1. Chapter 17 Thermochemistry

2. Thermochemistry: Study of energy changes that occur during chemical reactions and changes in state Section 17.1: The flow of energy

3. energy changes can either occur through heat transfer or work heat (q), energy is transferred from a warmer object to a cooler object (always) Adding heat increases temperature Section 17.1: The flow of energy

4. Kinetic energy vs. potential energy 17.1: Endothermic & Exothermic processes potential energy is determined by the strength of repulsive and attractive forces between atoms In a chemical reaction, atoms are recombined into new arrangements that have different potential energies change in potential energy: due to absorption and release of energy to and from the surroundings

5. Two parameters crucial in Thermochemistry: a) system--part of the universe attention is focused b) surroundings--everything else in the universe System + surrounding = universe Fundamental goal of Thermochem.: study the heat flow between the system and its surroundings 17.1: Endothermic & Exothermic processes

6. If System (energy) Surrounding (energy) by the same amount the total energy of the universe does not change Law of conservation of energy 17.1: Endothermic & Exothermic processes

7. Direction of heat flow is given from the point of view of the system So, endothermic process: system absorbs heat from the surroundings (system heats up) Heat flowing into the system = +q Exothermic process: heat is released into the surroundings (system cools down, -q)

8. 17.1: Endothermic & Exothermic processes Example 1: A container of melted paraffin wax is allowed to stand at room temperature (r.t.) until the wax solidifies. What is the direction of heat flow as the liquid wax solidifies? Is the process exothermic or endothermic? Answer: Heat flows from the system (paraffin) to the surroundings (air) Process: exothermic

9. 17.1: Endothermic & Exothermic processes Example 2: When solid Ba(OH) 2 ▪8H 2 O is mixed in a beaker with solid NH 4 SCN, a reaction occurs. The beaker quickly becomes very cold. Is the reaction exothermic or endothermic? Answer: Endothermic surrounding = beaker and air System = chemicals within beaker

10. 17.1: Units of heat flow Two units used: a) calorie (cal)—amount of heat required to raise the temperature of 1g of pure water by 1 o C b) joule (j)—1 joule of heat raises the temperature of 1 g of pure water 0.2390 o C Joule = SI unit of energy

11. 17.1 Heat capacity & specific heat Heat capacity is the quantity of heat needed to raise the temperature of an object exactly 1 o C Heat capacity depends on: a) mass b) chemical composition So, the greater the mass the greater the heat capacity eg.: cup of water vs. a drop of water (cup of water = greater heat capacity)

12. 17.1: Heat capacity & specific heat Specific heat: amount of heat required to raise the temperature of 1g of a substance by 1 o C Table 17.1 (p.508): List of specific heats of substances

13. Specific heat calculation C = q = heat (joules/calories) m * ΔT mass (g) * ΔTemp. ( o C) ΔT = T f –T i (T f = final temperature) (T i = initial temperature) C = jor cal (g * o C) (g * o C)

14. Example 1 1.The temperature of a 95.4 g piece of copper increases from 25.0 o C to 48.0 o C when the copper absorbs 849 j of heat. What is the specific heat of copper? unknown: C cu Know: mass copper = 95.4 g ΔT = T f – T i = (48.0 o C – 25.0 o C) = 23.0 o C q = 849 j

15. Example1 C = q m * ΔT C = 849 j 95.4 g * 23.0 o C C = 0.387 j/g * o C Sample problem 17.1, page 510

16. Example 2 2.How much heat is required to raise the temperature of 250.0 g of mercury (Hg) 52 o C? unknown: q Know: mass Hg = 250.0 g ΔT = 52.0 o C C Hg = 0.14 j/(g * o C)

17. Example 2 C = q m * ΔT q = C Hg * m * ΔT q = 0.14 (j/g * o C) * 250.0 g * 52 o C q = 1.8 x 10 3 j (1.8 kj) Problem #4 page 510

18. Section 17.2 Measuring Enthalpy Changes

19. 17.2: Enthalpy (measuring heat flow) Calorimetry: accurate measurement of heat flow into or out of a system in chemical and physical processes In calorimetry, heat released by a system is equal to the heat absorbed by its surroundings and vice versa Instrument used to measure absorption or release of heat is a calorimeter

20. Two types of calorimeters: a) Constant-Pressure calorimeter (eg. foam cups) As most reactions occur at constant pressure we can say that: A change in enthalpy (ΔH) = heat supplied (q) So, a release of heat (exothermic) corresponds to a decrease in enthalpy (at constant pressure) An absorption of heat (endothermic) corresponds to an increase in enthalpy (constant pressure) 17.2: Enthalpy (measuring heat flow)

21. b) Constant-Volume Calorimeters (eg. bomb calorimeters) Substance is burned (in the presence of O 2 ) inside a chamber surrounded by water (high pressure) Heat released warms the water Figure 17.6 (p. 512) Bomb calorimeter.

22. 17.2: Thermochemical equations A chemical equation that includes the enthalpy change is called a thermochemical equation Reactants and products at their usual physical state (at 25 o C) given at standard pressure (101.3 kPa) So, the heat of reaction (or ΔH) for the above equation is -65.2 kJ

23. 17.2: Thermochemical equations So, rewrite the equation as follows: Other reactions absorb heat from the surroundings, eg.: Rewrite to show heat of reaction

24. Amount of heat released/absorbed during a reaction depends on the number of moles of reactants involved eg.: 17.2: Thermochemical equations

25. Enthalpy Diagrams Diagram A: Enthalpy of reactants greater than of products Diagram B: Enthalpy of reactant less than of products

26. Physical states of reactants and products must be stated: 17.2: Thermochemical equations Vaporization of H 2 O (l) requires more heat (44.0 kJ)

27. Example1 1. Calculate the amount of heat in (kJ) required to decompose 2.24 mol NaHCO 3(S) Known: 2.24 mol NaHCO 3 decomposes ΔH = 129 kJ (2 mol NaHCO 3 ) Unknown: ΔH = ? Solve: 129 kJ = ΔH 2 mol NaHCO 3(s) 2.24 mol NaHCO 3(s) ΔH = (129 kJ) * 2.24 mol NaHCO 3(s) 2 mol NaHCO 3(s) ΔH = 144 kJ Sample problem 17.3; p. 516

28. Example 2 2. When carbon disulfide is formed from its elements, heat is absorbed. Calculate the amount of heat in (kJ) absorbed when 5.66 g of carbon disulfide is formed. Known: 5.66 g CS 2 is formed ΔH = 89.3 kJ (1 mol CS 2(l) ) Molar mass: CS 2(l) : C = 12.0 g/mol 2 *S = 32.1 g/mol = 64.2 g/mol 76.2 g/mol Unknown: ΔH = ?

29. Example 2 Solve: 1. Moles CS 2(l) = 5.66g CS 2 = 0.0743 mol CS 2(l) 76.2 g/mol CS 2(l) 2. 89.3 kJ = ΔH 1 mol CS 2(l) 0.074 mol CS 2(l) ΔH = (89.3 kJ) * 0.074 mol CS 2(l) 1 mol CS 2(l) ΔH = 6.63 kJ

30. 17.3: Heat in changes of state Objective: -Heats of Fusion and Solidification -Heats of Vaporization and Condensation -Heat of solution

31. The temperature remains constant when a change of state occurs via a gain/loss of energy Heat absorbed by 1 mole of a solid during melting (constant temperature) is the molar heat of fusion (ΔH fus ) Molar heat of solidification (ΔH solid ) is the heat lost by 1 mole of liquid as it solidifies (constant temperature) So, ΔH fus = ΔH solid 17.3: Heat of fusion and solidification

32. Figure 17.9: Enthalpy changes and changes of state 17.3: Heat of fusion and solidification

34. 17.3: Heats of Vaporization and Condensation Molar heat of vaporization (ΔH vap ): Amount of heat required to vaporize one mole of a liquid at the liquid’s normal boiling point Molar heat of condensation (ΔH cond ): heat released when 1 mole of vapor condenses So, ΔH vap = -ΔH cond

35. Figure 17.10: Heating curve of water 17.3: Heat of vaporization and condensation

36. 17.3: Heat of solution There is heat released/gained when a solute dissolves in a solvent The enthalpy change due to 1 mole of a substance dissolving: molar heat of solution (ΔH soln )

37. Applications: hot/cold packs Hot pack: 17.3: Heat of solution Cold pack: