Chapter 16 Spontaneity, Entropy, and Free Energy

Home , Entropy, Gibbs free energy, Spontaneous process

SpontaneousSpontaneous ProcessesProcesses andand Chapter 16 EntropyEntropy •Thermodynamics lets us predict whether a Spontaneity, process will occur but gives no information Entropy, about the amount of time required for the process. and Free Energy •A spontaneous process is one that occurs without outside intervention.

Figure 16.2: The rate of a reaction depends on the pathway First Law of Thermodynamics from reactants to products; • The First Law of Thermodynamics states this is the domain of that the energy of the universe is constant. kinetics. Thermodynamics Ea While energy may change in form (heat, tells us whether a reaction is spontaneous based only work, etc.) and be exchanged between the on the properties of the system & surroundings - the total energy reactants and products. remains constant. To describe the system: The predictions of thermodynamics do not Work done by the system is negative. require knowledge of the Work done on the system is positive. pathway between reactants and products. Heat evolved by the system is negative. Heat absorbed by the system is positive.

Enthalpy and Enthalpy Change Enthalpy and Enthalpy Change

• In Chapter 6, we tentatively defined • In Chapter 6, we tentatively defined enthalpy in terms of the relationship of ∆H enthalpy in terms of the relationship of ∆H to the heat at constant pressure. to the heat at constant pressure. • This means that at a given temperature and • In practice, we measure certain heats of pressure, a given amount of a substance has a reactions and use them to tabulate enthalpies of o definite enthalpy. formation, ∆H f. • Therefore, if you know the enthalpies of • Standard enthalpies of formation for selected substances, you can calculate the change in compounds are listed in Appendix 4. enthalpy, ∆H , for a reaction.

1 Spontaneous Processes and Enthalpy and Enthalpy Change Entropy • In Chapter 6, we tentatively defined •A spontaneous process is a physical or enthalpy in terms of the relationship of ∆H chemical change that occurs by itself . to the heat at constant pressure. • Examples include: • The standard enthalpy change for a reaction is A rock at the top of a hill rolls down.

o o o Heat flows from a hot object to a cold one. ∆H = ∑n∆Hf (products) − ∑m∆Hf (reactants) An iron object rusts in moist air. • These processes occur without requiring an outside force and continue until equilibrium is reached.

Entropy and the Second Law of Thermodynamics Examples of a • The second law of thermodynamics spontaneous and addresses questions about spontaneity in nonspontaneous terms of a quantity called entropy. process. • Entropy, S , is a thermodynamic quantity that is a measure of the randomness or disorder or the “available arrangements” for the system or surroundings. • The SI unit of entropy is joules per Kelvin (J/K) and, like enthalpy, is a state function.

TheThe SecondSecond LawLaw ofof Figure 16.3: The expansion of an ThermodynamicsThermodynamics ideal gas into an evacuated bulb.

• . . . in any spontaneous process there is always an increase in the entropy of the universe.

∆Suniv > 0

or )Ssystem + )Ssurroundings > 0 – for a spontaneous process.

2 For two molecules, Figure 16.4: Possible arrangements (states) A & B, there are of four molecules four microstates in a two-bulbed flask.

Nature spontaneously proceeds towards the states that have the highest probabilities of existing.

PositionalPositional EntropyEntropy Entropy and the Second Law of •A gas expands into a vacuum because the Thermodynamics expanded state has the highest positional • The second law of thermodynamics states probability of states available to the system. that the total entropy of the universe always increases for a spontaneous process. •Therefore,

• The net change in entropy of the system, ∆S , •Ssolid < Sliquid << Sgas equals the sum of the entropy created during Solid: Only a few Gas: Many allowed the spontaneous process and the change in “allowed” positions, positions, molecules energy associated with the heat flow. molecules or atoms are far apart. close together

3 Entropy and the Some examples of entropy changes: Second Law of Thermodynamics Does entropy of the system increase or decrease for the following? So for any process: )Suniverse > 0, process is spontaneous 2 H (g) + O (g) º 2 H O (g) at constant pressure and 25oC )Suniverse = 0, process tends not to occur, 2 2 2 at equilibrium Decreases – simple molecules form more complex molecule, )Suniverse < 0, reverse process occurs So )Ssystem is - spontaneously

Na (s) + heat º Na (l) at the m.p. temperature of Na We can determine )Ssystem – How can we determine )Ssurroundings? )S determined primarily by heat flow between system Increases – atoms of Na have more “available positions” in surroundings the liquid state. )S is + & surroundings. If heat flows into the surroundings (i.e., when system a reaction is exothermic) the random motions of the molecules

H2O in the surroundings increase. Thus, the entropy of the + - o NaCl (s) º Na (aq) + Cl (aq) for 10 g NaCl in 1L H2O at 25 C surroundings increases. Increases – ions formed from NaCl are more simple in

structure and have more available position. )Ssystem is +

Entropy and the Second Law of Entropy and the Second Law of Thermodynamics Thermodynamics • The second law of thermodynamics states • The second law of thermodynamics states that the total entropy of the universe always that the total entropy of the universe always increases for a spontaneous process. increases for a spontaneous process.

• We can say that for the surroundings • At constant T and P, q ∆ H sys ∆ S surr = ∆ S surr = − T T Heat flows into the surroundings during exothermic reactions and out of the surroundings for endothermic reactions

Entropy and the Second Law of Thermodynamics

The impact of the transfer of a quantity of heat energy to the surroundings will be greater when the temperature is low

• For a given reaction, the sign of )Ssurr depends on whether )Hsys is + or -. The heat energy transferred to the surroundings will have the opposite sign!

• The magnitude of )Ssurr will depend on the temperature as well as the magnitude of )Hsys.

4 Entropy Change for a Phase Entropy Change for a Phase Transition Transition • If during a phase transition, such as ice • If during a phase transition, such as ice melting, heat is slowly absorbed by the system, melting, heat is slowly absorbed by the system, it remains near equilibrium as the ice melts. it remains near equilibrium as the ice melts. • Under these conditions, no significant amount • Other phase changes, such as vaporization of a of entropy is created. liquid, also occur under equilibrium conditions. • The entropy results entirely from the absorption • Therefore, you can use the previous equation to

of heat. Therefore, obtain the )Ssys for a phase change. For the q melting of 1 mole of ice at 0oC, the )S is ∆S = (For an equilibrium process) sys ) T Ssys = +6.03 kJ/273 K = +0.0221 kJ/(mole K)

A Problem To Consider A Problem To Consider

• The heat of vaporization, ∆Hvap of carbon • The heat of vaporization, ∆Hvap of carbon o o tetrachloride, CCl4, at 25 C is 43.0 kJ/mol. If 1 mol tetrachloride, CCl4, at 25 C is 43.0 kJ/mol. If 1 mol of liquid CCl4 has an entropy of 214 J/K, what is the of liquid CCl4 has an entropy of 214 J/K, what is the entropy of 1 mol of the vapor at this temperature? entropy of 1 mol of the vapor at this temperature?

• When liquid CCl4 evaporates, it absorbs heat: • In other words, 1 mol of CCl4 increases in 3 o ∆Hvap = 43.0 kJ/mol (43.0 x 10 J/mol) at 25 C, entropy by 144 J/K when it vaporizes. or 298 K. The entropy change, ∆S, is • The entropy of 1 mol of vapor equals the entropy 3 ∆Hvap 43.0×10 J / mol of 1 mol of liquid (214 J/K) plus 144 J/K. ∆Ssys = = =144 J/(mol⋅ K) T 298 K Svap = Entropy of vapor = (214 + 144)J/K = 358 J/(mol ⋅K) Since )Ssys = Svap -Sliq

Figure 16.5: (a) A perfect crystal of hydrogen chloride at 0 K. Standard Entropies and the (b) As the temperature rises above 0 K, lattice vibrations allow some dipoles to change their orientations, producing Third Law of Thermodynamics some disorder and an increase in entropy. • The third law of thermodynamics states that a substance that is perfectly crystalline at 0 K has an entropy of zero. • When temperature is raised, however, the substance becomes more disordered as it absorbs heat. • The entropy of a substance is determined by measuring how much heat is required to change its temperature per Kelvin degree.

5 Standard entropy of methyl chloride, CH3Cl, at various temperatures.

Figure 16.6: The H2O molecule can vibrate and rotate in several ways, some of which are shown here.

Standard Entropies and the Third Law of Thermodynamics • The standard entropy of a substance or ion (Appendix 4), also called its absolute entropy, So, is the entropy value for the standard state of the species.

• Standard state implies 25 oC, 1 atm pressure, and 1 M for dissolved substances.

Standard Entropies and the Standard Entropies and the Third Law of Thermodynamics Third Law of Thermodynamics • The standard entropy of a substance or ion • The standard entropy of a substance or ion (Table 19.1), also called its absolute (Appendix 4), also called its absolute entropy, So, is the entropy value for the entropy, So, is the entropy value for the standard state of the species. standard state of the species. • Note that the elements have nonzero values, o o o unlike standard enthalpies of formation, ∆Hf , • The symbol S , rather than ∆S , is used for which by convention, are zero. standard entropies to emphasize that they originate from the third law.

6 Entropy Change for a Reaction Entropy Change for a Reaction

• You can calculate the entropy change for • You can calculate the entropy change for a reaction using a summation law, similar to a reaction using a summation law, similar to the way you obtained ∆Ho. the way you obtained ∆Ho. o o o • The entropy usually increases in the following ∆S = ∑ nS (products ) − ∑ mS (reactants) situations: • Even without knowing the values for the 1. A reaction in which a molecule is broken entropies of substances, you can sometimes into two or more smaller molecules. predict the sign of ∆So for a reaction.

Entropy Change for a Reaction Entropy Change for a Reaction

• You can calculate the entropy change for • You can calculate the entropy change for a reaction using a summation law, similar to a reaction using a summation law, similar to the way you obtained ∆Ho. the way you obtained ∆Ho. • The entropy usually increases in the following • The entropy usually increases in the following situations: situations: 2. A reaction in which there is an increase in 3. A process in which a solid changes to the moles of gases. liquid or gas or a liquid changes to gas.

A Problem To Consider A Problem To Consider

• Calculate the change in entropy, ∆So, at 25 oC for the • Calculate the change in entropy, ∆So, at 25 oC for the reaction in which urea is formed from NH3 and CO2. reaction in which urea is formed from NH3 and CO2. . . The standard entropy of NH2CONH2 is 174 J/(mol K). The standard entropy of NH2CONH2 is 174 J/(mol K). See Appendix 4 for other values. See Appendix 4 for other values.

2NH3 (g) + CO2 (g) → NH2CONH2 (aq) + H2O(l) 2NH3 (g) + CO2 (g) → NH2CONH2 (aq) + H2O(l) o • The calculation is similar to that used to obtain S : 2 x 193 214 174 70 ∆Ho from standard enthalpies of formation. • It is convenient to put the standard entropies (multiplied by their stoichiometric coefficients) below the formulas.

7 A Problem To Consider A Problem To Consider

2NH3 (g) + CO2 (g) → NH2CONH2 (aq) + H2O(l) 2NH3 (g) + CO2 (g) → NH2CONH2 (aq) + H2O(l) • We can now use the summation law to • We can now use the summation law to calculate the entropy change. calculate the entropy change. ∆S o = ∑ nS o (products) − ∑ mS o (reactants) ∆So = [(174 + 70) − (2×193 + 214)]J / K = −356 J/K

Free Energy Concept • The American physicist J. Willard Gibbs introduced the concept of free energy 3 )Ssys = (6.03 x 10 J/273 K) (sometimes called the Gibbs free energy), G, o At - 10 C, )Ssurr is negative and larger than )Ssys, so )Suniv is which is a thermodynamic quantity defined by negative and reverse process is spontaneous the equation G=H-TS, or for processes

o undergoing change: )Gsys = )Hsys -T)Ssys. At 0 C, )Ssys + )Ssurr = 0, so process is at equilibrium • This quantity gives a direct criterion for o ) ) ) At 10 C, Ssys is + and > Ssurr, so Suniv is + and the process is spontaneity of reaction. spontaneous

Since )G = )H -T )S sys sys sys Free Energy and Spontaneity Dividing by -T

-)G/T = -)Hsys/T + )Ssys • Changes in H an S during a reaction result in a change in free energy, ∆G , given by But )S = - )H /T so that surr sys the equation -)G/T = )S + )S , but )S = )S + )S surr sys universe surr sys ∆G = ∆H − T∆S -Therefore, when )G is negative, • Thus, if you can show that ∆G is negative at a given temperature and pressure, you can predict

)Suniv is + and process is spontaneous that the reaction will be spontaneous. • If )G = 0, the reaction is at equilibrium These equations apply to processes occurring at constant T & P, and in which no useful work is done! • If )G is positive, the reaction is nonspontaneous

8 EffectEffect ofof ∆∆HH andand ∆∆SS onon SpontaneitySpontaneity Standard Free-Energy Change Result ∆H ∆S •The standard free energy change, ∆Go, is − + spontaneous at all temps the free energy change that occurs when reactants and products are in their standard + + spontaneous at high temps states. −−spontaneous at low temps • The next example illustrates the calculation of the standard free energy change, ∆Go, from + − not spontaneous at any temp ∆Ho and ∆So. o o o ∆G = ∆H − T∆S ∆G = ∆H − T∆S

A Problem To Consider A Problem To Consider

• What is the standard free energy change, ∆Go , for • What is the standard free energy change, ∆Go , for the following reaction at 25 oC? Use values of the following reaction at 25 oC? Use values of o o o o ∆Hf and S , from Appendix 4. ∆Hf and S , from Appendix 4. N2 (g) + 3H2 (g) → 2NH3 (g) N2 (g) + 3H2 (g) → 2NH3 (g) o ∆Hf : 00 2 x (-45.9) kJ • You can calculate ∆Ho and ∆So using their So: 191.53 x 130.6 2 x 193 J/K respective summation laws. • Place below each formula the values of ∆H o o o o f ∆H = ∑n∆Hf (products) − ∑m∆Hf (reactants) and So multiplied by stoichiometric coefficients. = [2×(−45.9) − 0]kJ = −91.8 kJ

A Problem To Consider A Problem To Consider

• What is the standard free energy change, ∆Go , for • What is the standard free energy change, ∆Go , for the following reaction at 25 oC? Use values of the following reaction at 25 oC? Use values of o o o o ∆Hf and S , from Appendix 4. ∆Hf and S , from Appendix 4. N2 (g) + 3H2 (g) → 2NH3 (g) N2 (g) + 3H2 (g) → 2NH3 (g) • You can calculate ∆Ho and ∆So using their • Now substitute into our equation for ∆Go. Note respective summation laws. that ∆So is converted to kJ/K. o o o ∆S o = ∑ nS o (products) − ∑ mS o (reactants) ∆G = ∆H − T∆S = −91.8 kJ − (298 K)(−0.197 kJ/K) = [2×193 − (191.5 + 3×130.6)] J/K = -197 J/K = −33.1kJ

9 Standard Free Energies of Formation o •The standard free energy of formation, ∆Gf , of a substance is the free energy change that occurs when 1 mol of a substance is formed from its elements in their stablest states at 1 atm pressure and 25 oC. o • By tabulating ∆Gf for substances, as in Appendix 4, you can calculate the ∆Go for a reaction by using a summation law. o o o ∆G = ∑n∆Gf (products) − ∑m∆Gf (reactants)

A Problem To Consider A Problem To Consider

• Calculate ∆Go for the combustion of 1 mol of • Calculate ∆Go for the combustion of 1 mol of o o ethanol, C2H5OH, at 25 C. Use the standard free ethanol, C2H5OH, at 25 C. Use the standard free energies of formation given in Appendix 4. energies of formation given in Appendix 4.

C2H5OH(l) + 3O2 (g) → 2CO2 (g) + 3H2O(g) C2H5OH(l) + 3O2 (g) → 2CO2 (g) + 3H2O(g) o o ∆Gf : -174.8 0 2(-394.4) 3(-228.6)kJ ∆Gf : -174.8 0 2(-394.4) 3(-228.6)kJ o o • Place below each formula the values of ∆Gf • You can calculate ∆G using the summation multiplied by stoichiometric coefficients. law. o o o ∆G = ∑n∆Gf (products) − ∑m∆Gf (reactants) ∆Go = [2(−394.4) + 3(−228.6) − (−174.8)] kJ

∆Go as a Criteria for A Problem To Consider Spontaneity • Calculate ∆Go for the combustion of 1 mol of o • The following rules are useful in judging ethanol, C2H5OH, at 25 C. Use the standard free energies of formation given in Appendix 4. the spontaneity of a reaction. 1. When ∆Go is a large negative number (more C H OH(l) + 3O (g) → 2CO (g) + 3H O(g) 2 5 2 2 2 negative than about –10 kJ), the reaction is ∆G o: -174.8 0 2(-394.4) 3(-228.6)kJ f spontaneous as written, and the reactants o • You can calculate ∆G using the summation transform almost entirely to products when law. equilibrium is reached. o o o ∆G = ∑n∆Gf (products) − ∑m∆Gf (reactants) ∆Go = −1299.8 kJ

10 ∆Go as a Criteria for ∆Go as a Criteria for Spontaneity Spontaneity • The following rules are useful in judging • The following rules are useful in judging the spontaneity of a reaction. the spontaneity of a reaction. 2. When ∆Go is a large positive number (more 3. When ∆Go is a small negative or positive positive than about +10 kJ), the reaction is value (less than about 10 kJ), the reaction nonspontaneous as written, and reactants do gives an equilibrium mixture with not give significant amounts of product at significant amounts of both reactants and equilibrium. products.

Free Energy Change During Reaction • As a system approaches equilibrium, the instantaneous change in free energy approaches zero. • The next figure illustrates the change in free energy during a spontaneous reaction. • As the reaction proceeds, the free energy eventually reaches its minimum value. • At that point, ∆G = 0, and the net reaction stops; it comes to equilibrium.

Free-energy change during a spontaneous reaction. Free Energy Change During Reaction • As a system approaches equilibrium, the instantaneous change in free energy approaches zero. • The next figure illustrates the change in free energy during a nonspontaneous reaction. • Note that there is a small decrease in free energy as the system goes to equilibrium.

11 Free-energy change during a nonspontaneous reaction. Relating ∆Go to the Equilibrium Constant • The free energy change when reactants are in non-standard states (other than 1 atm pressure or 1 M) is related to the standard free energy change, ∆Go, by the following equation. ∆G = ∆Go + RTlnQ • Here Q is the thermodynamic form of the reaction quotient.

Relating ∆Go to the Relating ∆Go to the Equilibrium Constant Equilibrium Constant • The free energy change when reactants are • The free energy change when reactants are in non-standard states (other than 1 atm in non-standard states (other than 1 atm pressure or 1 M) is related to the standard pressure or 1 M) is related to the standard free energy change, ∆Go, by the following free energy change, ∆Go, by the following equation. equation. ∆G = ∆Go + RTlnQ ∆G = ∆Go + RTlnQ ∆G represents an instantaneous change in free • At equilibrium, ∆G=0 and the reaction quotient energy at some point in the reaction Q becomes the equilibrium constant K. approaching equilibrium.

Relating ∆Go to the Relating ∆Go to the Equilibrium Constant Equilibrium Constant • This result easily rearranges to give the basic • The free energy change when reactants are equation relating the standard free-energy in non-standard states (other than 1 atm change to the equilibrium constant. pressure or 1 M) is related to the standard o free energy change, ∆Go, by the following ∆G = −RTlnK equation. o o • When K > 1 , the ln K is positive and ∆G is 0 = ∆G + RTlnK negative. • At equilibrium, ∆G=0 and the reaction quotient • When K < 1 , the ln K is negative and ∆Go is Q becomes the equilibrium constant K. positive.

12 A Problem To Consider

• Find the value for the equilibrium constant, K, at 25 oC (298 K) for the following reaction. The standard free- energy change, ∆Go, at 25 oC equals –13.6 kJ.

2NH3 (g) + CO2 (g) NH2CONH2 (aq) + H2O(l) • Rearrange the equation ∆Go=-RTlnK to give

∆Go lnK = − RT

A Problem To Consider A Problem To Consider

• Find the value for the equilibrium constant, K, at 25 oC • Find the value for the equilibrium constant, K, at 25 oC (298 K) for the following reaction. The standard free- (298 K) for the following reaction. The standard free- energy change, ∆Go, at 25 oC equals –13.6 kJ. energy change, ∆Go, at 25 oC equals –13.6 kJ.

2NH3 (g) + CO2 (g) NH2CONH2 (aq) + H2O(l) 2NH3 (g) + CO2 (g) NH2CONH2 (aq) + H2O(l) • Substituting numerical values into the equation, • Hence, 5.49 2 −13.6×103 J K = e = 2.42×10 lnK = = 5.49 − 8.31 J/(mol ⋅K)× 298 K

Calculation of ∆Go at Various Temperatures • In this method you assume that ∆Ho and ∆So are essentially constant with respect to temperature. o • You get the value of ∆GT at any temperature T by substituting values of ∆Ho and ∆So at 25 oC into the following equation. o o o ∆GT = ∆H − T∆S

13 A Problem To Consider A Problem To Consider

• Find the ∆Go for the following reaction at 25 oC • Find the ∆Go for the following reaction at 25 oC and 1000 oC. Relate this to reaction spontaneity. and 1000 oC. Relate this to reaction spontaneity. CaCO (s) → CaO(s) + CO (g) 3 2 CaCO3 (s) → CaO(s) + CO2 (g) o o ∆Hf : -1206.9-635.1 -393.5 kJ ∆Hf : -1206.9-635.1 -393.5 kJ So: 92.938.2 213.7 J/K So: 92.938.2 213.7 J/K • You can calculate ∆Ho and ∆So using their o • Place below each formula the values of ∆Hf respective summation laws. and So multiplied by stoichiometric coefficients.

A Problem To Consider A Problem To Consider

• Find the ∆Go for the following reaction at 25 oC • Find the ∆Go for the following reaction at 25 oC and 1000 oC. Relate this to reaction spontaneity. and 1000 oC. Relate this to reaction spontaneity. CaCO (s) → CaO(s) + CO (g) CaCO3 (s) → CaO(s) + CO2 (g) 3 2 o o ∆Hf : -1206.9-635.1 -393.5 kJ ∆Hf : -1206.9-635.1 -393.5 kJ So: 92.938.2 213.7 J/K So: 92.938.2 213.7 J/K o o o o o o ∆H = ∑n∆Hf (products) − ∑m∆Hf (reactants) ∆S = ∑n∆S (products) − ∑m∆S (reactants) = [(−635.1− 393.5) − (−1206.9)]kJ = 178.3 kJ = [(38.2 + 213.7) − (92.9)] = 159.0 J / K

A Problem To Consider A Problem To Consider

CaCO3 (s) → CaO(s) + CO2 (g) CaCO3 (s) → CaO(s) + CO2 (g) o o ∆Hf : -1206.9-635.1 -393.5 kJ ∆Hf : -1206.9-635.1 -393.5 kJ So: 92.938.2 213.7 J/K So: 92.938.2 213.7 J/K • Now you substitute ∆Ho, ∆So (=0.1590 kJ/K), • Now you substitute ∆Ho, ∆So (=0.1590 kJ/K), o o and T (=298K) into the equation for ∆Gf . and T (=298K) into the equation for ∆Gf . o o o o ∆GT = ∆H − T∆S ∆GT = 178.3kJ − (298 K)(0.1590 kJ / K)

14 A Problem To Consider A Problem To Consider

• Find the ∆Go for the following reaction at 25 oC • Find the ∆Go for the following reaction at 25 oC and 1000 oC. Relate this to reaction spontaneity. and 1000 oC. Relate this to reaction spontaneity. CaCO (s) → CaO(s) + CO (g) 3 2 CaCO3 (s) → CaO(s) + CO2 (g) o o ∆Hf : -1206.9-635.1 -393.5 kJ ∆Hf : -1206.9-635.1 -393.5 kJ So: 92.938.2 213.7 J/K So: 92.938.2 213.7 J/K • Now you substitute ∆Ho, ∆So (=0.1590 kJ/K), • Now we’ll use 1000 oC (1273 K) along with o o o and T (=298K) into the equation for ∆Gf . our previous values for ∆H and ∆S . o So the reaction is o ∆GT = 130.9 kJ nonspontaneous ∆GT = 178.3kJ − (1273 K)(0.1590 kJ / K) at 25 oC.

A Problem To Consider A Problem To Consider

CaCO3 (s) → CaO(s) + CO2 (g) CaCO3 (s) → CaO(s) + CO2 (g) o o ∆Hf : -1206.9-635.1 -393.5 kJ ∆Hf : -1206.9-635.1 -393.5 kJ So: 92.938.2 213.7 J/K So: 92.938.2 213.7 J/K • Now we’ll use 1000 oC (1273 K) along with • To determine the minimum temperature for o o o our previous values for ∆H and ∆S . spontaneity, we can set ∆Gf =0 and solve for T. o o So the reaction is ∆H ∆G = −24.1 kJ spontaneous at T T = o 1000 oC. ∆S

A Problem To Consider A Problem To Consider

• To determine the minimum temperature for • Thus, CaCO3 should be thermally stable until o o spontaneity, we can set ∆Gf =0 and solve for T. its heated to approximately 848 C. 178.3 kJ T = = 1121 K (848 oC) 0.1590 kJ / K

15 TemperatureTemperature DependenceDependence ofof KK Maximum Work )Go = - RTlnK = )Ho -T )So and dividing both sides by T & R, multiply both sides by -1 • Often reactions are not carried out in a way that does useful work. • As a spontaneous precipitation reaction occurs, ∆ H° the free energy of the system decreases and ln(K ) =− (/1 TSR )+°∆ / entropy is produced, but no useful work is R obtained. •y = mx + b • In principle, if a reaction is carried out to obtain •(∆H° and S°≈independent of temperature the maximum useful work, no entropy is over a small temperature range) produced.

ReversibleReversible v.v. IrreversibleIrreversible Maximum Work ProcessesProcesses

• Often reactions are not carried out in a way •Reversible: The universe is exactly the same that does useful work. • It can be shown that the maximum useful work, as it was before the cyclic process.

wmax , for a spontaneous reaction is ∆G. •Irreversible: The universe is different after w = ∆G the cyclic process. max •All real processes are irreversible -- (some • The term free energy comes from this result. work is changed to heat).

Figure 16.10: A battery can do work by sending current to a starter motor.