Thermodynamics

So hot right now

Chemistry can be divided (somewhat arbitrarily) into the study of structures, equilibria, and rates. Chemical structure is ultimately described by the methods of quantum mechanics; equilibrium phenomena are studied by statistical mechanics and thermodynamics; and the study of rates constitutes the subject of kinetics.

Kinetics can be subdivided into physical kinetics, dealing with physical phenomena such as diffusion and viscosity, and chemical kinetics, which deals with the rates of chemical reactions (including both covalent and noncovalent bond changes).

– Connors, K. (Connors 1990)

Key concepts:

• Thermo

Entropy

Adapted from Wright

The 1st Law of Thermodynamics states that energy is conserved. It is neither created or destroyed; it is only transferred. This transfer of energy can result in work. For example, a fire heats water to create steam. The steam turns a turbine, performing work, to generate power or move an object.

Now imagine two objects of different temperature.

• Object 1: hot - T_H
• Object 2: cold - T_C

Now consider the two objects in contact with each other.

Case 1: Heat (Q) spontaneously flows from the hot object to the cold object until their temperatures reach equilibrium. Energy is conserved.

Case 2: Heat spontaneously flows from the cold object to the hot object. Energy would also be conserved.

In both cases, the 1st law is not violated. However, the second case is never observed in nature.

The 2nd Law of Thermodynamics, which states that “the total entropy of a system either remains constant or increases in any spontaneous process,” addresses the impossibility of Case #2 by introducing a new property of a system, entropy, which has many definitions depending on the context.

Entropy is energy that cannot be used to do work. Entropy is energy that must be present for something to exist in its state.

Consider the following combustion reaction.

\[\mathrm{CH_4}(g) + 2\mathrm{O_2}(g) \longrightarrow \mathrm{CO_2}(g) + 2\mathrm{H_2O}(g)\]

This reaction is very exothermic where the standard heat of reaction is -727.95 kJ mol^–1. Therefore, one might expect to be able to extract all this heat energy to do some other work. But, this is not the case. The entropy of this reaction is –0.1916 kJ mol^–1 K^–1.

We use the Gibbs free energy equation to determine the amount of heat energy that can be extracted from this process. If this reaction were carried out under standard conditions such as 25 °C, the maximum amount of heat that can be extracted to do work is

\[\begin{align*} \Delta G^{\circ} &= \Delta H^{\circ} - T\Delta S^{\circ} \\[1.5ex] &= -727.95~\mathrm{kJ~mol^{-1}} - (298.15~\mathrm{K})(-0.1916~\mathrm{kJ~mol^{-1}~K^{-1}})\\[1.5ex] &= -670.82~\mathrm{kJ~mol^{-1}} \end{align*}\]

For the products to exist in their states (at 25 °C), some amount of energy released by the chemical reaction is used for that. As the temperature of the system increased, the less energy that could be extracted from the reaction.

Entropy exists on an absolute scale. All matter possesses entropy and at 0 Kelvin, a perfect crystal of a perfect substance, entropy is defined to be zero. As temperature increases, entropy increases.

Uniform Distribution of Energy and Matter

An increase in entropy can be seen with

1. The uniform distribution of energy (such as the heat flow shown above)
2. The uniform distribution of matter (think of two gases mixing… they will never de-mix)

In both cases, the process of distribution is irreversible, at least in a spontaneous way. Two gases will never un-mix spontaneously. Two objects will never spontaneously un-mix heat (they remain the same temperature).

We say, in both cases, that entropy has increased. The more spread out the energy or matter, the higher the entropy. That means, for the spontaneous, irreversible processes,

\[S_{\mathrm{final}} > S_{\mathrm{initial}}\] Entropy is an extensive property and defined as an amount of heat divided by the temperature (and given as J K^–1) for a given amount of substance.

\[\Delta S = \dfrac{Q}{T}\]

Consider the heat (i.e. flow of energy) between the two objects. The hot object with a temperature of T_H will experience an average temperature of

\[T_{\mathrm{H,avg}} = \dfrac{T_{\mathrm{H}} + T_{\mathrm{final}}}{2}\]

whereas the cold object with a temperature of T_C will experience an average temperature of

\[T_{\mathrm{C,avg}} = \dfrac{T_{\mathrm{C}} + T_{\mathrm{final}}}{2}\] The final temperature for both objects are equal at equilibrium. Therefore,

\[T_{\mathrm{H,avg}} > T_{\mathrm{C,avg}}\]

since the hot object has a higher temperature than the cold object such that

\[T_{\mathrm{H}} > T_{\mathrm{C}}\]

The entropy for the hot object would then be

\[\Delta S_{\mathrm{H}} = \dfrac{-Q}{T_{\mathrm{H}}}\]

and the entropy for the cold object would be

\[\Delta S_{\mathrm{C}} = \dfrac{Q}{T_{\mathrm{C}}}\]

where

\[ \vert \Delta S_{\mathrm{H}} \vert < \vert \Delta S_{\mathrm{C}} \vert \]

Since both objects are the system, the entropy for the system would be given as

\[\begin{align*} S_{\mathrm{final}} &= S_{\mathrm{initial}} + \dfrac{-Q}{T_{\mathrm{H}}} + \dfrac{Q}{T_{\mathrm{C}}} \\[2ex] &= S_{\mathrm{initial}} - \Delta S_{\mathrm{H}} + \Delta S_{\mathrm{C}} \end{align*}\]

Therefore,

\[S_{\mathrm{final}} > S_{\mathrm{initial}}\]

and the entropy has increased for this spontaneous process.

If heat flowed from cold to hot,

\[S_{\mathrm{final}} < S_{\mathrm{initial}}\] which never occurs spontaneously, thereby violating the 2nd law of thermodynamics.

Gibbs Energy

Gibbs free energy (or Gibbs energy; G) is a thermodynamic potential that is minimized when a chemical system reaches chemical equilibrium. The Gibbs free energy change for a reaction, ΔG, is used to define if a reaction or process is spontaneous or non-spontaneous.

Sign	Spontaneity	Favor
ΔG < 0	spontaneous	product favored
ΔG > 0	non-spontaneous	reactant favored
ΔG = 0	at equilibrium	none

The Gibbs free energy equation determines the Gibbs energy for a reaction by taking into account the enthalpy, ΔH, and entropy, ΔS, of reaction.

\[\Delta G^{\circ} = \Delta H^{\circ} - T\Delta S^{\circ}\]

Note: The “degree” symbols indicate that the thermodynamic values were obtained for substances in their standard state (T = 25 °C, P = 1 bar, c = 1 M). Appendix G in Chemistry 2e reports these values for many substances. Also note that T in the Gibbs free energy equation can be any temperature, not simply 298.15 K.

Let us look up the thermodynamic values for the following reaction.

\[\mathrm{2NO_2}(g) \rightleftharpoons 2\mathrm{NO}(g) + \mathrm{O_2}(g)\]

All values in the table below are given in kJ mol^–1.

Species	ΔHf°	S°
NO2(g)	33.2	0.2401
NO(g)	90.25	0.2108
O2	0	0.2052

We can determine the enthalpy of reaction by

\[\Delta H_{\mathrm{rxn}} = \Sigma \left (\Delta H_{\mathrm{products}} \right ) - \Sigma \left (\Delta H_{\mathrm{reactants}} \right )\] Therefore,

\[\begin{align*} \Delta H_{\mathrm{rxn}} &= \left [ (2 \times 90.25) + (0) \right ] - \left [ (2\times 33.2) \right ] \\[1.5ex] &= 114.1~\mathrm{kJ~mol^{-1}} \end{align*}\]

This reaction is endothermic! We repeat this process for the entropy of reaction.

\[\Delta S_{\mathrm{rxn}} = \Sigma \left (\Delta S_{\mathrm{products}} \right ) - \Sigma \left (\Delta S_{\mathrm{reactants}} \right )\] to give

\[\begin{align*} \Delta S_{\mathrm{rxn}} &= \left [ (2 \times 0.2108) + (0.2052) \right ] - \left [ (2\times 0.2401) \right ] \\[1.5ex] &= 0.1466~\mathrm{kJ~mol^{-1}} \end{align*}\]

The reaction experiences an increase in entropy. This is expected becuase we go from 2 moles of gas to 3 moles of gas!

Now we determine the Gibbs free energy for the reaction using the Gibbs free energy equation.

\[\begin{align*} \Delta G^{\circ} &= \Delta H^{\circ} - T\Delta S^{\circ} \\ &= 114.1~\mathrm{kJ~mol^{-1}} - T(0.1466~\mathrm{kJ~mol^{-1}}) \end{align*}\]

I tabulate the Gibbs free energy (in kJ mol^–1 for this reaction at three different temperatures, T, in the table below.

T (°C)	T (K)	ΔG°
-200	73.15	103.38
25	298.15	70.39
600	873.15	-13.9

We see that at low temperatures, the Gibbs free energy is negative. The reaction is non-spontaneous at low temperatures. As we increase the temperature, the reaction eventually becomes spontaneous (ΔG < 0).

According to Le Chatelier’s principle, increasing the heat for an endothermic reaction should drive the equilibrium to the right. We can determine the equilibrium constant for this reaction at the three temperatures of interest using

\[\Delta G^{\circ} = -RT\ln K\]

where R is the gas constant, T is the temperature, and K is the equilibrium constant. Rearranging for K gives

\[K = e^{\frac{-\Delta G^{\circ}}{RT}}\]

I’ve added the equilibrium constants in our table below.

T (°C)	T (K)	ΔG°	K
-200	73.15	103.38	1.51e-74
25	298.15	70.39	4.65e-13
600	873.15	-13.9	6.79

We see for this endothermic reaction that the equilibrium constant gets larger (favors product) as the temperature goes up! This fits with Le Chatelier’s principle! Also, note how ΔG is decreasing with increasing temperature for this reaction. At high temperatures, this reaction is spontaneous and products is favored. We note the correlation

\[\downarrow \Delta G \propto K \uparrow\]

How fast does the reaction go at these temperatures? According to Collision theory, as temperature increases, the rate should increase. Let us revisit the Arrhenius equation

\[k = Ae^{\frac{-E_{\mathrm{a}}}{RT}}\]

For this reaction, A = 2×10⁹ and E_a = 112.55 kJ mol^–1 (from (D. L. Baulch 1973)). I update our table to include the rate constant, k in M^–1 s^–1, at our three chosen temperatures.

T (°C)	T (K)	ΔG°	K	k
-200	73.15	103.38	1.51e-74	8.49e-72
25	298.15	70.39	4.65e-13	3.82e-11
600	873.15	-13.9	6.79	369.6

We see the rate constant is very small at low temperatures and therefore, the reaction is very slow. At elevated temperatures, the rate constant is much larger and the reaction is much faster!

Let us now consider the following reaction

\[\mathrm{N_2O}(g) + \mathrm{O}(g) \rightleftharpoons \mathrm{N_2}(g) + \mathrm{O_2}(g)\] This reaction is exothermic and increasing the temperature should shift the reaction to the left. The thermodynamic values are presented below.

Species	ΔHf°	S°
N2O(g)	81.6	0.22
O	249.17	0.1611
N2	0	0.1916
O2	0	0.2052

The enthalpy of reaction is

\[\Delta H_{\mathrm{rxn}} = -330.77~\mathrm{kJ~mol^{-1}}\]

and the entropy change is

\[\Delta S_{\mathrm{rxn}} = 0.3379~\mathrm{kJ~K^{-1}}\] We see that this reaction experiences an increase in entropy. I compute our numbers again (ΔG, K, and k) and present them below for our three chosen temperatures plus an additional temperature (1000 °C).

T (°C)	T (K)	ΔG°	K	k
-200	73.15	-355.49	7.15E+253	2.82E-73
25	298.15	-431.51	4.00E+75	3.17E-10
600	873.15	-625.81	2.75E+37	1.00E+04
1500	1773.15	-929.92	1.79E+25	3.60E+07

For this exothermic reaction, the rate increases with increasing temperature. The equilibrium is driven to the left towards reactants at higher temperature (K increases) and the reaction becomes more spontaneous (ΔG decreases).

Driving Force

What drives a reaction forward? What force opposes the driving force to result in equilibrium mixtures of reactants and products? This “force” is a result of chemically reactive species and exists today as a topic of debate and study. However, here we focus on the thermodynamic perspective of this question.

Keep in mind the following key ideas:

1. Spontaneous reactions or processes are those that occur without needing an outside source of energy.
2. Non-spontaneous reactions or processes do not occur unless an external source of energy is supplied.
3. Heat spontaneously flows from hot to cold.
4. Energy tends to uniformly disperse.
5. Matter tends to uniformly disperse.
6. The universe tends toward a state of maximum entropy.

Some Spontaneous Processes

1. Heat flow from hot to cold (thermal potential)
2. Mass free falls or free flows from high to low elevation (gravitational potential)
3. Space expands from high to low pressure
4. Electricity flows from higher to lower electrical potential (voltage)
5. Solutes disperse from high to low concentration (chemical potential)

These processes can be carried out in reverse by supplying an external force. Some examples are:

1. Heat flow from cold to hot (refrigeration)
2. Mass transfer from low to high elevation (lifts or pumps)
3. Transfer electricity from low to higher voltage (transformers)
4. Create more concentrated systems (remove solvent from solution)

List from Entropy, 2020 (Kostic 2020).

Enthalpy

The enthalpy of reaction (ΔH; endothermic or exothermic) can affect the spontaneity of reaction and depends greatly on the temperature. For example, the melting of ice is an endothermic process. Ice will spontaneously melt at room temperature (under normal conditions) because the ice is colder than the surroundings and heat will spontaneously flow from the surroundings into the ice. The ice heats up and melts.

\[\mathrm{H_2O}(s) \longrightarrow \mathrm{H_2O}(l) \qquad \mathrm{spontaneous~at~>0~^{\circ}C}\]

However, ice will not spontaneously melt at temperatures below 0 °C. Heat must be absorbed by the ice to melt but the surroundings are colder than the ice and cannot provide heat to melt the ice. Ice will lose heat to the surroundings and the ice will become colder.

\[\mathrm{H_2O}(s) \longrightarrow \mathrm{H_2O}(l) \qquad \mathrm{not~spontaneous~at~<~0~^{\circ}C}\]

These examples illustrate how energy wants to be uniformly dispersed. Either the ice will lose heat to the colder surroundings or ice will absorb heat from the hotter surroundings. In both cases, heat flows from hot to cold and is uniformly dispersed.

Entropy

Thermodynamic entropy is measure of how dispersed energy (or matter) is and is a thermal property. It is often referred to as a measure of disorder (thermal disorder or randomness). That said, the concept of entropy can sometimes be extended to the disorder or randomness of matter.

Take H₂O for example. At 0 K, water will be in a state of the lowest possible energy. The particles can be arranged such that it adopts a perfect crystal without any defects (giving rise to a minimum possible energy). This configuration of water particles at 0 K is defined as having zero entropy. Note: Entropy typically has units of J K^–1.

\[S = 0~~\mathrm{J~K^{-1}} ~~\mathrm{at~0~K}\] Now we heat the ice up (T > 0 K). Heat is transferred to the ice and we begin to see thermal disorder or randomness. The heat disperses itself throughout the system of ice and the water molecules absorb some amount of this heat. The water molecules begin to vibrate more rapidly, though their translational motions are still frozen since the particles are locked into rigid positions (hence ice is a solid). Entropy is increasing. S > 0 and Δ S > 0.

Continue to heat ice up (e.g. displace more heat into the ice) and eventually the ice melts into a liquid. More thermal energy is displaced into the system (entropy is increasing) and the molecules begin to gain translational motion (they begin to move around). The vibrational motions increase. The thermal energy is dispersed throughout the sample of water.

Continue to heat the water up and eventually the water turns into a gas. More thermal energy is displaced into the system and entropy increases. The gaseous water molecules are now moving through space without restriction of neighboring molecules and intermolecular forces. The particles are free to traverse space and are vibrating in all sorts of manner. Matter is dispersed and as such, the thermal energy they carry.

As we move from solid → liquid → gas, the thermal energy in these phases increase as does the thermal disorder or randomness. Think about it this way. Entropy is an amount of energy that is required to have some substance exist in some state. Ice at 0 °C requires less energy to exist at that state than water vapor at 100 °C.

Changes in entropy can be measured as a change in heat, Δq (or ΔH if the process is reversible), at a certain temperature, T.

\[\Delta S = \dfrac{\Delta q}{T} \equiv \dfrac{\Delta H}{T}\] The temperature, T, is the temperature of the system. Let us do a thought experiment. Imagine displacing 50 kJ worth of heat (Δq) into an object. If the object was cold (at low T), the object will experience a larger entropy change than an object that was hot (at high T).

Recall the heat of fusion for ice (6.01 kJ mol^–1). Transferring 6.01 kJ worth of heat to 1 mole of ice (18.02 g) at 0 °C (273.15 K) and 1 atm melts the ice and a positive entropy change is experienced.

\[\begin{align*} \Delta S &= \dfrac{\Delta H}{T} \\[1.5ex] &= \dfrac{6.01~\mathrm{kJ~mol^{-1}}}{273.15~\mathrm{K}} \left ( 1~\mathrm{mol} \right ) \left ( \dfrac{10^3~\mathrm{J}}{\mathrm{kJ}}\right ) \\[1.5ex] &= 22.00~\mathrm{J~K^{-1}} \end{align*}\]

The ice experiences a positive entropy change as it converts to a liquid at its normal melting point. It absorbed 22 J worth of heat per Kelvin.

Another way to state this: “A change in entropy is the amount of thermal energy that is displaced by a process at some temperature.” We can rearrange this equation as

\[\Delta H = T\Delta S\]

which says “the thermal energy displaced by a process is equal to the change in entropy at some temperature.” For our melting of ice example, if 22 J per Kelvin is required to melt 1 mol of ice, it would require 6.01 kJ worth of heat.

Of course, ice also freezes at 0 °C where the heat of freezing is -6.01 kJ mol_–1. In fact, if the ice is kept at exactly 0 °C, the rate of melting and the rate of fusion will be equal (in a closed system) and will be in a state of equilibrium.

\[\mathrm{H_2O}(s) \rightleftharpoons \mathrm{H_2O}(l)\]

We will realize later on down the page that ΔG = 0 for this state.

Any reaction or process is spontaneous if the entropy of the universe, ΔS_univ, increases as a result of that process.

\[\Delta S_{\mathrm{univ}} = \Delta S_{\mathrm{sys}} + \Delta S_{\mathrm{surr}}\]

So, the spontaneous process of ice melting at 25 °C is a process that increases the entropy of the universe. A thermal displacement occurs where the surroundings loses energy and the system absorbs the energy. The positive entropy change of the ice (i.e. the system) is larger than the negative entropy change of the surroundings

\[\lvert \Delta S_{\mathrm{sys}} \rvert > \lvert \Delta S_{\mathrm{surr}} \rvert\]

and the entropy change of the universe is positive.

How? Well, solid water is an ordered system where water molecules are held together in fixed positions. Liquid water has water molecules that are loosely held together and the water molecules themselves are more “spread out” (i.e. more dispersed). The liquid water has a higher entropy than solid water. Therefore, the melting of ice is a process that increases the entropy of the system.

\[\mathrm{H_2O}(s) \longrightarrow \mathrm{H_2O}(l) \qquad \Delta S_{\mathrm{system}} > 0\]

Additionally, the ice is absorbing heat at some temperature, Therefore, Δq is positive giving rise to a positive change in entropy.

To melt, the system (the ice), absorbs heat from the surroundings. The surroundings, therefore, loses a bit of heat and the entropy of the surroundings decreases.

\[\mathrm{H_2O}(s) \longrightarrow \mathrm{H_2O}(l) \qquad \Delta S_{\mathrm{surr}} < 0\]

The heat that the system (the ice) absorbs causes a greater increase in the entropy gain of the system than the entropy that the surroundings lost.

\[\lvert \Delta S_{\mathrm{sys}} \rvert > \lvert \Delta S_{\mathrm{surr}} \rvert\]

Therefore, the entropy of the universe is positive for this process. We can rationalize this because heat was more uniformly dispersed when the hotter surroundings distributed some heat to the colder system.

Sign	Spontaneity
ΔSuniv > 0	spontaneous
ΔSuniv < 0	non-spontaneous

Gibbs Energy

Imagine combusting gasoline. Gasoline is a liquid and when combusted, a gas is formed. To simplify the example, we will consider octane (C₈H₁₈) instead of gasoline which is a complex mixture of molecules. When octane undergoes combustion, the entropy change is relatively large (≈ 500 J K^–1). The enthalpy of the reaction is, however, much larger (≈ -5400 kJ mol^–1). So, the combustion of octane produces a enormous amount of energy per mole, but not all of that energy is available for us to extract! Some amount of that energy is required/used to allow the products of the reaction to exist in their states.

It is very inconvenient, if not impossible, to take a measurement of the surroundings. It is much easier to take a measurement of the system. Boil a pot of water on the stove and try to figure out the temperature of the water by measuring the surroundings. It would be easier to just stick a thermometer in the water! Gibbs energy does just this! It it a thermodynamic property that allows us to measure the change in entropy of the universe by only having to measuring the system!

Gibbs free energy (ΔG), or just Gibbs energy, according to thermodynamics, represents the maximum amount of non-expansion work that can be extracted from a thermodynamically closed system. The relationship between enthalpy, temperature, and entropy is given below. All terms are with respect to the system.

\[\Delta G = \Delta H - T\Delta S\]

Let us tackle three thermodynamic principles here.

A reaction or process is spontaneous if the Gibbs free energy for that process is negative.

This principle is due to the following relationship

\[-\Delta G_{\mathrm{sys}} \propto \Delta S_{\mathrm{univ}}\]

As stated earlier, any process is spontaneous if it increases the entropy of the universe. Therefore, the change in free energy for that process must be negative.

A reaction or process is non-spontaneous if the Gibbs free energy for that process is positive.

\[\Delta G_{\mathrm{sys}} \propto -\Delta S_{\mathrm{univ}}\] Any process that is non-spontaneous has a positive change in Gibbs energy which correlates to a decrease in the entropy of the universe.

A reaction or process is at equilibrium if the Gibbs free energy for that process is zero.

\[\Delta G_{\mathrm{sys}} = \Delta S_{\mathrm{univ}} = 0 ~~\mathrm{at~equilibrium}\]

ΔG_sys and ΔS_univ relation

Begin with the entropy change of the universe.

\[\Delta S_{\mathrm{univ}} = \Delta S_{\mathrm{surr}} + \Delta S_{\mathrm{sys}}\]

We can express the entropy of the surroundings as a change in heat per unit temperature.

\[\Delta S_{\mathrm{univ}} = \dfrac{\Delta q_{\mathrm{surr}}}{T} + \Delta S_{\mathrm{sys}}\]

Furthermore, we can express the change in heat of surroundings as the change in heat of the system.

\[\Delta q_{\mathrm{surr}} = -\Delta q_{sys}\]

Therefore,

\[\Delta S_{\mathrm{univ}} = \dfrac{-\Delta q_{\mathrm{sys}}}{T} + \Delta S_{\mathrm{sys}}\]

We express heat (q) as enthalpy (ΔH).

\[\Delta S_{\mathrm{univ}} = \dfrac{-\Delta H_{\mathrm{sys}}}{T} + \Delta S_{\mathrm{sys}}\]

Next, we multiply every term by T and –1 give

\[-T\Delta S_{\mathrm{univ}} = \Delta H_{\mathrm{sys}} - T\Delta S_{\mathrm{sys}}\]

Finally, we define the Gibbs free energy of the system (ΔG_sys) as –TΔS_univ giving

\[\Delta G_{\mathrm{sys}} = \Delta H_{\mathrm{sys}} - T\Delta S_{\mathrm{sys}}\]

Since every term is defined from the perspective of the system, we drop the “sys” labels and simply have

\[\Delta G = \Delta H - T\Delta S\] It is important to note/remember that Gibbs energy is related to the change in entropy of the universe.

\[\Delta G = -T\Delta S_{\mathrm{universe}}\]

Let us return to our example of ice melting and freezing at 0 °C and in a state of equilibrium.

\[\begin{align*} \Delta G &= \Delta H - T\Delta S\\ &= 6.01~\mathrm{kJ~mol^{-1}} - 273.15~\mathrm{K} \left ( 22.0~\mathrm{J~K^{-1}} \right ) \left ( \dfrac{\mathrm{kJ}}{10^3~\mathrm{J}} \right ) \\ &= 0.00~\mathrm{kJ~mol^{-1}} \end{align*}\]

We see that ΔG = 0 at the equilibrium. What would ΔG be at 1 °C?

\[\begin{align*} \Delta G &= \Delta H - T\Delta S\\ &= 6.01~\mathrm{kJ~mol^{-1}} - 274.15~\mathrm{K} \left ( 22.0~\mathrm{J~K^{-1}} \right ) \left ( \dfrac{\mathrm{kJ}}{10^3~\mathrm{J}} \right ) \\ &= -0.02~\mathrm{kJ~mol^{-1}} \end{align*}\]

Here, ΔG is negative and ice will spontaneously melt (it favors products) at 1 °C! Let us repeat this once more for ice at -1 °C.

\[\begin{align*} \Delta G &= \Delta H - T\Delta S\\ &= 6.01~\mathrm{kJ~mol^{-1}} - 272.15~\mathrm{K} \left ( 22.0~\mathrm{J~K^{-1}} \right ) \left ( \dfrac{\mathrm{kJ}}{10^3~\mathrm{J}} \right ) \\ &= 0.02~\mathrm{kJ~mol^{-1}} \end{align*}\]

Here, ΔG is positive and ice will not spontaneously melt (it favors reactants) at –1 °C!

When ΔG < 0, the process is spontaneous.

Spontaneous reactions proceed until the change in Gibbs free energy is zero.

A reaction occurs meaning the initial change in free energy is negative. At some point, the reaction reaches equilibrium and the driving force behind the reaction is equally balanced with a reverse driving force. Here, ΔG = 0 and the process is at equilibrium.

Standard Gibbs free energy (where all reactants and products are in their standard states) can be related to an equilibrium constant for a system at equilibrium and is given as

\[\Delta G^{\circ} = -RT\ln K\]

For values of K < 1 (reactant favored), ln(K) < 0 and for values of K > 1 (product favored), ln(K) > 0.

For a system that is not at equilibrium, we use ΔG and Q to determine the direction that the reaction must proceed.

\[\Delta G = RT\ln \dfrac{Q}{K}\]

Here, if Q > K, then ln(Q/K) > 0 and Δ G is positive for the forward direction. The reaction must proceed left to reach equilibrium. If Q < K, then ΔG is negative and proceeds right to reach equilibrium.