1 Chapters 29 and 35 Thermochemistry and Chemical Thermodynamics Copyright (c) 2011 by Michael A. Janusa, PhD. All rights reserved.

2 Thermochemistry Thermochemistry is the study of the energy effects that accompany chemical reactions. Why do chemical reactions occur? What is the driving force of rxn? Answer: Stability, wants to get to lower E. For a rxn to take place spontaneously the products of reaction must be more stable (lower E) than the starting reactants. Nonspontaneous means never happen by self. E R P release E, spon higher E, less stable, more reactive E R P absorb E, nonspon lower E, more stable, less reactive

Reaction Enthalpy In chemical reactions, heat is often transferred from the “system or reaction” to its “surroundings,” or vice versa. system - the substance or mixture of substances under study in which a change occurs. The surroundings are everything in the vicinity of the thermodynamic system. system or rxn surroundings ( + into system - out system

4 Heat of Reaction Heat flow is defined as the energy that flows into or out of a system. We follow heat flow by watching the difference in temperature between the system and its surroundings. Often we follow the surroundings temp (solvent) and must realize that the opposite is happening to the system. If system is absorbing heat from the surroundings than the temp of the surroundings must be decreasing. T system (+) T surr (-)

5 Heat of Reaction Heat flow or heat of reaction is denoted by the symbol q and is the amount of heat required to return a system to the given temperature at the completion of the reaction.  For an endothermic rxn the sign of q is positive; heat is absorbed by the system from the surroundings. E P absorb heat, nonspon (endo) R  q > 0 Surroundings +q T system T surr System

6 Heat of Reaction  q < 0 -q System Surroundings  For an exothermic rxn, the sign of q is negative; heat is evolved (released) by the system to the surroundings. T system T surr E R P release heat, spon (exo)

7 Enthalpy and Enthalpy Change The heat absorbed or evolved by a reaction depends on the conditions under which it occurs. ex. pressure Usually, a reaction takes place in an open vessel, and therefore under the constant pressure of the atmosphere. heat of this type of reaction is denoted q p ; this heat at constant pressure is named enthalpy and given symbol H. H is the heat flow at constant pressure.

8 –an extensive property - depends on the quantity of substance. –Enthalpy is a state function, a property of a system that depends only on its present state and is independent of any previous history of the system. Enthalpy and Enthalpy Change Enthalpy, denoted H, is an extensive property of a substance that can be used to obtain the heat absorbed or evolved in a chemical reaction at constant pressure. ooo

9 The reaction enthalpy for a reaction at a given temperature and pressure Enthalpy and Enthalpy Change

10 As we already stated the reaction enthalpy is equal to the heat of reaction at constant pressure. This represents the entire change in internal energy (  U) minus any expansion “work” done by the system; therefore we can define enthalpy and internal work by the 1st law of thermodynamics: In any process, the total change in energy of the system,  U, is equal to the sum of the heat absorbed, q, and the work, w, done by the system.  U = q p + w =  H + w Enthalpy and Enthalpy Change

11 –Changes in E manifest themselves as exchanges of energy between the system and surroundings. –These exchanges of energy are of two kinds; heat and work - must account for both. –Heat is energy that moves into or out of a system because of a temperature difference between system and surroundings. –Work is the energy exchange that results when a force F moves an object through a distance d; work (w) = F  d In chemical systems, work is defined as a change in volume at a given pressure, that is:

12 negative sign is to keep sign correct in terms of system. For expansion,  V, will be a positive value but expansion involves the system doing work on the surroundings and a decrease in internal energy -- negative keeps it neg. For contraction work,  V, will be a negative value but contraction involves the surroundings doing work on the system and an increase in internal energy -- negative keeps it positive (- x - = +). Giving us the 1st law of thermo is more useful form: realize absorb heat (+) release or evolved heat (-) HW 44 code: first

Thermochemical Equations A thermochemical equation is the chemical equation for a reaction (including phase labels {important}) in which the equation is given a molar interpretation, and the enthalpy of reaction for these molar amounts is written directly after the equation. If  H has a superscript like  H o, means thermo standard conditions o C (298K) and 1 atm.

14 The following are two important rules for manipulating thermochemical equations: –1.) When a thermochemical equation is multiplied by any factor, the value of  H for the new equation is obtained by multiplying the  H in the original equation by that same factor. –2.) When a chemical equation is reversed, the value of  H is reversed in sign. Thermochemical Equations exo endo

15 Hess’s law of heat summation states that for a chemical equation that can be written as the sum of two or more steps, the enthalpy change for the overall equation is the sum of the enthalpy changes for the individual steps. Basically, R & P in individual steps can be added like algebraic quantities in determining overall equation and enthalpy change Hess’s Law

16 simple example : Given: A + D  E + C  H = X kJ 2A + B  2C  H = Y kJ Question: 2D  B + 2E  H = ? 2A + 2D  2E + 2C  H = 2X kJ 2C  2A + B  H = -Y kJ 2D  B + 2E  H = 2X – Y kJ _______________________________________ 1.Correct side? 2.Correct # moles?

17 For example, suppose you are given the following GIVEN data: Hess’s Law use these data to obtain the enthalpy change for the following reaction? x2 flip

18 If we multiply the first equation by 2 and reverse the second equation, they will sum together to become the third. HW 45 x2 flip code: ten

19 The standard enthalpy of formation of a substance, denoted  H f o, is the enthalpy change for the formation of one mole of a substance in its standard state from its component elements in their standard state (298K & 1 atm). –Note that the standard enthalpy of formation for a pure element in its standard state and H + is zero. This means elements in their standard state has  H f o = 0: metals - solids, diatomic gases, H + ion Standard Enthalpies of Formation (molecular scale) Ag (s) + ½ Cl 2 (g)  AgCl (s)  H f o AgCl

20 Another way to determine heat of reaction is the The law of summation of heats of formation which states that the enthalpy of a reaction is equal to the total formation energy of the products minus that of the reactants..  is the mathematical symbol meaning “the sum of”, and n is the coefficients of the substances in the chemical equation. Standard Enthalpies of Formation

21 Ex. Generic Law of Summation aA + bB  cC + dD

22 A Problem to Consider –What is the standard reaction enthalpy,  H o rxn, for this reaction?

23 Using the summation law: –Be careful of arithmetic signs as they are a likely source of mistakes. HW 46 code: formation

24 –Entropy, S, is a thermodynamic quantity that is a measure of the randomness or disorder of a system. –The SI unit of entropy is joules per Kelvin (J/K) and, like enthalpy, is a state function The Second Law of Thermodynamics The second law of thermodynamics addresses questions about spontaneity in terms of a quantity called entropy.

25 E R P release E, spon (exo) E R P absorb E, nonspon (endo) Most soluble salts dissolve in water spontaneously; however, most soluble salts dissolve by an endothermic process. NH 4 NO 3 (s)  NH 4 + (aq) + NO 3 - (aq)  H = 28.1 kJ There is an increase in molecular disorder or randomness of the system. Solids: high order/low disorder, high energy Liquids: middle order/low disorder, medium energy Gases: low order/high disorder, low energy

26 entropy (S) - is a thermodynamic quantity that is a measure of how dispersed the energy of a system is among the different possible ways that system can contain energy, typically in J/K units. One example of entropy is the amount of molecular disorder or randomness in the system. S increases as disorder increases and energy decreases gases have high disorder, low energy solids have low disorder, high energy We typically follow the change in entropy in the system so we treat it as a state property and measure  S = S final - S initial +  S = increase in entropy, i.e. disorder increased; -  U -  S = decrease in entropy, ie. disorder decreased ; +  U This gets us to the second law of thermo

27 Entropy and the Second Law of Thermodynamics The second law of thermodynamics states that the total entropy of a system and its surroundings increases for a spontaneous process.

28 The tendency of a system to increase its entropy (+  S) is the second important factor in determining the spontaneity of a chemical or physical change in addition to  H. recap: spontaneous process: (system goes to lower energy state) favored by -  H (exo) favored by +  S (ie. increase disorder) nonspontaneous process: (system goes to higher energy state) favored by +  H (endo) favored by -  S (ie. decrease in disorder) Do both need to be true for spon rxn? No, remember soluble salt dissolving example. The larger term will dictate overall process.

29 –As temperature is raised the substance becomes more disordered as it absorbs heat and becomes a liquid then a gas, where entropy > 0; S increases as temp increase. –The entropy of a substance is determined by measuring how much heat is required to change its temperature per Kelvin degree (J/K) Third Law of Thermodynamics The third law of thermodynamics states that the entropy of all perfect crystalline substances approaches zero as the temperature approaches absolute zero (Kelvin).

30 –Standard state implies 25 o C (298K), 1 atm pressure, and 1 M for dissolved substances.(Thermo standard state) 35.5 Standard Reaction Entropy The standard entropy of a substance or ion, also called its absolute entropy, S o, is the entropy value for the standard state of the species. Similar to heats of formation,  H f o, except on absolute not relative scale.

31 –This means that elements have nonzero values for entropy (absolute scale), unlike standard enthalpies of formation,  H f o, which by convention, are zero (relative scale). Standard Entropies and the Third Law of Thermodynamics –The symbol S o, rather than  S o, is used for standard entropies to emphasize that they originate from the third law and absolute not relative values.

32 –Even without knowing the values for the entropies of substances, you can sometimes predict the sign of  S o for a reaction. Entropy Change for a Reaction You can calculate the entropy change for a reaction using a summation law, similar to the way you obtained  H f o.

33 1.A reaction in which a molecule is broken into two or more smaller molecules. The entropy usually increases in the following situations: Entropy Change for a Reaction 2.A reaction in which there is an increase in the moles of gases. 3.A process in which a solid changes to liquid or gas, or a liquid changes to gas. AB  A + B +  S A(g)  B(g) + C(g) +  S A(s)  B(l) or B(g) +  S B(l)  C(g) +  S

34 Predict  S and spon/nonspon based only on entropy for the following rxns: C 2 H 4 (g) + Br 2 (g)  BrCH 2 CH 2 Br (l) 2 C 2 H 6 (g) + 7 O 2 (g)  4 CO 2 (g) + 6 H 2 O (g) C 6 H 12 O 6 (s)  2 C 2 H 5 OH (l) + 2 CO 2 (g) HW 47 gas to liquid; decrease in disorder; -  S; nonspon based on S only 9 mols gas to 10 mols of gas; increase in disorder; +  S; spon based on S only solid to liquid/gas (decompose); increase in disorder; +  S; spon based on S only code: ben

35 –The calculation is similar to that used to obtain  H o from standard enthalpies of formation. A Problem To Consider Calculate the change in entropy,  S o, at 25 o C for the reaction in which urea is formed from NH 3 and CO 2. Gas to liquid; decrease in disorder; predict -  S

36 So:So:193 J/mol. K A Problem To Consider decrease in disorder as predicted

37 –This quantity gives a direct criterion for spontaneity of reaction Gibbs Free Energy The question arises as to how do we decide if enthalpy or entropy dictates the spontaneity of a reaction. What is the relationship between  H and  S? The American physicist J. Willard Gibbs introduced the concept of free energy (sometimes called the Gibbs free energy), G, which is a thermodynamic quantity defined by the equation   G=  H-T  S T – Kelvin scale

38 At a given temperature and pressure  G = 0, the reaction gives an equilibrium mixture with significant amounts of both reactants and products (Temp transfer point where reaction switches spon/nonspon)  G > 0, the reaction is nonspontaneous as written, and reactants do not give significant amounts of product at equilibrium.  G < 0, the reaction is spontaneous as written, and the reactants transform almost entirely to products when equilibrium is reached. Free Energy and Spontaneity Changes in H an S during a reaction result in a change in free energy,  G, given by the equation

39 HH SS GG Description – (exo) spon +disorder spon – spon Spontaneous at all T Lets look at relationship among the signs of  H,  S and  G and spontaneity. Note that temperature will dictate which will rule. Also realize T is in K meaning no negative temp. enthalpy rules at low temp but entropy at very high T + (endo) non –disorder non + non Nonspontaneous at all T – – (exo) Spon –disorder non + or – Spontaneous at low T (room);  H > T  S; -  G Nonspontaneous at high T (1000K);  H < T  S +  G + (endo) Non +disorder spon + or – Nonspontaneous at low T;  H > T  S; +  G Spontaneous at high T;  H < T  S; -  G

40 –The next example illustrates the calculation of the standard free energy change,  G o, from  H o and  S o Gibbs Energy and Equilibrium The standard free energy change,  G o, is the free energy change that occurs when reactants and products are in their standard states.

41 So:So: J/mol K Hfo:Hfo: kJ/mol A Problem To Consider What is the standard free energy change,  G o, for the following reaction at 25 o C? predict  H, spon  S, nonspon  G, spon

42 –Now substitute into our equation for  G o. Note that  S o is converted to kJ/K and Kelvin for temp. spon rxn as written So:So: J/mol K Hfo:Hfo: kJ/mol

43 –By tabulating  G f o for substances, you can calculate the  G o for a reaction by using a summation law. Standard Free Energies of Formation The standard free energy of formation,  G f o, 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 o C.

44 Gfo:Gfo: kJ/mol A Problem To Consider Calculate  G o for the following reaction at 25 o C using std. free energies of formation. spon rxn

45 –Here Q is the thermodynamic form of the reaction quotient ([products]/[reactants] not necessarily at equil); T in kelvin; R=8.31 J/molK. Relating  G o to the Equilibrium Constant The free energy change (  G) when reactants are in non-standard states (meaning other than 298K, 1 atm pressure or 1 M) is related to the standard free energy change,  G o, by the following equation.

46 –  G represents an instantaneous change in free energy at some point in the reaction approaching equilibrium  G=0. Relating  G o to the Equilibrium Constant –At equilibrium,  G=0 and the reaction quotient Q becomes the equilibrium constant K.

47 –When K > 1 (meaning equil lies to the right), the ln K is positive and  G o is negative (spon). –When K < 1 (meaning equil lies to the left), the ln K is negative and  G o is positive (nonspon). This result easily rearranges to give the basic equation relating the standard free-energy change to the equilibrium constant. Relating  G o to the Equilibrium Constant

48 –Rearrange the equation  G o = -RTlnK to give A Problem To Consider Find the value for the equilibrium constant, K, at 25 o C (298 K) for the following reaction. The standard free-energy change,  G o, at 25 o C equals –13.6 kJ/mol.

49 –Substituting numerical values into the equation, A Problem To Consider

50 –You get the value of  G T o at any temperature T by substituting values of  H o and  S o at 25 o C into the following equation. Calculation of  G o at Various Temperatures We typically assume that  H o and  S o are essentially constant with respect to temperature.

51 So:So: J/mol K Hfo:Hfo: kJ/mol A Problem To Consider Find the  G o for the following reaction at 25 o C and 1000 o C. Relate this to reaction spontaneity.

52 –Now you substitute  H o,  S o (= kJ/K), and T (=298K) into the equation for  G f o. So the reaction is nonspontaneous at 25 o C. So:So: J/mol K Hfo:Hfo: kJ/mol

53 A Problem To Consider Find the  G o for the following reaction at 1000 o C. –Now we’ll use 1000 o C (1273 K) along with our previous values for  H o and  S o because assume does not change much. So the reaction is spontaneous at 1000 o C. You see that this reaction change from nonspon to spon somewhere between 25 o C to 1000 o C. How can we determine at what temp this switch occurred?  G=0 is equil, switch point

54 –To determine the minimal temperature for spontaneity, we can set  Gº=0 and solve for T. –Thus, CaCO 3 should be thermally stable until its heated to approximately 848 o C. –This is way you could calculate the normal boiling point of a liquid. At  G=0, the liquid phase and gas phase will be at equilibrium; temperature at which switch from liquid to gaseous phase. HW 48 nonspon < 848 o C; CaCO 3 stable spon > 848 o C; CaCO 3 decomposes easily l  g code: six