# Chapter 17 Free Energy and Thermodynamics

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Chapter 17 Free Energy and Thermodynamics
Chemistry: A Molecular Approach, 2nd Ed. Nivaldo Tro Chapter 17 Free Energy and Thermodynamics Roy Kennedy Massachusetts Bay Community College Wellesley Hills, MA

First Law of Thermodynamics
You can’t win! First Law of Thermodynamics: Energy cannot be created or destroyed the total energy of the universe cannot change though you can transfer it from one place to another DEuniverse = 0 = DEsystem + DEsurroundings Tro: Chemistry: A Molecular Approach, 2/e

First Law of Thermodynamics
Conservation of Energy For an exothermic reaction, “lost” heat from the system goes into the surroundings Two ways energy is “lost” from a system converted to heat, q used to do work, w Energy conservation requires that the energy change in the system equal the heat released + work done DE = q + w DE = DH + PDV DE is a state function internal energy change independent of how done Tro: Chemistry: A Molecular Approach, 2/e

The Energy Tax You can’t break even!
To recharge a battery with 100 kJ of useful energy will require more than 100 kJ because of the Second Law of Thermodynamics Every energy transition results in a “loss” of energy an “Energy Tax” demanded by nature and conversion of energy to heat which is “lost” by heating up the surroundings Tro: Chemistry: A Molecular Approach, 2/e

fewer steps generally results in a lower total heat tax
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Thermodynamics and Spontaneity
Thermodynamics predicts whether a process will occur under the given conditions processes that will occur are called spontaneous nonspontaneous processes require energy input to go Spontaneity is determined by comparing the chemical potential energy of the system before the reaction with the free energy of the system after the reaction if the system after reaction has less potential energy than before the reaction, the reaction is thermodynamically favorable. Spontaneity ≠ fast or slow Tro: Chemistry: A Molecular Approach, 2/e

Comparing Potential Energy
The direction of spontaneity can be determined by comparing the potential energy of the system at the start and the end Tro: Chemistry: A Molecular Approach, 2/e

Reversibility of Process
Any spontaneous process is irreversible because there is a net release of energy when it proceeds in that direction it will proceed in only one direction A reversible process will proceed back and forth between the two end conditions any reversible process is at equilibrium results in no change in free energy If a process is spontaneous in one direction, it must be nonspontaneous in the opposite direction Tro: Chemistry: A Molecular Approach, 2/e

Thermodynamics vs. Kinetics
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Diamond → Graphite Graphite is more stable than diamond, so the conversion of diamond into graphite is spontaneous – but don’t worry, it’s so slow that your ring won’t turn into pencil lead in your lifetime (or through many of your generations) Tro: Chemistry: A Molecular Approach, 2/e

Spontaneous Processes
Spontaneous processes occur because they release energy from the system Most spontaneous processes proceed from a system of higher potential energy to a system at lower potential energy exothermic But there are some spontaneous processes that proceed from a system of lower potential energy to a system at higher potential energy endothermic How can something absorb potential energy, yet have a net release of energy? Tro: Chemistry: A Molecular Approach, 2/e

Melting Ice Melting is an Endothermic process, yet ice will spontaneously melt above 0 °C. When a solid melts, the particles have more freedom of movement. More freedom of motion increases the randomness of the system. When systems become more random, energy is released. We call this energy, entropy Tro: Chemistry: A Molecular Approach, 2/e

Factors Affecting Whether a Reaction Is Spontaneous
There are two factors that determine whether a reaction is spontaneous. They are the enthalpy change and the entropy change of the system The enthalpy change, DH, is the difference in the sum of the internal energy and PV work energy of the reactants to the products The entropy change, DS, is the difference in randomness of the reactants compared to the products Tro: Chemistry: A Molecular Approach, 2/e

Enthalpy Change DH generally measured in kJ/mol
Stronger bonds = more stable molecules A reaction is generally exothermic if the bonds in the products are stronger than the bonds in the reactants exothermic = energy released, DH is negative A reaction is generally endothermic if the bonds in the products are weaker than the bonds in the reactants endothermic = energy absorbed, DH is positive The enthalpy change is favorable for exothermic reactions and unfavorable for endothermic reactions Tro: Chemistry: A Molecular Approach, 2/e

Entropy Entropy is a thermodynamic function that increases as the number of energetically equivalent ways of arranging the components increases, S S generally J/mol S = k ln W k = Boltzmann Constant = 1.38 x 10−23 J/K W is the number of energetically equivalent ways a system can exist unitless Random systems require less energy than ordered systems Tro: Chemistry: A Molecular Approach, 2/e

W These are energetically equivalent states for the expansion of a gas. It doesn’t matter, in terms of potential energy, whether the molecules are all in one flask, or evenly distributed But one of these states is more probable than the other two Tro: Chemistry: A Molecular Approach, 2/e

Macrostates → Microstates
These microstates all have the same macrostate So there are six different particle arrangements that result in the same macrostate This macrostate can be achieved through several different arrangements of the particles Tro: Chemistry: A Molecular Approach, 2/e

Macrostates and Probability
There is only one possible arrangement that gives State A and one that gives State B There are six possible arrangements that give State C The macrostate with the highest entropy also has the greatest dispersal of energy Therefore State C has higher entropy than either State A or State B There is six times the probability of having the State C macrostate than either State A or State B Tro: Chemistry: A Molecular Approach, 2/e

Changes in Entropy, DS DS = Sfinal − Sinitial
Entropy change is favorable when the result is a more random system DS is positive Some changes that increase the entropy are reactions whose products are in a more random state solid more ordered than liquid more ordered than gas reactions that have larger numbers of product molecules than reactant molecules increase in temperature solids dissociating into ions upon dissolving Tro: Chemistry: A Molecular Approach, 2/e

Increases in Entropy Tro: Chemistry: A Molecular Approach, 2/e

DS For a process where the final condition is more random than the initial condition, DSsystem is positive and the entropy change is favorable for the process to be spontaneous For a process where the final condition is more orderly than the initial condition, DSsystem is negative and the entropy change is unfavorable for the process to be spontaneous DSsystem = DSreaction = Sn(S°products) − Sn(S°reactants) Tro: Chemistry: A Molecular Approach, 2/e

Entropy Change in State Change
When materials change state, the number of macrostates it can have changes as well the more degrees of freedom the molecules have, the more macrostates are possible solids have fewer macrostates than liquids, which have fewer macrostates than gases Tro: Chemistry: A Molecular Approach, 2/e

Entropy Change and State Change
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Practice – Predict whether DSsystem is + or − for each of the following
A hot beaker burning your fingers Water vapor condensing Separation of oil and vinegar salad dressing Dissolving sugar in tea 2 PbO2(s)  2 PbO(s) + O2(g) 2 NH3(g)  N2(g) + 3 H2(g) Ag+(aq) + Cl−(aq)  AgCl(s) DS is + DS is − DS is − DS is + DS is + DS is + DS is − Tro: Chemistry: A Molecular Approach, 2/e

The 2nd Law of Thermodynamics
The 2nd Law of Thermodynamics says that the total entropy change of the universe must be positive for a process to be spontaneous for reversible process DSuniv = 0 for irreversible (spontaneous) process DSuniv > 0 DSuniverse = DSsystem + DSsurroundings If the entropy of the system decreases, then the entropy of the surroundings must increase by a larger amount when DSsystem is negative, DSsurroundings must be positive and big for a spontaneous process Tro: Chemistry: A Molecular Approach, 2/e

Heat Flow, Entropy, and the 2nd Law
When ice is placed in water, heat flows from the water into the ice According to the 2nd Law, heat must flow from water to ice because it results in more dispersal of heat. The entropy of the universe increases. Tro: Chemistry: A Molecular Approach, 2/e

Heat Transfer and Changes in Entropy of the Surroundings
The 2nd Law demands that the entropy of the universe increase for a spontaneous process Yet processes like water vapor condensing are spontaneous, even though the water vapor is more random than the liquid water If a process is spontaneous, yet the entropy change of the process is unfavorable, there must have been a large increase in the entropy of the surroundings The entropy increase must come from heat released by the system – the process must be exothermic! Tro: Chemistry: A Molecular Approach, 2/e

Entropy Change in the System and Surroundings
When the entropy change in system is unfavorable (negative), the entropy change in the surroundings must be favorable (positive), and large to allow the process to be spontaneous Tro: Chemistry: A Molecular Approach, 2/e

Heat Exchange and DSsurroundings
When a system process is exothermic, it adds heat to the surroundings, increasing the entropy of the surroundings When a system process is endothermic, it takes heat from the surroundings, decreasing the entropy of the surroundings The amount the entropy of the surroundings changes depends on its original temperature the higher the original temperature, the less effect addition or removal of heat has Tro: Chemistry: A Molecular Approach, 2/e

Temperature Dependence of DSsurroundings
When heat is added to surroundings that are cool it has more of an effect on the entropy than it would have if the surroundings were already hot Water freezes spontaneously below 0 °C because the heat released on freezing increases the entropy of the surroundings enough to make DS positive above 0 °C the increase in entropy of the surroundings is insufficient to make DS positive Tro: Chemistry: A Molecular Approach, 2/e

Quantifying Entropy Changes in Surroundings
The entropy change in the surroundings is proportional to the amount of heat gained or lost qsurroundings = −qsystem The entropy change in the surroundings is also inversely proportional to its temperature At constant pressure and temperature, the overall relationship is Tro: Chemistry: A Molecular Approach, 2/e

Example 17.2a: Calculate the entropy change of the surroundings at 25ºC for the reaction below C3H8(g) + 5 O2(g)  3 CO2(g) + 4 H2O(g) DHrxn = −2044 kJ Given: Find: DHsystem = −2044 kJ, T = 25 ºC = 298 K DSsurroundings, J/K Conceptual Plan: Relationships: DS T, DH Solution: Check: combustion is largely exothermic, so the entropy of the surroundings should increase significantly Tro: Chemistry: A Molecular Approach, 2/e

Practice – The reaction below has DHrxn = +66. 4 kJ at 25 °C
Practice – The reaction below has DHrxn = kJ at 25 °C. (a) Determine the Dssurroundings, (b) the sign of DSsystem, and (c) whether the process is spontaneous 2 O2(g) + N2(g)  2 NO2(g) Tro: Chemistry: A Molecular Approach, 2/e

DSsurr, J/K, DSreact + or −, DSuniverse + or −
Practice – The reaction 2 O2(g) + N2(g)  2 NO2(g) has DHrxn = kJ at 25 °C. Calculate the entropy change of the surroundings. Determine the sign of the entropy change in the system and whether the reaction is spontaneous. Given: Find: DHsys = kJ, T = 25 ºC = 298 K DSsurr, J/K, DSreact + or −, DSuniverse + or − Conceptual Plan: Relationships: DS T, DH Solution: the major difference is that there are fewer product molecules than reactant molecules, so the DSreaction is unfavorable and (−) both DSsurroundings and DSreaction are (−), DSuniverse is (−) and the process is nonspontaneous because DSsurroundings is (−) it is unfavorable Tro: Chemistry: A Molecular Approach, 2/e 34

Gibbs Free Energy and Spontaneity
It can be shown that −TDSuniv = DHsys−TDSsys The Gibbs Free Energy, G, is the maximum amount of work energy that can be released to the surroundings by a system for a constant temperature and pressure system the Gibbs Free Energy is often called the Chemical Potential because it is analogous to the storing of energy in a mechanical system DGsys = DHsys−TDSsys Because DSuniv determines if a process is spontaneous, DG also determines spontaneity DSuniv is + when spontaneous, so DG is − Tro: Chemistry: A Molecular Approach, 2/e

Gibbs Free Energy, DG A process will be spontaneous when DG is negative DG will be negative when DH is negative and DS is positive exothermic and more random DH is negative and large and DS is negative but small DH is positive but small and DS is positive and large or high temperature DG will be positive when DH is + and DS is − never spontaneous at any temperature When DG = 0 the reaction is at equilibrium Tro: Chemistry: A Molecular Approach, 2/e

DG, DH, and DS Tro: Chemistry: A Molecular Approach, 2/e

Free Energy Change and Spontaneity
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Example 17.3a: The reaction CCl4(g)  C(s, graphite) + 2 Cl2(g) has DH = kJ and DS = J/K at 25 °C. Calculate DG and determine if it is spontaneous. Given: Find: DH = kJ, DS = J/K, T = 298 K DG, kJ Conceptual Plan: Relationships: DG T, DH, DS Solution: Because DG is +, the reaction is not spontaneous at this temperature. To make it spontaneous, we need to increase the temperature. Answer: Tro: Chemistry: A Molecular Approach, 2/e

Practice – The reaction Al(s) + Fe2O3(s)  Fe(s) + Al2O3(s) has DH = −847.6 kJ and DS = −41.3 J/K at 25 °C. Calculate DG and determine if it is spontaneous. Tro: Chemistry: A Molecular Approach, 2/e

Practice – The reaction Al(s) + Fe2O3(s)  Fe(s) + Al2O3(s) has DH = −847.6 kJ and DS = −41.3 J/K at 25 °C. Calculate DG and determine if it is spontaneous. Given: Find: DH = −847.6 kJ and DS = −41.3 J/K, T = 298 K DG, kJ Conceptual Plan: Relationships: DG T, DH, DS Solution: Because DG is −, the reaction is spontaneous at this temperature. To make it nonspontaneous, we need to increase the temperature. Answer: Tro: Chemistry: A Molecular Approach, 2/e 41

Example 17.3b: The reaction CCl4(g)  C(s, graphite) + 2 Cl2(g) has DH = kJ and DS = J/K. Calculate the minimum temperature it will be spontaneous. Given: Find: DH = kJ, DS = J/K, DG < 0 T, K Conceptual Plan: Relationships: T DG, DH, DS Solution: the temperature must be higher than 673K for the reaction to be spontaneous Answer: Tro: Chemistry: A Molecular Approach, 2/e

Practice – The reaction Al(s) + Fe2O3(s)  Fe(s) + Al2O3(s) has DH = −847.6 kJ and DS = −41.3 J/K. Calculate the maximum temperature it will be spontaneous. Tro: Chemistry: A Molecular Approach, 2/e

DH = −847.6 kJ and DS = −41.3 J/K, DG > 0 T, C
Practice – The reaction Al(s) + Fe2O3(s)  Fe(s) + Al2O3(s) has DH = −847.6 kJ and DS = −41.3 J/K at 25 °C. Determine the maximum temperature it is spontaneous. Given: Find: DH = −847.6 kJ and DS = −41.3 J/K, DG > 0 T, C Conceptual Plan: Relationships: T DG, DH, DS Solution: 2048 − 273 = 1775 C Answer: any temperature above 1775 C will make the reaction nonspontaneous Tro: Chemistry: A Molecular Approach, 2/e 44

Standard Conditions The standard state is the state of a material at a defined set of conditions Gas = pure gas at exactly 1 atm pressure Solid or Liquid = pure solid or liquid in its most stable form at exactly 1 atm pressure and temperature of interest usually 25 °C Solution = substance in a solution with concentration 1 M Tro: Chemistry: A Molecular Approach, 2/e

The 3rd Law of Thermodynamics: Absolute Entropy
The absolute entropy of a substance is the amount of energy it has due to dispersion of energy through its particles The 3rd Law states that for a perfect crystal at absolute zero, the absolute entropy = 0 J/mol∙K therefore, every substance that is not a perfect crystal at absolute zero has some energy from entropy therefore, the absolute entropy of substances is always + Tro: Chemistry: A Molecular Approach, 2/e

Standard Absolute Entropies
Extensive Entropies for 1 mole of a substance at 298 K for a particular state, a particular allotrope, particular molecular complexity, a particular molar mass, and a particular degree of dissolution Tro: Chemistry: A Molecular Approach, 2/e

Standard Absolute Entropies
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Relative Standard Entropies: States
The gas state has a larger entropy than the liquid state at a particular temperature The liquid state has a larger entropy than the solid state at a particular temperature Substance S°, (J/mol∙K) H2O (l) 70.0 H2O (g) 188.8 Tro: Chemistry: A Molecular Approach, 2/e

Relative Standard Entropies: Molar Mass
The larger the molar mass, the larger the entropy Available energy states more closely spaced, allowing more dispersal of energy through the states Tro: Chemistry: A Molecular Approach, 2/e

Relative Standard Entropies: Allotropes
The less constrained the structure of an allotrope is, the larger its entropy The fact that the layers in graphite are not bonded together makes it less constrained Tro: Chemistry: A Molecular Approach, 2/e

Relative Standard Entropies: Molecular Complexity
Larger, more complex molecules generally have larger entropy More available energy states, allowing more dispersal of energy through the states Substance Molar Mass S°, (J/mol∙K) CO (g) 28.01 197.7 C2H4 (g) 28.05 219.3 Substance Molar Mass S°, (J/mol∙K) Ar (g) 39.948 154.8 NO (g) 30.006 210.8 Tro: Chemistry: A Molecular Approach, 2/e

Relative Standard Entropies Dissolution
Dissolved solids generally have larger entropy Distributing particles throughout the mixture Substance S°, (J/mol∙K) KClO3(s) 143.1 KClO3(aq) 265.7 Tro: Chemistry: A Molecular Approach, 2/e

The Standard Entropy Change, DS
The standard entropy change is the difference in absolute entropy between the reactants and products under standard conditions DSºreaction = (∑npSºproducts) − (∑nrSºreactants) remember: though the standard enthalpy of formation, DHf°, of an element is 0 kJ/mol, the absolute entropy at 25 °C, S°, is always positive Tro: Chemistry: A Molecular Approach, 2/e

Substance S, J/molK NH3(g) 192.8 O2(g) 205.2 NO(g) 210.8 H2O(g) 188.8 Example 17.4: Calculate DS for the reaction 4 NH3(g) + 5 O2(g)  4 NO(g) + 6 H2O(g) Given: Find: standard entropies from Appendix IIB DS, J/K Conceptual Plan: Relationships: DS SNH3, SO2, SNO, SH2O Sol’n: Check: DS is +, as you would expect for a reaction with more gas product molecules than reactant molecules Tro: Chemistry: A Molecular Approach, 2/e

Practice – Calculate the DS for the reaction 2 H2(g) + O2(g)  2 H2O(g)
Substance S, J/molK H2(g) 130.7 O2(g) 205.2 H2O(g) 188.8 Tro: Chemistry: A Molecular Approach, 2/e

Substance S, J/molK H2(g) 130.6 O2(g) 205.2 H2O(g) 188.8 Example 17.4: Calculate DS for the reaction 2 H2(g) + O2(g)  2 H2O(g) Given: Find: standard entropies from Appendix IIB DS, J/K Conceptual Plan: Relationships: DS SH2, SO2, SH2O Solution: Check: DS is −, as you would expect for a reaction with more gas reactant molecules than product molecules Tro: Chemistry: A Molecular Approach, 2/e

Calculating DG At 25 C DGoreaction = SnDGof(products) - SnDGof(reactants) At temperatures other than 25 C assuming the change in DHoreaction and DSoreaction is negligible DGreaction = DHreaction – TDSreaction or DGtotal = DGreaction 1 + DGreaction Tro: Chemistry: A Molecular Approach, 2/e

Example 17. 6: The reaction SO2(g) + ½ O2(g)  SO3(g) has DH = −98
Example 17.6: The reaction SO2(g) + ½ O2(g)  SO3(g) has DH = −98.9 kJ and DS = −94.0 J/K at 25 °C. Calculate DG at 125 C and determine if it is more or less spontaneous than at 25 °C with DG° = −70.9 kJ/mol SO3. Given: Find: DH = −98.9 kJ, DS = −94.0 J/K, T = 398 K DG, kJ Conceptual Plan: Relationships: DG T, DH, DS Solution: because DG is −, the reaction is spontaneous at this temperature, though less so than at 25 C Answer: Tro: Chemistry: A Molecular Approach, 2/e

Practice – Determine the free energy change in the following reaction at 298 K 2 H2O(g) + O2(g)  2 H2O2(g) Substance DH, kJ/mol S, J/mol H2O2(g) −136.3 232.7 O2(g) 205.2 H2O(g) −241.8 188.8 Tro: Chemistry: A Molecular Approach, 2/e

Practice – Determine the free energy change in the following reaction at 298 K 2 H2O(g) + O2(g)  2 H2O2(g) Given: Find: standard energies from Appendix IIB, T = 298 K DG, kJ DH = kJ, DS = −117.4 J/K, T = 298 K DG, kJ Conceptual Plan: Relationships: DG DHf° Sº of prod & react DH, DSº Solution: Tro: Chemistry: A Molecular Approach, 2/e

Standard Free Energies of Formation
The free energy of formation (DGf°) is the change in free energy when 1 mol of a compound forms from its constituent elements in their standard states The free energy of formation of pure elements in their standard states is zero Tro: Chemistry: A Molecular Approach, 2/e

Substance DGf°, kJ/mol CH4(g) −50.5 O2(g) 0.0 CO2(g) −394.4 H2O(g) −228.6 O3(g) +163.2 Example 17.7: Calculate DG at 25 C for the reaction CH4(g) + 8 O2(g)  CO2(g) + 2 H2O(g) + 4 O3(g) Given: Find: standard free energies of formation from Appendix IIB DG, kJ Conceptual Plan: Relationships: DG DGf° of prod & react Solution: Tro: Chemistry: A Molecular Approach, 2/e

Practice – Determine the free energy change in the following reaction at 298 K 2 H2O(g) + O2(g)  2 H2O2(g) Substance DG, kJ/mol H2O2(g) −105.6 O2(g) H2O(g) −228.6 Tro: Chemistry: A Molecular Approach, 2/e

Practice – Determine the free energy change in the following reaction at 298 K 2 H2O(g) + O2(g)  2 H2O2(g) Given: Find: standard free energies of formation from Appendix IIB DG, kJ Conceptual Plan: Relationships: DG DGf° of prod & react Solution: Tro: Chemistry: A Molecular Approach, 2/e

DG Relationships If a reaction can be expressed as a series of reactions, the sum of the DG values of the individual reaction is the DG of the total reaction DG is a state function If a reaction is reversed, the sign of its DG value reverses If the amount of materials is multiplied by a factor, the value of the DG is multiplied by the same factor the value of DG of a reaction is extensive Tro: Chemistry: A Molecular Approach, 2/e

Example 17.8: Find DGºrxn for the following reaction using the given equations 3 C(s) + 4 H2(g)  C3H8(g) Given: Find: C3H8(g) + 5 O2(g)  3 CO2(g) + 4 H2O(g) DGº = −2074 kJ C(s) + O2(g)  CO2(g) DGº = −394.4 kJ 2 H2(g) + O2(g)  2 H2O(g) DGº = −457.1 kJ DGº of 3 C(s) + 4 H2(g)  C3H8(g) Rel: Manipulate the given reactions so they add up to the reaction you wish to find. The sum of the DGº’s is the DGº of the reaction you want to find. Solution: 3 CO2(g) + 4 H2O(g)  C3H8(g) + 5 O2(g) DGº = kJ 3 C(s) + 3 O2(g)  3 CO2(g) DGº = − kJ 4 H2(g) + 2 O2(g)  4 H2O(g) DGº = −914.2 kJ 3 C(s) + 4 H2(g)  C3H8(g) DGº = −23 kJ Given: Find: C3H8(g) + 5 O2(g)  3 CO2(g) + 4 H2O(g) DGº = −2074 kJ C(s) + O2(g)  CO2(g) DGº = −394.4 kJ 2 H2(g) + O2(g)  2 H2O(g) DGº = −457.1 kJ DGº of 3 C(s) + 4 H2(g)  C3H8(g) Rel: Manipulate the given reactions so they add up to the reaction you wish to find. The sum of the DGº’s is the DGº of the reaction you want to find. Solution: 3 CO2(g) + 4 H2O(g)  C3H8(g) + 5 O2(g) DGº = kJ 3 x [C(s) + O2(g)  CO2(g) ] DGº = 3(−394.4 kJ) 2 x [2 H2(g) + O2(g)  2 H2O(g)] DGº = 2(−457.1 kJ) Tro: Chemistry: A Molecular Approach, 2/e

Practice – Determine the free energy change in the following reaction at 298 K 2 H2O(g) + O2(g)  2 H2O2(g) Substance DG, kJ/mol H2 (g) + O2(g)  H2O2(g) −105.6 2 H2 (g) + O2(g)  2 H2O(g) −457.2 Tro: Chemistry: A Molecular Approach, 2/e

Practice – Determine the free energy change in the following reaction at 298 K 2 H2O(g) + O2(g)  2 H2O2(g) Given: Find: H2(g) + O2(g)  H2O2(g) DGº = −105.6 kJ 2 H2(g) + O2(g)  2 H2O(g) DGº = −457.2 kJ DGº of 2 H2O(g) + O2(g)  2 H2O2(g) Rel: Manipulate the given reactions so they add up to the reaction you wish to find. The sum of the DGº’s is the DGº of the reaction you want to find. Solution: 2 H2(s) + 2 O2(g)  2 H2O2(g) DGº = −211.2 kJ 2 H2O(g)  2 H2(g) + O2(g) DGº = kJ 2 H2O(g) + O2(g)  2 H2O2(g) DGº = kJ Tro: Chemistry: A Molecular Approach, 2/e

The free energy is the maximum amount of energy released from a system that is available to do work on the surroundings For many exothermic reactions, some of the heat released due to the enthalpy change goes into increasing the entropy of the surroundings, so it is not available to do work And even some of this free energy is generally lost to heating up the surroundings Tro: Chemistry: A Molecular Approach, 2/e

Free Energy of an Exothermic Reaction
C(s, graphite) + 2 H2(g) → CH4(g) DH°rxn = −74.6 kJ = exothermic DS°rxn = −80.8 J/K = unfavorable DG°rxn = −50.5 kJ = spontaneous DG° is less than DH° because some of the released heat energy is lost to increase the entropy of the surroundings Tro: Chemistry: A Molecular Approach, 2/e

Free Energy and Reversible Reactions
The change in free energy is a theoretical limit as to the amount of work that can be done If the reaction achieves its theoretical limit, it is a reversible reaction Tro: Chemistry: A Molecular Approach, 2/e

Real Reactions In a real reaction, some of the free energy is “lost” as heat if not most Therefore, real reactions are irreversible Tro: Chemistry: A Molecular Approach, 2/e

DG under Nonstandard Conditions
DG = DG only when the reactants and products are in their standard states their normal state at that temperature partial pressure of gas = 1 atm concentration = 1 M Under nonstandard conditions, DG = DG + RTlnQ Q is the reaction quotient At equilibrium DG = 0 DG = −RTlnK Tro: Chemistry: A Molecular Approach, 2/e

Tro: Chemistry: A Molecular Approach, 2/e

non-standard conditions, DGº, T DG, kJ
Example 17.9: Calculate DG at 298 K for the reaction under the given conditions 2 NO(g) + O2(g)  2 NO2(g) DGº = −71.2 kJ Substance P, atm NO(g) 0.100 O2(g) NO2(g) 2.00 Given: Find: non-standard conditions, DGº, T DG, kJ Conceptual Plan: Relationships: DG DG, PNO, PO2, PNO2 Solution: Tro: Chemistry: A Molecular Approach, 2/e

Practice – Calculate DGrxn for the given reaction at 700 K under the given conditions N2(g) + 3 H2(g) ® 2 NH3(g) DGº = kJ Substance P, atm N2(g) 33 H2(g) 99 NH3(g) 2.0 Tro: Chemistry: A Molecular Approach, 2/e

non-standard conditions, DGº, T DG, kJ
Substance P, atm N2(g) 33 H2(g) 99 NH3(g) 2.0 Practice – Calculate DGrxn for the given reaction at 700 K under the given conditions N2(g) + 3 H2(g) ® 2 NH3(g) DGº = kJ Given: Find: non-standard conditions, DGº, T DG, kJ Conceptual Plan: Relationships: DG DG, PNO, PO2, PNO2 Solution: Tro: Chemistry: A Molecular Approach, 2/e

DGº and K Because DGrxn = 0 at equilibrium, then DGº = −RTln(K)
When K < 1, DGº is + and the reaction is spontaneous in the reverse direction under standard conditions nothing will happen if there are no products yet! When K > 1, DGº is − and the reaction is spontaneous in the forward direction under standard conditions When K = 1, DGº is 0 and the reaction is at equilibrium under standard conditions Tro: Chemistry: A Molecular Approach, 2/e

Tro: Chemistry: A Molecular Approach, 2/e

Substance DGf°, kJ/mol N2O4(g) +99.8 NO2(g) +51.3
Example 17.10: Calculate K at 298 K for the reaction N2O4(g)  2 NO2(g) Given: Find: standard free energies of formation from Appendix IIB K Conceptual Plan: Relationships: DG DGf of prod & react K DGº = −RTln(K) Solution: Tro: Chemistry: A Molecular Approach, 2/e

Practice – Estimate the equilibrium constant for the given reaction at 700 K N2(g) + 3 H2(g) ® 2 NH3(g) DGº = kJ Tro: Chemistry: A Molecular Approach, 2/e

Practice – Estimate the Equilibrium Constant for the given reaction at 700 K N2(g) + 3 H2(g) ® 2 NH3(g) DGº = kJ Given: Find: DGº K Conceptual Plan: Relationships: DG K DGº = −RTln(K) Solution: Tro: Chemistry: A Molecular Approach, 2/e

Why Is the Equilibrium Constant Temperature Dependent?
Combining these two equations DG° = DH° − TDS° DG° = −RTln(K) It can be shown that This equation is in the form y = mx + b The graph of ln(K) versus inverse T is a straight line with slope and y-intercept Tro: Chemistry: A Molecular Approach, 2/e