becomes large and positive, so \(\Delta G\) becomes positive.

The processes spontaneous at low temperature are therefore freezing, condensation and deposition (exothermic with decrease in entropy). The reverse processes, melting, evaporation and sublimation (\(\Delta H>0\), \(\Delta S>0\)), are spontaneous at high temperature.

Use Hess’s law. Reverse the first equation (sign changes), and keep the other two:

On adding, \(C\), \(H_2\) and \(O_2\) cancel and the result is \(C_4H_{10}O_{(l)}\rightarrow C_4H_{8(g)}+H_2O_{(g)}\):

The reaction is endothermic (absorbs 76 kJ per mole). The same answer comes from \(\Delta H=\sum\Delta H_f(\text{products})-\sum\Delta H_f(\text{reactants})=[-17+(-242)]-(-335)=+76\) kJ.

Note: the formation of \(C_4H_{10}O_{(l)}\) from its elements, \(4C+5H_2+\frac12O_2\rightarrow C_4H_{10}O_{(l)}\), is the first given equation itself, \(\Delta H=-335\) kJ mol^-1.

Arrange the equations so that they add up to the target equation:

Adding: \(2H_2\) and \(O_2\) on the right cancel the same amounts on the left, and \(2H_2O_{(g)}\) cancels, leaving the required equation.

Adding, \(KCl\), \(3MgO\) and \(3Mg\) cancel and the required equation is obtained:

This is the enthalpy of formation of \(KClO_3\) (exothermic).

\(BaCl_{2(aq)}\) can be obtained from \(BaCl_{2(s)}\) by two routes: directly (\(\Delta H=-20.6\) kJ), or in two steps, first forming the dihydrate (\(\Delta H_1\)) and then dissolving it (\(\Delta H_2=+8.8\) kJ). Hess’s law says that the enthalpy change is the same by both routes because it depends only on the initial and final states:

The enthalpy of solution is the enthalpy change when 1 mole of the solute dissolves in water, so it is the direct step: \(\Delta H_{sol}=-20.6\) kJ mol^-1 (exothermic).

The enthalpy of hydration of \(BaCl_{2(s)}\) to \(BaCl_2\cdot2H_2O_{(s)}\) is \(-29.4\) kJ mol^-1. The hydration is exothermic, and the dissolution of the hydrate is endothermic.

The reaction is endothermic. In the graph the products lie at a higher enthalpy level than the reactants, so \(\Delta H=H_{products}-H_{reactants}\) is positive. The system absorbs heat from the surroundings, and the surroundings become cooler.

1. Thermal decomposition of calcium carbonate: \(CaCO_3\rightarrow CaO+CO_2\), \(\Delta H=+178\) kJ/mol.

2. Photosynthesis: \(6CO_2+6H_2O\rightarrow C_6H_{12}O_6+6O_2\) (energy of sunlight absorbed). Other examples are the formation of \(NO\) from \(N_2\) and \(O_2\) (\(+180\) kJ), the dissolution of \(NH_4NO_3\) in water (\(+25.7\) kJ/mol) and the melting of ice.

The change in the internal energy of a system is equal to the heat absorbed by the system plus the work done on the system. (Written with work done by the system, \(\Delta E=Q-W\).) For example, a gas absorbs 600 J of heat and does 200 J of work in expanding: \(\Delta U=600-200=+400\) J.

Isothermal process (\(\Delta T=0\)): the internal energy of an ideal gas depends on temperature only, so \(\Delta U=0\) and \(q=-w\). All the heat absorbed is changed into work.

Adiabatic process (\(q=0\)): there is no heat exchange, so \(\Delta U=w\). Work done on the system raises its internal energy and the temperature, and work done by the system lowers it.

The internal energy of a closed system (no exchange of matter) increases (i) when heat is absorbed from the surroundings (\(q>0\)), and (ii) when work is done on the system by the surroundings, as in compression (\(w>0\)).

\(q\) is negative when the system releases heat to the surroundings (exothermic process). \(w\) is negative when the system does work on the surroundings, for example when a gas expands against external pressure.

Allotropes are the different forms of the same element in the same physical state, which have different arrangements of atoms and therefore different physical properties. Graphite, diamond and fullerene are allotropes of carbon.

Graphite is the more stable allotrope. Diamond releases more heat on burning (\(-395.4\) kJ against \(-393.5\) kJ), so diamond starts at a higher enthalpy than graphite, and the one with the lower enthalpy is more stable. This is why graphite is the standard state of carbon (\(\Delta H_f^\circ=0\)).

Keep the first equation and reverse the second:

By the Laplace and Lavoisier law, \(C_{(diamond)}\rightarrow C_{(graphite)}\) has \(\Delta H=-1.9\) kJ. Although this conversion is spontaneous, diamond does not change to graphite in practice since the activation energy is very high.

Reaction (2) is easier. It is exothermic (\(\Delta H=-142.3\) kJ), so it releases energy and goes to the more stable \(O_2\), whereas reaction (3) is strongly endothermic (\(\Delta H=+495.0\) kJ) because the strong \(O=O\) double bond has to be broken, which needs a large amount of energy.

Adding the three equations: \(O_3\) appears on both sides and cancels. The left side has \(\frac32O_2\) and the right side has \(1+\frac12=\frac32O_2\), so \(O_2\) cancels as well. What remains is \(NO+O\rightarrow NO_2\).

\(\Delta S\) is positive. The ordered crystal lattice breaks up into free, mobile ions in solution, so the disorder increases.

\(\Delta S\) is negative. The molecules of liquid water, which move randomly, become fixed in the regular hydrogen-bonded lattice of ice, so the disorder decreases.

\(\Delta S\) is positive. Dry ice sublimes to a gas whose molecules move freely and occupy a large volume, which is a large increase in randomness.

\(\Delta S\) is positive. One solid forms a solid and a gas, and the gaseous product has far greater entropy than any solid (the number of moles of gas increases from 0 to 1).

Step 1: enthalpy change

Step 2: entropy change

Step 3: at equilibrium \(\Delta G=0\), so \(\Delta H=T\Delta S\):

Above 370.5 K, \(\Delta G<0\) and the evaporation is spontaneous; below it the process is non-spontaneous. The value is close to the boiling point of water (373 K), the small difference being due to the use of \(\Delta H\) and \(S^\circ\) at 298 K.

The \(C=O\) bond in \(CO_2\) (+805 kJ mol^-1) is the most difficult to break because it has the highest bond enthalpy. It is a double bond between the small, strongly attracting \(C\) and \(O\) atoms.

It means that 413 kJ of energy is required to break 1 mole of \(C-H\) bonds in gaseous molecules to give gaseous atoms (the average value of \(C-H\) bonds in different compounds). The same amount of energy is released when 1 mole of \(C-H\) bonds is formed.

\(CH_{4(g)}+2O_{2(g)}\rightarrow CO_{2(g)}+2H_2O_{(g)}\). Bonds broken: 4 \(C-H\) and 2 \(O=O\). Bonds formed: 2 \(C=O\) and 4 \(O-H\).

The negative sign shows that the combustion is exothermic. The value is for water as a gas; it is close to the value \(-802\) kJ obtained from enthalpies of formation, because bond enthalpies are average values.

Enthalpy (\(H\)) is the total heat content of a system, equal to its internal energy plus the pressure-volume term, \(H=E+PV\). The change in enthalpy \(\Delta H=H_{products}-H_{reactants}\) is equal to the heat absorbed or released at constant pressure.

This is the complete combustion of methane (the products in the figure are \(CO_2\) and liquid water; the carbon atom of methane must appear as \(CO_2\) to balance the equation).

The reaction is exothermic. When 1 mole of methane (16 g) burns completely in oxygen to form carbon dioxide and liquid water, 890 kJ of heat is given out to the surroundings. The enthalpy of the products is 890 kJ per mole lower than that of the reactants, and this value is the enthalpy of combustion of methane.

The calorific value is the heat released when 1 g of the fuel burns completely. Molar mass of \(CH_4=12+4=16\) g/mol.

That is about 55,600 kJ/kg, in agreement with the table of fuels in the notes (methane, 55,500 kJ/kg).

In the reaction 3 moles of gas become 1 mole of gas and 2 moles of liquid, so the entropy of the system decreases (about \(-243\) J K^-1 per mole of methane). But the 890 kJ of heat released goes into the surroundings and increases the random motion of their particles:

The gain in entropy of the surroundings is much larger than the loss in the system, so \(\Delta S_{univ}=\Delta S_{sys}+\Delta S_{surr}\approx-243+2987>0\). By the second law the reaction is spontaneous (once it is started).

Entropy (\(S\)) is the thermodynamic quantity that measures the disorder or randomness of a system. A more ordered system has lower entropy. For a reversible process \(\Delta S=q_{rev}/T\). Unit: J K^-1 (J K^-1 mol^-1 per mole). Entropy increases in the order solid < liquid < gas, and when the number of moles of gas increases.

\(\Delta S\) is negative: the liquid (random) changes to an ordered crystalline solid.

\(Cl_{2(g)}\rightarrow2Cl_{(g)}\): \(\Delta S\) is positive, as one mole of gas gives two moles of gas (more particles and more ways of arrangement).

\(CO_{(g)}+2H_{2(g)}\rightarrow CH_3OH_{(l)}\): \(\Delta S\) is negative, since 3 moles of gas change to 1 mole of liquid, so the disorder decreases greatly.

\(3Ca(NO_3)_{2(aq)}+2(NH_4)_3PO_{4(aq)}\rightarrow Ca_3(PO_4)_{2(s)}+6NH_4NO_{3(aq)}\): \(\Delta S\) is negative. Free ions \(Ca^{2+}\) and \(PO_4^{3-}\) leave the solution and settle as an ordered solid, so the disorder of the system decreases.

The enthalpy of combustion (\(\Delta H_c\)) is the enthalpy change when 1 mole of a substance burns completely in oxygen under standard conditions. It is always negative because combustion is exothermic.

Molar mass of \(C_2H_6=2(12)+6(1)=30\) g/mol.

Required: \(2C_{(graphite)}+3H_{2(g)}\rightarrow C_2H_{6(g)}\), \(\Delta H_f=?\)

Adding, \(2CO_2\), \(3H_2O\) and the \(O_2\) (\(2+\frac32=\frac72\)) cancel:

The standard enthalpy change (\(\Delta H^\circ\)) and standard entropy change (\(\Delta S^\circ\)) are the enthalpy and entropy changes of a reaction when all the reactants and products are in their standard states (pressure 1 atm, temperature 298 K, and 1 mol/L for solutions), the amounts being those of the balanced equation.

\(\Delta H_f^\circ(CO_2)=-393\) kJ mol^-1. The reaction forms 1 mole of \(CO_2\) from carbon (graphite, the standard state) and oxygen gas, both elements in their standard states, so the reaction is the formation of \(CO_2\) itself. Hence the enthalpy of combustion of graphite and the enthalpy of formation of \(CO_2\) are the same (the tabulated value is \(-393.5\) kJ/mol).

\(\Delta G^\circ=\Delta H^\circ-T\Delta S^\circ\). Here \(\Delta H^\circ\) is negative and \(\Delta S^\circ\) is positive, so \(-T\Delta S^\circ\) is also negative and \(\Delta G^\circ\) is negative at every temperature. The process is spontaneous at all temperatures.

So \(\Delta G^\circ=-394\) kJ at about 345 K. (At 298 K, \(\Delta G^\circ=-393-298(0.0029)=-393.9\) kJ, which is about \(-394\) kJ as well.)

For \(H_2O_{(s)}\rightleftharpoons H_2O_{(l)}\) the melting has \(\Delta H>0\) (heat absorbed) and \(\Delta S>0\) (more disorder); freezing has \(\Delta H<0\) and \(\Delta S<0\). In \(\Delta G=\Delta H-T\Delta S\):

At the melting point \(T_m\) (273 K for ice), \(\Delta G=0\) and the two phases are in equilibrium: \(T_m=\Delta H_{fusion}/\Delta S_{fusion}=6010/22.0=273\) K. So ice melts spontaneously above 0 degrees Celsius and water freezes spontaneously below it.

The standard enthalpy of formation (\(\Delta H_f^\circ\)) is the enthalpy change when 1 mole of a compound is formed from its elements in their standard states under standard conditions (1 atm and 298 K). For example \(H_2+\frac12O_2\rightarrow H_2O_{(l)}\), \(\Delta H_f^\circ=-285.8\) kJ/mol. The more negative the value, the more stable the compound.

Oxygen gas, \(O_2\), is the element in its most stable (standard) form. No change is needed to form it from itself, so no energy is absorbed or released, and by convention \(\Delta H_f^\circ\) of every element in its standard state (\(O_2, N_2, H_2\), graphite) is zero.

\(NO\) is the difficult one to form, since its \(\Delta H_f^\circ=+90.3\) kJ/mol is positive: energy has to be supplied to form it from \(N_2\) and \(O_2\), because the strong \(N\equiv N\) and \(O=O\) bonds must be broken (it forms only at very high temperature, such as in a lightning flash or an engine). The others release energy when they form.

The most stable is \(H_2O_{(l)}\) with the most negative value (\(-285.8\) kJ/mol), as it releases the most energy on formation. The unstable one is \(NO_{(g)}\) with a positive value (\(+90.3\) kJ/mol), its energy is higher than that of its elements.

Ammonia. Complete combustion: \(4NH_{3(g)}+3O_{2(g)}\rightarrow2N_{2(g)}+6H_2O_{(l)}\)

Nitric oxide. \(NO\) is already an oxide and no data for \(NO_2\) are given, so its combustion cannot be calculated from the table. The energy it can release is that of its decomposition into elements, \(2NO\rightarrow N_2+O_2\), \(\Delta H=-180.6\) kJ, that is \(90.3\) kJ per mole:

The value is very small compared with ammonia and with methane (55.6 kJ/g). When ammonia is oxidised to \(NO\) (Ostwald process, \(4NH_3+5O_2\rightarrow4NO+6H_2O_{(g)}\)), \(\Delta H=4(90.3)+6(-241.8)-4(-45.9)=-906\) kJ.

It is the enthalpy change when 1 mole of a compound is formed from its elements in their standard states under standard conditions (1 atm, 298 K), for example \(2C_{(s)}+3H_{2(g)}\rightarrow C_2H_{6(g)}\), \(\Delta H_f^\circ=-84.7\) kJ/mol.

A compound with a more negative \(\Delta H_f^\circ\) has released more energy when it formed, lies lower in energy, and is more stable and less reactive. Ethane (\(-84.7\) kJ/mol) is stable and unreactive. Acetylene (\(+226.7\) kJ/mol) is formed by absorption of energy, so it is at a high energy level, unstable and very reactive; it burns with a very hot flame and can decompose explosively into its elements.

The reaction is the hydrogenation of ethene to ethane. The standard enthalpy of formation of ethene, \(+52.5\) kJ/mol, is given in the table of the notes, and \(\Delta H_f^\circ(H_2)=0\):

By Hess’s law this is the sum of the decomposition of ethene into elements (\(-52.5\)), which is the Laplace and Lavoisier reverse of its formation (\(+52.5\)), and the formation of ethane (\(-84.7\)). The reaction is exothermic.

With the acetylene value given in the question, the hydrogenation of acetylene, \(C_2H_2+2H_2\rightarrow C_2H_6\), has \(\Delta H=-84.7-226.7=-311.4\) kJ/mol.

Bond enthalpy (bond energy) is the amount of energy required to break 1 mole of a particular bond in gaseous molecules to form gaseous atoms (the same energy is released when 1 mole of the bond is formed). Unit: kJ mol^-1.

Bond order: the higher the bond order, the stronger the bond (\(C-C\) 347 < \(C=C\) 612 < \(C\equiv C\) 837 kJ/mol).

Bond length: the shorter the bond, the stronger it is (\(F-F\) 158 kJ/mol is weak and long, \(N\equiv N\) 945 kJ/mol is short and strong).

Electronegativity (polarity): the higher the electronegativity of the bonded atom, the stronger the bond (\(C-F\) 485 kJ/mol is stronger than \(C-I\) 240 kJ/mol). The size of the bonded atoms also matters, since larger atoms form longer and weaker bonds.

The \(H-Cl\) bond (427 kJ/mol) is the most difficult to break because it has the highest bond enthalpy. It is a short, polar bond, as chlorine is much more electronegative than hydrogen. The \(Cl-Cl\) bond (239 kJ/mol) is the weakest, since it is long and non-polar.

One \(C-H\) bond and the \(Cl-Cl\) bond are broken, and one \(C-Cl\) bond and one \(H-Cl\) bond are formed (the other bonds are unchanged).

The chlorination of ethane is exothermic.

Spontaneity: a process is spontaneous when it takes place by itself, in the given conditions, without continuous help from outside (water flowing downhill, melting of ice at room temperature, rusting of iron). A non-spontaneous process occurs only if energy is supplied (electrolysis of water).

Gibbs free energy (\(G\)): the part of the total energy of the system that is available (free) to do useful work, \(G=H-TS\). The change in free energy equals the maximum useful work, \(\Delta G=-W_{max}\).

It follows from \(\Delta S_{univ}=\Delta S_{sys}-\dfrac{\Delta H_{sys}}{T}\), that is, \(-T\Delta S_{univ}=\Delta H-T\Delta S_{sys}=\Delta G\).

A process is spontaneous when \(\Delta G\) is negative. In any spontaneous process at constant temperature and pressure the free energy decreases.

Summary of the sign rules: \(\Delta H<0,\ \Delta S>0\): spontaneous at all temperatures; \(\Delta H>0,\ \Delta S<0\): non-spontaneous at all temperatures; \(\Delta H<0,\ \Delta S<0\): spontaneous at low temperature; \(\Delta H>0,\ \Delta S>0\): spontaneous at high temperature.