States and State Functions

The state of a system is described by the macroscopic properties of the system. Examples of macroscopic properties include temperature (T), pressure (P), and volume (V). When the state of a system changes, the values of the properties also change. If the change in the value of a property depends only on the initial and final states of the system, and not on the path of the change (how the change was accomplished), that property is known as a state function. Pressure, temperature, and volume are important state functions. Other examples are enthalpy (H), entropy (S), Gibbs free energy (G) (all discussed below), and internal energy (E or U).

A set of standard conditions (25°C and 1 atm) is normally used for measuring the enthalpy, entropy, and Gibbs free energy of a reaction. A substance in its most stable form under standard conditions is said to be in its standard state. Examples of substances in their standard states include hydrogen as H2 (g), water as H2O (l), and salt as NaCl (s). The changes in enthalpy, entropy, and Gibbs free energy that occur when a reaction takes place under standard conditions are symbolized by ΔH°, ΔS°, and ΔG°, and are known as the standard change in enthalpy (or standard enthalpy change), et cetera.

Enthalpy

Most reactions in the lab occur under constant pressure (at 1 atm, in open containers). (Reactions carried out under a constant pressure are said to be isobaric.) To express heat changes at constant pressure, chemists use the term enthalpy (H), often thought of as the “heat content” of a system. The change in enthalpy (∆H) of a process is equal to the heat absorbed or evolved by the system at constant pressure. Since enthalpy is a state function, the enthalpy change of a process depends only on the enthalpies of the initial and final states, not on the path. Thus, to find the enthalpy change of a reaction, ∆Hrxn, one may subtract the enthalpy of the reactants from the enthalpy of the products:

Hrxn = HproductsHreactants

A positive ∆H corresponds to an endothermic process (absorbs heat), and a negative ∆H corresponds to an exothermic process (releases heat). The value of enthalpy change (or of enthalpies in general) is dependent on external conditions such as temperature and pressure. As we shall see, enthalpy plays an important role in determining the favorability of a reaction, but first we need to discuss ways in which one can calculate or obtain values for changes in enthalpy.

Hess’s Law

Hess’s law is simply the application of the concept of path-independence to enthalpy. It states that if a reaction can be broken down into a series of steps, the enthalpy change for the overall net reaction is just the sum of the enthalpy changes of each step. The steps need not even correspond to actual processes carried out in the real world or in the lab, but can be purely hypothetical. For example, consider the reaction:

Br2 (l) Br2 (g) ∆H = 31 kJ

The enthalpy change of the above reaction will always be 31 kJ/mol provided that the same initial and final states Br2 (l) and Br2 (g) are operative. Instead of direct vaporization, Br2 (l) could first be decomposed to Br atoms and then recombined to form Br2 (g); since the net reaction is the same (the two possible sequences share the same initial state and the same final state), the change in enthalpy will be the same.

Example: Given the following thermochemical equations:
  1. Calculate ΔH for the reaction:

  2. 3C (graphite) + 4H2 (g) C3H8 (g)
Solution: Equations a, b, and c must be combined to obtain equation d. Since equation d contains only C, H2, and C3H8, we must eliminate O2, CO2, and H2O from the first three equations. Equation a is reversed to get C3H8 on the product side.
  1. e)
(Note that when we reverse the reaction, the sign of the enthalpy change is reversed as well.)

Next, equation b is multiplied by 3 (this gives equation f) and c by 4 (this gives equation g). The following addition is done to obtain the required equation d: 3b + 4c + e.
  1. f)
  2. g)

where ∆Hd = ∆He + ∆Hf + ∆Hg.

Standard Enthalpy of Formation

Hess’s law is useful because measuring the change in enthalpy for a process directly can be challenging experimentally, but by taking advantage of the fact that this quantity is independent of path, one can calculate ∆H for any process if the values for certain other reactions are known. The most common approach is to express the enthalpy change of a reaction in terms of the standard enthalpies of formation of the products and the reactants.

The standard enthalpy of formation of a compound, ∆H°f , is the enthalpy change that would occur if one mole of a compound were formed directly from its elements in their standard states. For example, the standard enthalpy of formation of H2O is just the enthalpy change for the reaction:

H2 (g) + 1/2 O2 (g) H2O (l)

if the reaction were carried out under standard conditions. We have picked the gaseous forms of hydrogen and oxygen because that is the most stable form in which they exist under such conditions.

Note that ∆H°f of an element in its standard state is zero. The ∆H°f’s of most known substances are tabulated. The enthalpy of formation is also often referred to as the heat of formation.

Standard Enthalpy of Reaction

The standard enthalpy (or heat) of a reaction, ∆H°rxn, is the hypothetical enthalpy change that would occur if the reaction were carried out under standard conditions; that is, when reactants in their standard states are converted to products in their standard states at 298 K. It can be expressed as:

H°rxn = (sum of ∆H°f of products) – (sum of ∆H°f of reactants)

Earlier we gave a general definition ∆Hrxn = HproductsHreactants. This expression, however, is not very useful. The actual values of the enthalpies of the species cannot be measured (and are in fact more a matter of definition); what we can measure is only enthalpy changes, and those often only with difficulty. But from our discussion on Hess’s law, we know we can concoct a scheme consisting of a series of steps that yield the same net reaction, and the sums of enthalpy changes would be the same as the net enthalpy change of the reaction. The equation above, H°rxn = (sum of ∆H°f of products) – (sum of ∆H°f of reactants), establishes a common scheme for all reactions: We first break the reactants down to give elements in their standard states, then we combine these rudimentary “building blocks” in new ways to give the products of the reaction. For example, when we express the enthalpy change for the reaction

CaCO3 (s) → CaO (s) + CO2 (g)

as ∆H°rxn = ∆H°f of CaO + ∆H°f of CO2 – ∆H°f of CaCO3, we are essentially reporting the reaction as follows: First we break down each mole of CaCO3 into 1 mole of Ca, 1 of C, and 3/2 of O2, then we combine the Ca with half a mole of O2 to form a mole of CaO, and use the remaining oxygen to form CO2 with carbon. The enthalpy changes for the last two steps are the standard enthalpies of formation of CaO and CO2, while the first step is the reverse of the formation of CaCO3, from which we get the minus sign in front of the enthalpy of formation of CaCO3, and which explains why we subtract the enthalpies of formation of the reactants in the general equation.

It should be emphasized again that this scheme is not actually carried out, but merely used for “accounting conveniences”: Instead of tabulating an enthalpy change for every reaction imaginable, we need only a list of enthalpies of formation of different compounds which, while still numerous, do not possess this almost infinite possible number of combinations.

Bond Dissociation Energy

Enthalpies or heats of reaction are related to changes in energy associated with the breakdown and formation of chemical bonds. The reason why bonds are formed in the first place is because it is energetically favorable for the atoms to come together—it corresponds to a state of lower energy. This implies, therefore, that energy needs to be supplied to break a bond and separate the atoms (i.e., bond breaking is endothermic). Bond energy, or bond dissociation energy, is an average of the energy required to break a particular type of bond in one mole of gaseous molecules. It is tabulated as the magnitude of the energy absorbed as the bonds are broken. For example:

H2 (g) 2H (g) ∆H = 436 kJ

A molecule of H2 gas is cleaved to produce two gaseous, unassociated hydrogen atoms. For each mole of H2 gas cleaved, 436 kJ of energy is absorbed by the system. This is the bond energy of the H—H bond. For other types of bonds, the energy requirements are averaged by measuring the enthalpy of cleaving many different compounds with that bond. The averaging is needed because the energy required to break a bond is not uniquely determined by what atoms are being separated; it also depends on what else may be bonded to those atoms. For example, the energy needed to break a C—H bond is different on going from CH4 to CH3Cl to CCl3H, et cetera. The C—H bond dissociation energy one would find in a table (415 kJ/mol) was compiled from measurements on thousands of different organic compounds.

Bond energies can be used to estimate enthalpies of reactions. The enthalpy change of a reaction is given by:

  1. Example: Calculate the enthalpy change for the following reaction:
C (s) + 2H2 (g) CH4 (g)
  1. Bond dissociation energies of H—H and C—H bonds are 436 kJ/mol and 415 kJ/mol, respectively.
Hf of C (g) = 715 kJ/mol
  1. Solution: For each mole of CH4 formed, 2 moles of H—H bonds are broken and 4 moles of C—H bonds are formed (each molecule of CH4 contains 4 C—H bonds). The one additional thing we have to take into consideration is that bond dissociation energies are always in reference to the gas phase, yet here we have carbon in its solid form. This is why we need the enthalpy of formation of gaseous carbon: The first step in our hypothetical scheme is to convert carbon from its solid to its gaseous form. Thus the enthalpy change is:

Again, we subtract the bond dissociation energy to reflect the fact that the C—H bonds are being formed rather than broken.

Heats of Combustion

One more type of standard enthalpy change that is often used is the standard heat of combustion, ∆H°comb. A combustion reaction is one in which the reactant reacts with (excess) oxygen to yield (in most cases) carbon dioxide and water, producing a flame during the reaction. (Excess oxygen is specified because inadequate oxygen leads to the generation of carbon monoxide rather than carbon dioxide.) The burning of a log, for example, is a combustion reaction. These reactions are exothermic (release energy, have a negative enthalpy change). The reactions used in the C3H8 (g) example earlier were combustion reactions, and the corresponding values ∆Ha, ∆Hb, and ∆Hc in that example were thus heats of combustion.

Entropy

Entropy (S) is a measure of the disorder, or randomness, of a system. The units of entropy are energy/temperature, commonly J/K or cal/K. The greater the order in a system, the lower the entropy; the greater the disorder or randomness, the higher the entropy. At any given temperature, a solid will have lower entropy than a gas, because individual molecules in the gaseous state are moving randomly, while individual molecules in a solid are constrained in place. Entropy is a state function, so a change in entropy depends only on the initial and final states:

S = SfinalSinitial

A change in entropy is also given by:

where qrev is the heat added to the system undergoing a reversible process (a process that proceeds with infinitesimal changes in the system’s conditions) and T is the absolute temperature.

A standard entropy change for a reaction, ∆S°, is calculated using the standard entropies of reactants and products:

S°rxn = (sum of S°products) − (sum of S°reactants)

The Second Law of Thermodynamics

Entropy is an important concept because it determines whether a process will occur spontaneously. The second law of thermodynamics states that all spontaneous processes proceeding in an isolated system lead to an increase in entropy. Since the universe as a whole is one big isolated system, we can also rephrase this law in a way that is perhaps more stimulating to our imagination—the entropy of the universe either increases (spontaneous, irreversible processes) or stays the same (reversible processes). It can never decrease.

Suniverse = ∆Ssystem + ∆Ssurroundings ≥ 0

Note that the entropy of a system can decrease, as long as it is compensated for by a larger increase in entropy in the surroundings. The mechanism of refrigeration decreases the entropy within the refrigerator by maintaining a low temperature that would not persist if left to nature; heat is, however, generated and dumped outside into the kitchen that increases the entropy. Our cells are constantly engaging in biochemical reactions that increase the order locally: Synthesizing large biomolecules from disordered “building blocks,” sequestering ions in different compartments when it would be more “natural” for them to diffuse over a larger volume, et cetera. All these processes come at the expense of entropy increases elsewhere: for example, disorder that was generated when we digested our meal the previous evening. A system will spontaneously tend toward an equilibrium state (one of maximum entropy) if left alone.

The Third Law of Thermodynamics

Instead of just working with changes or relative magnitudes (as in the case of enthalpy), there is a standard with which one can assign the actual value of entropy of a substance. The third law of thermodynamics states that the entropy of a pure crystalline substance at absolute zero is zero. This corresponds to a state of “perfect order” because all the atoms in this hypothetical state possess no kinetic energy and do not vibrate at all; thus, there is absolutely no randomness and no disorder in the spatial arrangement of the atoms.

Gibbs Free Energy

What makes a reaction favorable? In the quantity of entropy, we have an unambiguous criterion of whether a reaction would occur spontaneously: The total entropy of the universe has to increase. The only problem with this is that it is not very practical—who knows how to keep track of the entropy of the entire universe? It would be nice to have a quantity that deals only with the system itself that we can examine to determine the favorability of a reaction. The thermodynamic state function, G (known as the Gibbs free energy), is just such a quantity. It combines the two factors that affect the spontaneity of a reaction—changes in enthalpy, ∆H, and changes in entropy, ∆S, of the system. The change in the free energy of a system, ∆G, represents the maximum amount of energy released by a process, occurring at constant temperature and pressure, that is available to perform useful work. ∆G is defined by the equation:

G = ∆HTS

where T is the absolute temperature.

Spontaneity of Reaction

In the equilibrium state, free energy is at a minimum. A process can occur spontaneously if the Gibbs function decreases, i.e., ∆G < 0.

  1. If ΔG is negative, the reaction is spontaneous.
  2. If ΔG is positive, the reaction is not spontaneous.
  3. If ΔG is zero, the system is in a state of equilibrium; thus, ΔG = 0 and ΔH = TΔS at equilibrium.

Because the temperature is always positive, i.e., in Kelvins, the effects of the signs of ΔH and ΔS and the effect of temperature on spontaneity can be summarized as follows:

H S Outcome
+ spontaneous at all temperatures
+ nonspontaneous at all temperatures
+ + spontaneous only at high temperatures
spontaneous only at low temperatures

Qualitatively, the equation ΔG = ΔHTΔS, and more generally the whole notion of free energy, tell us that there are two factors that favor a reaction: a decrease in energy in the form of enthalpy and an increase in disorder. If these two factors are working against each other in a reaction, then temperature will determine which is the more dominant factor—the higher the temperature, the easier entropic considerations override enthalpic ones. Note also the implication that a reaction that is favorable at one temperature may not be favorable at another. This, actually, should not be a surprise; after all, for example, the melting of ice is expected to occur at 15 ºC, but not at −100 ºC (assuming atmospheric pressure).

Standard Free Energy Change

Standard free energy change, ΔGº, is defined as the ΔG of a process occurring under standard conditions, and for which the concentrations of any solutions involved are 1 M. The standard free energy of formation of a compound, ΔGºf, is the free-energy change that occurs when 1 mole of a compound in its standard state is formed from its elements in their standard states. The standard free energy of formation of any element in its most stable form (and, therefore, its standard state) is zero. The standard free energy of a reaction, ΔGºrxn, is the free energy change that occurs when that reaction is carried out under standard state conditions; i.e., when the reactants in their standard states are converted to the products in their standard states, at standard conditions of T and P. Bearing in mind what we did for enthalpy changes, we can write:

ΔG°rxn = (sum of ΔG°f of products) – (sum of ΔG°f of reactants).

Relation Between Free Energy and the Equilibrium Constant

The value of the free energy change in general (under nonstandard conditions) is related to the standard free energy change by the following equation:

ΔG = ΔG° + RT ln Q

where R is the gas constant, T is the temperature in Kelvins, ln stands for the natural logarithm function, and Q is the reaction quotient we mentioned briefly in passing in the last chapter. For the reaction a A + b B ⇄ c C + d D, the reaction quotient is

Unlike the mass action expression, the values for the concentrations are not necessarily those at equilibrium. Because the concentrations of the species change as the reaction progresses towards equilibrium, the value of Q, and hence also the value of ΔG, will change as the reaction progresses. As pointed out above, when the system is at equilibrium ΔG is zero, and in that case we would obtain from the equation

0 = ΔGº + RT ln Qeq

where Qeq is the value of the reaction quotient at equilibrium. But that is just K, the equilibrium constant! Hence we can rewrite and rearrange to get

ΔGº = −RT ln K

or equivalently,

K = eG˚/RT

Thus, we have obtained the final missing link to put it all together: the more negative ΔG° is, the larger the equilibrium constant, and hence the more the products are favored at equilibrium.

Examples

  1. Vaporization of water at one atmosphere pressure
H2O (l) + heat H2O (g)

When water boils, hydrogen bonds (H-bonds) are broken. Energy is absorbed (the reaction is endothermic), and thus ΔH is positive. Entropy increases as the closely packed molecules of the liquid become the more randomly moving molecules of a gas; thus, TΔS is also positive. Since ΔH and TΔS are both positive, the reaction will proceed spontaneously only if TΔS > ΔH. For the particular values of ΔS and ΔH applicable for this reaction, this condition is true only at temperatures above 100°C. Below 100°C, ΔG is positive and the water remains a liquid. At 100°C, ΔH = TΔS and ΔG = 0: an equilibrium is established between water and water vapor. The opposite is true when water vapor condenses: H-bonds are formed, and energy is released; the reaction is exothermic (ΔH is negative) and entropy decreases, since a liquid is forming from a gas (TΔS is negative). Condensation will be spontaneous only if ΔH < TΔS. This is the case at temperatures below 100°C; above 100°C, TΔS is more negative than ΔH, ΔG is positive, and condensation is not spontaneous. Again, at 100°C, an equilibrium is established.

  1. The combustion of C6H6 (benzene)
2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (g) + heat

In this case, heat is released (ΔH is negative) as the benzene burns and the entropy is increased (TΔS is positive), because two gases (18 moles total) have greater entropy than a gas and a liquid (15 moles gas and 2 liquid). ΔG is negative and the reaction is spontaneous.