CHAPTER 16. REACTION ENERGY SECTION 1. THERMOCHEMISTRY

1. CHAPTER 16. REACTION ENERGY SECTION 1. THERMOCHEMISTRY • Thermochemistryis the study of the transfers of energy as heat that accompany chemical reactions and physical changes. • Heatis energy transferred between samples of matter because of a difference in their temperature.

2. Temperatureis a measure of the average kinetic energy of the particles in a sample of matter. • Heat will move from a sample at higher temperature to one at lower temperature. • Like other forms of energy, the SI unit of heat is the joule.

3. Enthalpy of Reaction (commonly referred to as “heat of reaction”) Enthalpy can be considered as stored energy in a substance. It is not measured directly. Rather, we measure changes in enthalpy during chemical reactions. Change is indicated using the Greek letter delta = Δ H is the symbol for enthalpy.

4. ΔH = enthalpy change = amount of heat energy absorbed by a system during a process at constant pressure. Enthalpy change is the difference between the enthalpies of products and reactants. ∆H = Hproducts– Hreactants

5. Exothermic Reactions If the products have less enthalpy than the reactants, then ΔH < 0, and heat is released. Heat can be indicated as if it were a product of the reaction. This is an exothermic reaction.

6. Ex.: • 2H2(g) + O2(g) → 2H2O(g) + 483.6 kJ • If 2 mol H2 react with 1 mol O2 to produce 2 mol H2O, 483.6 kJ of heat are produced. • If different numbers of moles react, the energy released will be changed in proportion.

7. Energy Diagram of Exothermic Reaction:

8. Endothermic Reactions If the products have more enthalpy than the reactants, then ΔH > 0, and heat is absorbed. Heat can be indicated as if it were a reactant. This is an endothermic reaction. Ex. – cold pack in which NH4NO3 dissolves in water. The reaction absorbs 26 kJ per mol. NH4NO3(s) + 26 kJ → NH4NO3(aq)

9. Energy Diagram of Endothermic Reaction

10. Thermochemical equations are usually written by giving a ∆H value rather than writing the energy as a reactant or product. For the two examples given previously: 2H2(g) + O2(g) → 2H2O(g) ΔH = -483.6 kJ NH4NO3(s) → NH4NO3(aq) ΔH = +26 kJ Note that ∆H is negative for an exothermic reaction and positive for an endothermic reaction.

11. Enthalpy of Formation • Themolar enthalpy of formationis the enthalpy change that occurs when one mole of a compound is formed from its elements in their standard state at 25°C and 1 atm. • Standard state = state found at atmospheric pressure and 25° C. Ex. – O2 is a gas, Hg is liquid.

12. To signify standard states, a 0 sign is added to the enthalpy symbol, and the subscript f indicates a standard enthalpy of formation: Elements in their standard states by definition have = 0

13. Appendix Table A-14, p. 862, lists standard enthalpies of formation of some compounds. • Many of these are negative, indicating compounds that are stable compared to the elements (energy is released when they are formed).

14. Ex.: the of carbon dioxide is • –393.5 kJ per mol of gas produced. • This means that for the reaction • C(s) + O2(g) → CO2(g) at 25° C, • ΔH = -393.5 kJ/mol • Ethyne (acetylene) has • = +226.7 kJ/mol. It is very unstable.

15. Calculating Enthalpies of Reaction • Hess’s law:the overall enthalpy change in a reaction is equal to the sum of enthalpy changes for the individual steps in the process. It is independent of the path by which reactants go to products.

16. This lets you calculate H for a reaction that is difficult to measure. It also lets you calculate H from tables of enthalpies of formation (Hf0).

17. In using this law: • If a reaction is reversed, the sign of H is reversed. • If it is necessary to change all the coefficients of a reaction, change H in the same proportion.

18. Ex. 1 (p. 540): calculate the enthalpy of formation of methane gas, CH4, from its elements, hydrogen gas and solid carbon: C(s) + 2H2(g) → CH4(g) The calculation can be done using the enthalpies of combustion of C, H2, and CH4.

19. These reactions are: C(s) + O2(g) → CO2(g) H2(g) + ½O2(g) → H2O(l) CH4(g) + 2O2(g) → CO2(g) + 2H2O(l)

20. We can get the desired overall reaction if we double the second reaction (and also double H), reverse the third reaction (and reverse the sign of H), and add them together. To get the overall H, we add the three H values.

21. C(s) + O2(g) → CO2(g) H = -393.5 2H2(g) + O2(g) → 2H2O(l) H = 2(-285.8) = -571.6 CO2(g) + 2H2O(l) → CH4(g) + 2O2(g) H = +890.8 C(s) + 2H2(g) → CH4(g) H = -74.3 kJ

22. If you know the Hf0 for all reactants and products, H can be calculated for any reaction by subtracting the total Hf0s of the reactants from those of the products (all adjusted for the number of moles).

23. Ex. 2 (p. 541) Calculate H for NO(g) + 2O2(g) → NO2(g) Using Tab. A14: nitrogen monoxide Hf0 = +90.29 kJ/mol nitrogen dioxide Hf0 = +33.2 kJ/mol

24. This means that: ½N2(g) + ½O2(g) → NO(g) H = +90.29 ½ N2(g) + O2(g) → NO2(g) H = +33.2 Rewrite these so that the NO is on the left; keep NO2 on the right: NO(g) → ½N2(g) + ½O2(g) H = -90.29 ½ N2(g) + O2(g) → NO2(g) H = +33.2 Net: NO(g) + 2O2(g) → NO2(g) H = -90.29 + 33.2 = - 57.1

25. Note that H = Hf0(NO2) -Hf0(NO) product reactant

26. SECTION 2. DRIVING FORCE OF REACTIONS • In this section we consider the question of whether a reaction will proceed spontaneously. • For a reaction at constant temperature and pressure, this is determined by two factors: • enthalpy, which we have discussed • entropy

27. Entropy (abbreviation: S) can be considered a measure of randomness of the particles of a system. The higher the entropy, the more disordered the system is, and the more difficult to describe the location of the particles.

28. Entropy – 75329

29. In general, a gas has the most entropy, followed by a liquid, then a solid. A pure crystalline solid has zero entropy at absolute zero ( = ). The unit of entropy are joules per kelvin (J/K) or kJ/K. For one mole, it is kJ/(mol•K).

30. In general, a gas has the most entropy, followed by a liquid, then a solid. A pure crystalline solid has zero entropy at absolute zero ( = 0 K = -273˚C ). The unit of entropy are joules per kelvin (J/K) or kJ/K. For one mole, it is kJ/(mol•K).

31. Entropy and Reaction Tendency • Some processes that result in an increase in entropy (S is positive): • melting of ice • forming a solution • mixing gases • An increase in entropy tends to make a reaction favorable, since systems spontaneously go to more disordered states.

32. Ex.: The decomposition of ammonium nitrate: 2NH4NO3(s) 2N2(g) + 4H2O(l) + O2(g) • On the left side are 2 mol of solid ammonium nitrate. • The right-hand side of the equation shows 3 mol of gaseous molecules plus 4 mol of a liquid. • The arrangement of particles on the right-hand side of the equation is more random than the arrangement on the left side.

33. Standard Entropy Changes for Some Reactions

34. A decrease in enthalpy (H is negative) tends to make a reaction favorable, since it means moving to more stable chemicals. If the S and H effects are acting in opposite directions, how do we decide if the reaction is favorable?

35. Free Energy The combined effects of enthalpy and entropy are expressed in the function free energy (abbreviated G, for Gibbs free energy). G = H - TS (T = Kelvin temp.) units: kJ or kJ/mol

36. J. Willard Gibbs en.wikipedia.org

37. If G < 0, the reaction is favorable (it could spontaneously go to the right). Here are the possible combinations of H and S:

38. G = H - TS Note that as T increases, the effect of entropy becomes more important.

39. Relating Enthalpy, Entropy, and Free-Energy Changes75328

40. Ex. 1: C2H4(g) + H2(g) → C2H6(g) H = -136.9 kJ/mol (favorable) S = -0.121 kJ/(mol•K) (unfavorable) At 298 K: G = H - TS = -136.9 – 298(-0.121) = -136.9 + 36.1 = -100.8 kJ/mol Since this is negative, the reaction is favorable.

41. Ex. 2: CH4(g) + H2O(g) → CO(g) + 3H2(g) H = +206.1 kJ/mol (unfavorable) S = +0.215 kJ/(mol•K) (favorable) At 298 K: G = H - TS = +206.1 – 298(0.215) = +206.1 – 64.1 = +142.0 kJ/mol Since this is positive, the reaction is unfavorable.