Thermodynamics: Spontaneity, Entropy and Free Energy

Thermodynamics Thermodynamics studies how changes in energy, entropy and temperature affect the spontaneity of a process or chemical reaction. Using thermodynamics we can predict the direction a reaction will go, and also the driving force of a reaction or system to go to equilibrium.

Spontaneity A spontaneous process is one that occurs without outside intervention. Examples include: - a ball rolling downhill - ice melting at temperatures above 0oC - gases expanding to fill their container - iron rusts in the presence of air and water - two gases mixing

Spontaneity Spontaneous processes can release energy (a ball rolling downhill), require energy (ice melting at temperatures above 0oC), or involve no energy change at all (two gases mixing) . Spontaneity is independent of the speed or rate of a reaction. A spontaneous process may proceed very slowly.

Spontaneity There are three factors that combine to predict spontaneity. They are: 1. Energy Change 2. Temperature 3. Entropy Change

A measure of randomness or disorder Entropy A measure of randomness or disorder

Entropy Entropy, S, is a measure of randomness or disorder. The natural tendency of things is to tend toward greater disorder. This is because there are many ways (or positions) that lead to disorder, but very few that lead to an ordered state.

Entropy The entropy of a system is defined by the Boltzmann equation: S = k ln W k is the Boltzmann constant, and W is the number of energetically equivalent ways to arrange the components of the system.

Entropy Gases will spontaneously and uniformly mix because the mixed state has more possible arrangements (a larger value of W and higher entropy) than the unmixed state.

Entropy The driving force for a spontaneous process is an increase in the entropy of the universe.

ΔSo and Phase Changes Gases have more entropy than liquids or solids.

ΔSo and Mixtures Mixtures have more entropy than pure substances.

Entropy Values of Common Substances

The 2nd Law of Thermodynamics In any spontaneous process there is always an increase in the entropy of the universe.

The 2nd Law of Thermodynamics Water spontaneously freezes at a temperature below 0oC. Therefore, the process increases the entropy of the universe. The water molecules become much more ordered as they freeze, and experience a decrease in entropy. The process also releases heat, and this heat warms gaseous molecules in air, and increases the entropy of the surroundings.

The 2nd Law of Thermodynamics Since the process is spontaneous below 0oC, ΔSsurr, which is positive, must be greater in magnitude than ΔS of the water molecules.

Entropy Entropy can be viewed as the dispersal or randomization of energy. The freezing of water (an exothermic process) releases heat to the surroundings, and thus increases the entropy of the surroundings. The process is spontaneous at or below 0oC because the increase in entropy of the surroundings is greater than the decrease in entropy of the water as it freezes.

Δ S and Spontaneity

Spontaneity Entropy, temperature and heat flow all play a role in spontaneity. A thermodynamic quantity, the Gibbs Free Energy (G), combines these factors to predict the spontaneity of a process. ΔG = ΔH - TΔS

Spontaneity ΔG = ΔH - TΔS If a process releases heat (ΔH is negative) and has an increase in entropy (ΔS is positive), it will always be spontaneous. The value of ΔG for spontaneous processes is negative.

Spontaneity ΔG = ΔH - TΔS

Spontaneity and ΔG If ΔG is negative, the process is spontaneous (and the reverse process is non-spontaneous). If ΔG is positive, the process is non-spontaneous, and the reverse process is spontaneous. If ΔG = 0, the system is at equilibrium.

ΔG Although ΔG can be used to predict in which direction a reaction will proceed, it does not predict the rate of the reaction. For example, the conversion of diamond to graphite has a ΔGo = -3 kJ, so diamonds should spontaneously change to graphite at standard conditions. However, kinetics shows that the reaction is extremely slow.

The Significance of ΔG ΔG represents the driving force for the reaction to proceed to equilibrium.

The Significance of ΔG If negative, the value of ΔG in KJ is the maximum possible useful work that can be obtained from a process or reaction at constant temperature and pressure. In practice, some energy is always lost, so the actual work produced will be less than the calculated value.

The Significance of ΔG If positive, the value of ΔG in KJ is the minimum work that must be done to make the non-spontaneous process or reaction proceed. In practice, some additional work is required to make the non-spontaneous process or reaction proceed.

Reversibility A reversible reaction is a reaction that achieves the theoretical limit with respect to free energy. That is, there is no loss of energy (usually as heat) to the surroundings. All real reactions are irreversible, and do not achieve he theoretical limit of available free energy.

Predicting the sign of ΔSo For many chemical reactions or physical changes, it is relatively easy to predict if the entropy of the system is increasing or decreasing. If a substance goes from a more ordered phase (solid) to a less ordered phase (liquid or gas), its entropy increases.

Predicting the sign of ΔSo For chemical reactions, it is sometimes possible to compare the randomness of products versus reactants. 2 KClO3(s)  2 KCl(s) + 3 O2(g) The production of a gaseous product from a solid reactant will have a positive value of ΔSo.

Calculating Entropy Changes Since entropy is a measure of randomness, it is possible to calculate absolute entropy values. This is in contrast to enthalpy values, where we can only calculate changes in enthalpy. A perfect crystal at absolute zero has an entropy value (S) =0. All other substances have positive values of entropy due to some degree of disorder.

Calculating Entropy Changes Fortunately, the entropy values of most common elements and compounds have been tabulated. Most thermodynamic tables, including the appendix in the textbook, include standard entropy values, So.

Entropy Values of Common Substances

Entropy Values For comparable structures, the entropy increases with increasing mass

Entropy Values For molecules with similar masses, the more complex molecule has greater entropy. The molecule with more bonds has additional ways to absorb energy, and thus greater entropy.

Calculating Entropy Changes For any chemical reaction, Δ Soreaction= Σmolprod Soproducts- Σmolreact Soreactants The units of entropy are joules/K-mol.

Calculation of ∆Go ∆Go, the standard free energy change, can be calculated in several ways. ∆Go = ∆Ho - T ∆So It can be calculated directly, using the standard enthalpy change and entropy change for the process.

Calculation of ∆Go ∆Go = ∆Ho - T ∆So ∆Ho is usually calculated by using standard enthalpies of formation, ∆Hfo. ∆Horxn = Σnprod ∆Hoproducts- Σnreact ∆Horeactants

Calculation of ∆Go ∆Go = ∆Ho - T ∆So Once ∆Ho and ∆So have been calculated, the value of ∆Go can be calculated, using the temperature in Kelvins.

Calculation of ∆Go ∆Go can also be calculated by combining the free energy changes of related reactions. This is the same method used in Hess’ Law to calculate enthalpy changes. If the sum of the reactions gives the reaction of interest, the sum of the ∆Go values gives ∆Go for the reaction.

∆Gorxn = Σmolprod ∆Gfo prod - Σmolreact ∆Gfo react Calculation of ∆Go Lastly, ∆Go can be calculated using standard free energies of formation, ∆Gfo. Some tables of thermodynamic data, including the appendix of your textbook, include values of ∆Gfo. ∆Gorxn = Σmolprod ∆Gfo prod - Σmolreact ∆Gfo react

Calculation of ∆Go When calculating ∆Go from standard free energies of formation, keep in mind that ∆Gfo for any element in its standard state is zero. As with enthalpies of formation, the formation reaction is the reaction of elements in their standard states to make compounds (or allotropes).

Calculation of ∆Go

Calculation of ∆Go Note the values of zero for nitrogen, hydrogen and graphite.

CaCO3(s) ↔CaO(s) + CO2(g) at 25oC. Spontaneity Problem Consider the reaction:  CaCO3(s) ↔CaO(s)  +  CO2(g)  at 25oC.  Calculate ∆Go using the tables in the appendix of your textbook.  Is the process spontaneous at this temperature?  Is it spontaneous at all temperatures?  If not, at what temperature does it become spontaneous?

CaCO3(s) ↔CaO(s) + CO2(g) at 25oC. Spontaneity Problem Consider the reaction:  CaCO3(s) ↔CaO(s)  +  CO2(g)  at 25oC.  Calculate ∆Go using the tables in the appendix of your textbook.  Is the process spontaneous at this temperature?  Calculation of ∆Grxno will indicate spontaneity at 25oC. It can be calculated using ∆Gfo values or from ∆Hfo and ∆So values.

Calculation of ∆Go CaCO3(s) ↔CaO(s) + CO2(g) ∆Grxno = Σnprod ∆Gfo prod - Σnreact ∆Gfo react

CaCO3(s) ↔CaO(s) + CO2(g) Calculation of ∆Go CaCO3(s) ↔CaO(s)  +  CO2(g) ∆Grxno =[(1 mol) (-604.0 kJ/mol) + (1 mol)(-394.4 kJ/mol)] –[1 mol(-1128.8 kJ/mol)]

CaCO3(s) ↔CaO(s) + CO2(g) Calculation of ∆Go CaCO3(s) ↔CaO(s)  +  CO2(g) ∆Grxno =[(1 mol) (-604.0 kJ/mol) + (1 mol)(-394.4 kJ/mol)] –[1 mol(-1128.8 kJ/mol)] = +130.4 kJ

CaCO3(s) ↔CaO(s) + CO2(g) at 25oC. Spontaneity Problem Consider the reaction:  CaCO3(s) ↔CaO(s)  +  CO2(g)  at 25oC.  Calculate ∆Go using the tables in the appendix of your textbook.  Is the process spontaneous at this temperature?  Since ∆Grxno =+130.4 kJ, the reaction is not spontaneous at 25oC.

CaCO3(s) ↔CaO(s) + CO2(g) at 25oC. Spontaneity Problem Consider the reaction:  CaCO3(s) ↔CaO(s)  +  CO2(g)  at 25oC.    Is it spontaneous at all temperatures?  If not, at what temperature does it become spontaneous?

CaCO3(s) ↔CaO(s) + CO2(g) at 25oC. Spontaneity Problem Consider the reaction:  CaCO3(s) ↔CaO(s)  +  CO2(g)  at 25oC.    Is it spontaneous at all temperatures?  If not, at what temperature does it become spontaneous? At 25oC, ∆Grxno is positive, and the reaction is not spontaneous in the forward direction.

CaCO3(s) ↔CaO(s) + CO2(g) at 25oC. Spontaneity Problem Consider the reaction:  CaCO3(s) ↔CaO(s)  +  CO2(g)  at 25oC.    Is it spontaneous at all temperatures?  If not, at what temperature does it become spontaneous? Inspection of the reaction shows that it involves an increase in entropy due to production of a gas from a solid.

CaCO3(s) ↔CaO(s) + CO2(g) at 25oC. Spontaneity Problem Consider the reaction:  CaCO3(s) ↔CaO(s)  +  CO2(g)  at 25oC.    Is it spontaneous at all temperatures?  If not, at what temperature does it become spontaneous? We can calculate the entropy change and the enthalpy change, and then determine the temperature at which spontaneity will occur.

CaCO3(s) ↔CaO(s) + CO2(g) Since ∆Go = ∆Ho - T∆So, and there is an increase in entropy, the reaction will become spontaneous at higher temperatures. To calculate ∆So, use the thermodynamic tables in the appendix.

CaCO3(s) ↔CaO(s) + CO2(g) ∆Srxno =[1mol(213.6J/K-mol)+1mol(39.7J/K-mol)] -[1mol(92.9J/K-mol)] = 160.4 J/K

CaCO3(s) ↔CaO(s) + CO2(g) ∆Go = ∆Ho - T∆So Since we know the value of ∆Go (+130.4 kJ) and ∆So (160.4 J/K), we can calculate the value of ∆Ho at 25oC. 130.4 kJ = ∆Ho –(298K) (160.4 J/K) ∆Ho = 130.4 kJ + (298K) (.1604 kJ/K) ∆Ho = + 178.2 kJ

CaCO3(s) ↔CaO(s) + CO2(g) ∆Go = ∆Ho - T∆So If we assume that the values of ∆Ho and ∆So don’t change much with temperature, we can estimate the temperature at which the reaction will become spontaneous.

CaCO3(s) ↔CaO(s) + CO2(g) ∆Go = ∆Ho - T∆So ∆Go is positive at lower temperatures, and will be negative at higher temperatures. Set ∆Go equal to zero, and solve for temperature. 0 = ∆Ho - T∆So T = ∆Ho ∆So

CaCO3(s) ↔CaO(s) + CO2(g) ∆Go = ∆Ho - T∆So 0 = ∆Ho - T∆So T = ∆Ho ∆So T = (178.2 kJ)/(160.4 J/K)(10-3kJ/J) =1111K or 838oC The reaction will be spontaneous in the forward direction at temperatures above 838oC.

∆G for Non-Standard Conditions The thermodynamic tables are for standard conditions. This includes having all reactants and products present initially at a temperature of 25oC. All gases are at a pressure of 1 atm, and all solutions are 1 M.

∆G for Non-Standard Conditions For non-standard temperature, concentrations or gas pressures: ∆G = ∆Go + RTlnQ Where R = 8.314 J/K-mol T is temperature in Kelvins Q is the reaction quotient

∆G for Non-Standard Conditions For non-standard temperature, concentrations or gas pressures: ∆G = ∆Go + RTlnQ For Q, gas pressures are in atmospheres, and concentrations of solutions are in molarity, M.

∆Go and Equilibrium A large negative value of ∆Go indicates that the forward reaction or process is spontaneous. That is, there is a large driving force for the forward reaction. This also means that the equilibrium constant for the reaction will be large.

∆Go and Equilibrium A large positive value of ∆Go indicates that the reverse reaction or process is spontaneous. That is, there is a large driving force for the reverse reaction. This also means that the equilibrium constant for the reaction will be small. When a reaction or process is at equilibrium, ∆Go = zero.

∆Go and Equilibrium

∆Go and Equilibrium ∆G = ∆Go + RT lnQ At equilibrium, ∆G is equal to zero, and Q = K. 0 = ∆Go + RT lnK ∆Go = - RT lnK

C (s, diamond) ↔ C (s, graphite) ∆Go and Equilibrium Calculate, ∆Go and K at 25oC for: C (s, diamond) ↔ C (s, graphite)

∆Go and Equilibrium Calculate, ∆Go and K at 25oC for: C (s, diamond) ↔ C (s, graphite) ∆Go = (1 mol) ∆Gof (graphite) - (1 mol) ∆Gof (diamond) = 0 -(1 mol)(2.900 kJ/mol) = -2.900 kJ The reaction is spontaneous at 25oC.

∆Go and Equilibrium Calculate, ∆Go and K at 25oC for: C (s, diamond) ↔ C (s, graphite) ∆Go = -2.900 kJ ∆Go = -2.900 kJ = -RT ln K -2.900 kJ = -(8.314J/mol-K) (298.2K)ln K ln K = 1.170 K= e1.170 = 3.22

C (s, diamond) ↔ C (s, graphite) The negative value of ∆Go and the equilibrium constant >1 suggest that diamonds can spontaneously react to form graphite. Although the reaction is thermodynamically favored, the rate constant is extremely small due to a huge activation energy. The disruption of the bonding in the diamond to form planar sp2 hybridized carbon atoms is kinetically unfavorable.