V. Determining SpontaneityV. Determining Spontaneity

• It would be easier if we could determine spontaneity by just considering changes in the system.

• We derive an equation from the relationships we have learned and define a new quantity – the Gibbs free energy.

V. Gibbs Free Energy EquationV. Gibbs Free Energy Equation

V. Gibbs Free EnergyV. Gibbs Free Energy

• This equation calculates the change in the Gibbs free energy.

• Gibbs free energy (G) is formally defined as G = H – TS.

• From the derivation, ΔG = -TΔSuniv. Since ΔSuniv is the criterion for spontaneity, ΔG can

be used as a criterion for spontaneity.

V. Using ΔGV. Using ΔG

• The change in free energy can be calculated w/ respect to the system, and the result can be used to determine spontaneity. If ΔG is negative, the process is

spontaneous. If ΔG is positive, the process is

nonspontaneous.

V. Possible Combos in ΔGV. Possible Combos in ΔG

• There are 3 factors that determine the outcome of the sign on ΔG.

• Sometimes ΔH and ΔS work together; sometimes they don’t.

V. Sample ProblemV. Sample Problem

• The reaction C2H4(g) + H2(g) C2H6(g) has ΔH = -137.5 kJ and ΔS = -120.5 J/K. Calculate ΔG at 25 °C and determine whether the reaction is spontaneous at this temperature. Does ΔG become more negative or more positive as temperature increases?

V. Calculating ΔS°V. Calculating ΔS°rxnrxn

• Recall that the “not” symbol means standard conditions – 25 °C and 1 atm. Standard state would be the state the substance

exists as at these conditions. For a solution, [ ] must be exactly 1 M.

• The standard entropy change for a reaction (ΔS°rxn) is the change in entropy for a process in which all reactants and products are in their standard states.

V. 3V. 3rdrd Law of Thermodynamics Law of Thermodynamics

• In order to calculate changes in entropy, we need to have a reference point.

• 3rd Law of Thermodynamics: the entropy of a perfect crystal at absolute zero (0 K) is zero.

• Standard molar entropies can thus be tabulated.

V. Standard Molar EntropiesV. Standard Molar Entropies

V. Factors Influencing S°V. Factors Influencing S°

• The standard entropy of a substance is basically the amount of energy dispersed into one mole of that substance at 25 °C.

• Thus, the more “places” the compound can put the energy, the more “entropic” that compound will be.

V. S° and Physical StateV. S° and Physical State

• As mentioned before, entropy increases as a substance goes from solid to liquid, to gas. Liquid H2O has S° = 70.0 J/mole.K.

Gaseous H2O has S° = 188.8 J/mole.K.

V. S° and Molar MassV. S° and Molar Mass

• As the molar mass increases, the entropy increases.

• The “why” is outside the scope of this course, but it has to do with energy levels.

V. S° and AllotropesV. S° and Allotropes

• Recall that allotropes are different forms of the same element.

• Less constrained allotropes have more entropy than more constrained ones.

V. S° and Molecular ComplexityV. S° and Molecular Complexity

• The more complex the compound, the more places it can put energy due to its different modes of motion.

• e.g. Ar(g) vs. NO(g).

V. S° and DissolutionV. S° and Dissolution

• Dissolved ionic solids have more entropy than the nondissolved solid.

• Energy that was concentrated in the crystalline solid becomes dispersed when dissolved in a solution.

• e.g. KClO3(s) with S° = 143.1 J/mole.K vs. KClO3(aq) with S° = 265.7 J/mole.K.

V. Calculating ΔS°V. Calculating ΔS°rxnrxn

• Calculating changes in standard entropies for a reaction is done via a Hess’s Law type of calculation.

V. Rxn Free Energy ChangesV. Rxn Free Energy Changes

• So, to determine whether or not a process is spontaneous, we can calculate ΔG°rxn.

• Of course, this is the standard free energy change of a reaction.

• There are three ways to calculate ΔG°rxn, depending on what information is given.

V. Calculating ΔG°V. Calculating ΔG°rxnrxn with with

ΔH°ΔH°rxnrxn and ΔS° and ΔS°rxnrxn

• The first method involves using the Gibbs free energy equation. ΔG°rxn = ΔH°rxn - TΔS°rxn

• To use this method, typically need to do Hess’s Law type calculations for enthalpy and entropy.

V. Sample ProblemV. Sample Problem

• Determine whether the reaction NO(g) + ½ O2(g) NO2(g) is spontaneous at 25 °C. Note that the enthalpies of formation for NO(g) and NO2(g) are 91.3 kJ/mole and 33.2 kJ/mole, respectively and the standard entropies for NO(g), O2(g), and NO2(g) are 210.8 J/mole.K, 205.2 J/mole.K, and 240.1 J/mole.K, respectively.

V. Calculating ΔG°V. Calculating ΔG°rxnrxn Via Via

Hess’s LawHess’s Law• standard free energy of formation, ΔG°f:

change in free energy when 1 mole of a compound forms from its constituent elements in their standard states.

• If a table of ΔG°f’s is available, ΔG°rxn can be calculated using a Hess’s Law type of calculation.

V. Sample ProblemV. Sample Problem

• Calculate ΔG°rxn for the reaction 2CO(g) + 2NO(g) 2CO2(g) + N2(g) given that the standard free energies of formation for CO(g), NO(g), and CO2(g) are -137.2 kJ/mole, 87.6 kJ/mole, and -394.4 kJ/mole, respectively.

V. Calculating ΔG°V. Calculating ΔG°rxnrxn Stepwise Stepwise

• Just like thermochemical equations, reactions w/ associated free energies can be manipulated with the same rules applying.1) If the equation is multiplied by a factor, then

ΔGrxn is multiplied by the same factor.

2) If an equation is reversed, then ΔGrxn changes sign.

3) If a series of reactions adds up to an overall reaction, the ΔGrxn for the overall process is the sum of ΔGrxn for each step.

V. Sample ProblemV. Sample Problem

• Using the equations below, find the ΔG°rxn for the reaction N2O(g) + NO2(g) 3NO(g).

2NO(g) + O2(g) 2NO2(g) ΔG°rxn = -71.2 kJ N2(g) + O2(g) 2NO(g) ΔG°rxn = 175.2 kJ 2N2O(g) 2N2(g) + O2(g) ΔG°rxn = -207.4 kJ

V. Why Is It “Free” Energy?V. Why Is It “Free” Energy?

• The energy of a reaction available to do work is the free energy.

• The free energy is the theoretical maximum; usually, it will be less.

• The only way to get the maximum is by using an infinitesimally slow reaction known as a reversible reaction.

V. Nonstandard Free EnergiesV. Nonstandard Free Energies

• Of course, most of the time, a reaction will not be under standard conditions.

• We need a way to calculate spontaneity under a variety of conditions.

V. Sample ProblemV. Sample Problem

• The reaction 2H2S(g) + SO2(g) 3S(s,

rhombic) + 2H2O(g) has a standard free energy change of -102 kJ. Calculate ΔGrxn when the partial pressures of H2S, SO2, and H2O are 2.00 atm, 1.50 atm, and 0.0100 atm. Is the reaction more or less spontaneous under these conditions?

V. Relationship Between V. Relationship Between ΔG°ΔG°rxnrxn and K and K

• Recall that K is a measure of how far a reaction goes towards products which is another way of describing spontaneity.

• Thus, there must be a relationship between free energy and equil. constants.

• If ΔGrxn < 0, reaction is spontaneous. If ΔGrxn > 0, reaction is nonspontaneous. If ΔGrxn = 0, reaction is at equilibrium.

V. Relationship Between V. Relationship Between ΔG°ΔG°rxnrxn and K and K

V. Temp. Dependence of KV. Temp. Dependence of K

• Previously, we discovered that equilibrium constants depend on temperature.

• With the relationship between standard free energy and the equilibrium constant, we can see how temperature is affected by temperature.

V. Temp. Dependence of KV. Temp. Dependence of K