4/19/2016. Chapter 17 Free Energy and Thermodynamics. First Law of Thermodynamics. First Law of Thermodynamics. The Energy Tax.

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1 Chemistry: A Molecular Approach, 2nd Ed. Nivaldo Tro First Law of Thermodynamics Chapter 17 Free Energy and 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 DE universe = 0 = DE system + DE surroundings Roy Kennedy Massachusetts Bay Community College Wellesley Hills, MA 2 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 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 3 4 Heat Tax fewer steps generally results in a lower total heat tax 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 5 6 1

2 Comparing Potential Energy The direction of spontaneity can be determined by comparing the potential energy of the system at the start and the end 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 7 8 Thermodynamics vs. Kinetics 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) 9 10 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? 11 Melting Ice 12 Melting is an Endothermic When a solid process, melts, the yet particles ice will have more spontaneously freedom of movement. melt above 0 C. More freedom of motion increases the randomness of the system. When systems become more random, energy is released. We call this energy, entropy 2

3 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 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 Entropy W 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 J/K W is the number of energetically equivalent ways a system can exist unitless Random systems require less energy than ordered systems 15 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 16 Macrostates Microstates These microstates all have the same macrostate So there This are macrostate six can be achieved through different several particle different arrangements of the particles arrangements that result in the same macrostate 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 Therefore macrostate State with C has the highest higher entropy entropy also than has the greatest either State dispersal A or of State energy B There is six times the probability of having the State C macrostate than either State A or State B

4 Changes in Entropy, DS DS = S final S initial 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 Increases in Entropy DS For a process where the final condition is more random than the initial condition, DS system 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, DS system is negative and the entropy change is unfavorable for the process to be spontaneous DS system = DS reaction = Sn(S products ) Sn(S reactants ) 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 Entropy Change and State Change Practice Predict whether DS system is + or for each of the following A hot beaker burning your fingers DS is + Water vapor condensing DS is Separation of oil and vinegar salad dressing DS is Dissolving sugar in tea DS is + 2 PbO 2 (s) 2 PbO(s) + O 2 (g) DS is + 2 NH 3 (g) N 2 (g) + 3 H 2 (g) Ag + (aq) + Cl (aq) AgCl(s) DS is + DS is

5 The 2 nd Law of Thermodynamics The 2 nd Law of Thermodynamics says that the total entropy change of the universe must be positive for a process to be spontaneous for reversible process DS univ = 0 for irreversible (spontaneous) process DS univ > 0 DS universe = DS system + DS surroundings If the entropy of the system decreases, then the entropy of the surroundings must increase by a larger amount when DS system is negative, DS surroundings must be positive and big for a spontaneous process 25 Heat Flow, Entropy, and the 2 nd Law When ice is placed in water, heat flows from the water into the ice According to the 2 nd Law, heat must flow from water to ice because it results in more dispersal of heat. The entropy of the universe increases. 26 Heat Transfer and Changes in Entropy of the Surroundings The 2 nd 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! 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 Heat Exchange and DS surroundings 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 Temperature Dependence of DS surroundings 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

6 Quantifying Entropy Changes in Surroundings The entropy change in the surroundings is proportional to the amount of heat gained or lost q surroundings = q system The entropy change in the surroundings is also inversely proportional to its temperature At constant pressure and temperature, the overall relationship is Example 17.2a: Calculate the entropy change of the surroundings at 25ºC for the reaction below C 3 H 8(g) + 5 O 2(g) 3 CO 2(g) + 4 H 2 O (g) DH rxn = 2044 kj DH system = 2044 kj, T = 25 ºC = 298 K DS surroundings, J/K T, DH DS Check: combustion is largely exothermic, so the entropy of the surroundings should increase significantly Practice The reaction below has DH rxn = kj at 25 C. (a) Determine the Ds surroundings, (b) the sign of DS system, and (c) whether the process is spontaneous 2 O 2 (g) + N 2 (g) 2 NO 2 (g) Practice The reaction 2 O 2 (g) + N 2 (g) 2 NO 2 (g) has DH rxn = 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. DH sys = kj, T = 25 ºC = 298 K DS surr, J/K, DS react + or, DS universe + or T, DH DS the major difference is that there are fewer product molecules than reactant molecules, so the DS reaction is unfavorable and ( ) because DS surroundings is ( ) it is unfavorable both DS surroundings and DS reaction are ( ), DS universe is ( ) and the process is nonspontaneous Gibbs Free Energy and Spontaneity It can be shown that TDS univ = DH sys TDS sys 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 DG sys = DH sys TDS sys Because DS univ determines if a process is spontaneous, DG also determines spontaneity DS univ is + when spontaneous, so DG is 35 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 36 6

7 DG, DH, and DS Free Energy Change and Spontaneity Example 17.3a: The reaction CCl 4(g) C (s, graphite) + 2 Cl 2(g) has DH = kj and DS = J/K at 25 C. Calculate DG and determine if it is spontaneous. DH = kj, DS = J/K, T = 298 K Practice The reaction Al (s) + Fe 2 O 3(s) Fe (s) + Al 2 O 3(s) has DH = kj and DS = 41.3 J/K at 25 C. Calculate DG and determine if it is spontaneous. DG, kj T, DH, DS DG Answer: Because DG is +, the reaction is not spontaneous at this temperature. To make it spontaneous, we need to increase the temperature Practice The reaction Al (s) + Fe 2 O 3(s) Fe (s) + Al 2 O 3(s) has DH = kj and DS = 41.3 J/K at 25 C. Calculate DG and determine if it is spontaneous. DH = kj and DS = 41.3 J/K, T = 298 K Example 17.3b: The reaction CCl 4(g) C (s, graphite) + 2 Cl 2(g) has DH = kj and DS = J/K. Calculate the minimum temperature it will be spontaneous. DH = kj, DS = J/K, DG < 0 DG, kj T, K T, DH, DS DG DG, DH, DS T Answer: Because DG is, the reaction is spontaneous at this temperature. To make it nonspontaneous, we need to increase the temperature. Answer: the temperature must be higher than 673K for the reaction to be spontaneous

8 Practice The reaction Al (s) + Fe 2 O 3(s) Fe (s) + Al 2 O 3(s) has DH = kj and DS = 41.3 J/K. Calculate the maximum temperature it will be spontaneous. Practice The reaction Al (s) + Fe 2 O 3(s) Fe (s) + Al 2 O 3(s) has DH = kj and DS = 41.3 J/K at 25 C. Determine the maximum temperature it is spontaneous. DH = kj and DS = 41.3 J/K, DG > 0 T, C DG, DH, DS T = 1775 C Answer: any temperature above 1775 C will make the reaction nonspontaneous 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 The 3 rd 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 3 rd 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 Standard Absolute Entropies Standard Absolute Entropies S 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

9 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 S, (J/mol K) H 2 O (l) 70.0 H 2 O (g) 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 Relative Standard Entropies: Allotropes Relative Standard Entropies: Molecular Complexity 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 Larger, more complex molecules generally have larger entropy More available energy states, allowing more dispersal of energy through the states Molar S, Mass (J/mol K) CO Ar (g) CNO 2 H 4 (g) Relative Standard Entropies Dissolution The Standard Entropy Change, DS Dissolved solids generally have larger entropy Distributing particles throughout the mixture S, (J/mol K) KClO 3 (s) KClO 3 (aq) The standard entropy change is the difference in absolute entropy between the reactants and products under standard conditions DSº reaction = ( n p Sº products ) ( n r Sº reactants ) remember: though the standard enthalpy of formation, DH f, of an element is 0 kj/mol, the absolute entropy at 25 C, S, is always positive

10 Example 17.4: Calculate DS for the reaction 4 NH 3(g) + 5 O 2(g) 4 NO (g) + 6 H 2 O (g) H 2 O(g) standard entropies from Appendix IIB DS, J/K S NH3, S O2, S NO, S H2O DS Sol n: S, J/mol K NH 3 (g) O 2 (g) NO(g) Practice Calculate the DS for the reaction 2 H 2(g) + O 2(g) 2 H 2 O (g) S, J/mol K H 2 (g) O 2 (g) H 2 O(g) Check: DS is +, as you would expect for a reaction with more gas product molecules than reactant molecules Example 17.4: Calculate DS for the reaction 2 H 2(g) + O 2(g) 2 H 2 O (g) standard entropies from Appendix IIB DS, J/K S H2, S O2, S H2O Check: DS is, as you would expect for a reaction with more gas reactant molecules than product molecules 57 S, J/mol K H 2 (g) O 2 (g) H 2 O(g) DS Calculating DG At 25 C DG o reaction = SnDG o f (products) - SnDGo f (reactants) At temperatures other than 25 C assuming the change in DH o reaction and DS o reaction is negligible DG reaction = DH reaction TDS reaction or DG total = DG reaction 1 + DG reaction Example 17.6: The reaction SO 2(g) + ½ O 2(g) SO 3(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 SO 3. DH = 98.9 kj, DS = 94.0 J/K, T = 398 K DG, kj T, DH, DS DG Practice Determine the free energy change in the following reaction at 298 K 2 H 2 O (g) + O 2(g) 2 H 2 O 2(g) DH, kj/mol S, J/mol H 2 O 2 (g) O 2 (g) H 2 O(g) Answer: because DG is, the reaction is spontaneous at this temperature, though less so than at 25 C

11 Practice Determine the free energy change in the following reaction at 298 K 2 H 2 O (g) + O 2(g) 2 H 2 O 2(g) standard DH = energies kj, DS from = Appendix J/K, IIB, T = T 298 = 298 K K DG, kj kj DH f Sº of prod & react DH, DSº DG Standard Free Energies of Formation The free energy of formation (DG f ) 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 Example 17.7: Calculate DG at 25 C for the reaction CH 4(g) + 8 O 2(g) CO 2(g) + 2 H 2 O (g) + 4 O 3(g) standard free energies of formation from Appendix IIB DG, kj DG f of prod & react DG DG f, kj/mol CH 4 (g) 50.5 O 2 (g) 0.0 CO 2 (g) H 2 O(g) O 3 (g) Practice Determine the free energy change in the following reaction at 298 K 2 H 2 O (g) + O 2(g) 2 H 2 O 2(g) DG, kj/mol H 2 O 2 (g) O 2 (g) 0 H 2 O(g) Practice Determine the free energy change in the following reaction at 298 K 2 H 2 O (g) + O 2(g) 2 H 2 O 2(g) standard free energies of formation from Appendix IIB DG, kj DG f of prod & react DG 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

12 Example 17.8: Find DGº rxn for the following reaction using the given equations 3 C (s) + 4 H 2(g) C 3 H 8(g) C 3 H 8(g) + 5 O 2(g) 3 CO 2(g) + 4 H 2 O (g) DGº = 2074 kj C (s) + O 2(g) CO 2(g) DGº = kj 2 H 2(g) + O 2(g) 2 H 2 O (g) DGº = kj DGº of 3 C (s) + 4 H 2(g) C 3 H 8(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. 3 CO 2(g) + 4 H 2 O (g) C 3 H 8(g) + 5 O 2(g) DGº = kj 3 x C (s) [C (s) O 2(g) 3 CO CO 2(g) 2(g) ] DGº = 3( kj) 24 x H 2(g) [2 H+ 2(g) 2 O+ 2(g) O 2(g) 4 H2 2 HO 2(g) O (g) ] DGº DGº = = 2( kj) 3 C (s) + 4 H 2(g) C 3 H 8(g) DGº = 23 kj 67 Practice Determine the free energy change in the following reaction at 298 K 2 H 2 O (g) + O 2(g) 2 H 2 O 2(g) DG, kj/mol H 2 (g) + O 2 (g) H 2 O 2 (g) H 2 (g) + O 2 (g) 2 H 2 O(g) Practice Determine the free energy change in the following reaction at 298 K 2 H 2 O (g) + O 2(g) 2 H 2 O 2(g) H 2(g) + O 2(g) H 2 O 2(g) DGº = kj 2 H 2(g) + O 2(g) 2 H 2 O (g) DGº = kj DGº of 2 H 2 O (g) + O 2(g) 2 H 2 O 2(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. 2 H 2(s) + 2 O 2(g) 2 H 2 O 2(g) DGº = kj 2 H 2 O (g) 2 H 2(g) + O 2(g) DGº = kj 2 H 2 O (g) + O 2(g) 2 H 2 O 2(g) DGº = kj What s Free About Free Energy? 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 Free Energy of an Exothermic Reaction C (s, graphite) + 2 H 2(g) CH 4(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 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

13 Real Reactions In a real reaction, some of the free energy is lost as heat if not most Therefore, real reactions are irreversible 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 Example 17.9: Calculate DG at 298 K for the reaction under the given conditions 2 NO (g) + O 2(g) 2 NO 2(g) DGº = 71.2 kj P, atm NO(g) O 2 (g) NO 2 (g) 2.00 non-standard conditions, DGº, T DG, kj DG, P NO, P O2, P NO2 DG Practice Calculate DG rxn for the given reaction at 700 K under the given conditions N 2(g) + 3 H 2(g) 2 NH 3(g) DGº = kj P, atm N 2 (g) 33 H 2 (g) 99 NH 3 (g) 2.0 Practice Calculate DG rxn for the given reaction at 700 K under the given conditions N 2(g) + 3 H 2(g) 2 NH 3(g) DGº = kj non-standard conditions, DGº, T DG, kj DG, P NO, P O2, P NO2 DG P, atm N 2 (g) 33 H 2 (g) 99 NH 3 (g)

14 DGº and K Because DG rxn = 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 Example 17.10: Calculate K at 298 K for the reaction N 2 O 4(g) 2 NO 2(g) standard free energies of formation from Appendix IIB K DG f of prod & react DG K DGº = RTln(K) DG f, kj/mol N 2 O 4 (g) NO 2 (g) Practice Estimate the equilibrium constant for the given reaction at 700 K N 2(g) + 3 H 2(g) 2 NH 3(g) DGº = kj Practice Estimate the Equilibrium Constant for the given reaction at 700 K N 2(g) + 3 H 2(g) 2 NH 3(g) DGº = kj DGº K DG K DGº = RTln(K) 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