1. Pressure Volume Work Heats of reaction can be measured either at constant pressure, giving q P or ∆H values, or at constant volume, giving q V or ∆U values. ∆H and ∆U values can be appreciably different if there are gaseous reactants or products. The difference arises from gases doing work on the surroundings or the surroundings doing work on a system (normally compressing it). We need to consider PV (pressure- volume) work.

2. Work and Motion In work calculations we need usually to consider both a force(s) and motion. For gas expansion or compression there is obviously motion. One can easily show that, for expansion of a gas in a regular cylinder, the product P∆V has work units. We’ll do this as a class exercise. The easy P∆V calculations have volume in L and pressure in kPa. 1 joule = 1 J = 1 L. kPa = 1 m 3. Pa.

3. System Expanding Does PV Work An expanding system does work on its surroundings. A system can expand because of a chemical reaction or due to being heated. Heating can cause gas expansion or a phase change. He(l) → He(g) V System increases. Why? 2 KClO 3 (s) →2 KCl(s) + 3 O 2 (g) In this 2 nd case V System also increases. Why?

4. Slide 4 of 57 Work During a Chemical Reaction Copyright © 2011 Pearson Canada Inc. General Chemistry: Chapter 7 FIGURE 7-7 Illustrating work (expansion) during the chemical reaction 2 KClO 3 (s) 2 KCl(s) + 3 O 2 (g) In addition to heat effects chemical reactions may also do work. Gas formed pushes against the atmosphere. The volume changes. Pressure-volume work.

5. PV Work With No Chemistry The next slide shows a compressed gas. The system is at mechanical equilibrium unless the pressure on the gas is reduced by removing weights. The expanding gas does work on the surroundings and we write w System = - P External ∆V System or w = -P∆V Aside: If P external varies we will need an integration (not in Chemistry 1050).

6. Slide 6 of 57 Pressure-volume work FIGURE 7-8 Copyright © 2011 Pearson Canada Inc. General Chemistry: Chapter 7 (m x g) = P  V w = -P ext  V A  h  h = x A x h x h w = F x d = (m x g)

7. PV Calculation Using Previous Slide The previous slide can be used to do a simple PV “work term” calculation. This “work term” will be one of two considered for First Law of Thermodynamics calculations. The text provides an example problem (next slide) which we can also tackle slightly differently (Class example).

8. Copyright  2011 Pearson Canada Inc. 7 - 8

9. First Law & Kinetic Theory During the discussion of ideal gases it was suggested that gas molecules are point masses whose kinetic energy increased with T due to increasing translational velocity. Real molecules are not point masses and polyatomic gaseous molecules can also rotate and vibrate. Average rotational and vibrational energies increase with T. Condensed phases are more complicated. Why?

10. Slide 10 of 57 The First Law of Thermodynamics Copyright © 2011 Pearson Canada Inc. General Chemistry: Chapter 7 FIGURE 7-9 Some contributions to the internal energy of a system Internal Energy, U. Total energy (potential and kinetic) in a system. Translational kinetic energy. Molecular rotation. Bond vibration. Intermolecular attractions. Chemical bonds. Electrons.

11. Slide 11 of 57 The First Law of Thermodynamics A system contains only internal energy. – A system does not contain heat or work. – These only occur during a change in the system. Law of Conservation of Energy – The energy of an isolated system is constant Copyright © 2011 Pearson Canada Inc. General Chemistry: Chapter 7  U = q + w

12. Slide 12 of 57 The First Law of Thermodynamics An isolated system is unable to exchange either heat or work with its surroundings, so that  U isolated system = 0, and we can say: Copyright © 2011 Pearson Canada Inc. General Chemistry: Chapter 7 The energy of an isolated system is constant.

13. Thermodynamic Sign Conventions All of us are familiar with the fact that our bank balances increase when we make a deposit (a “positive” event) and decrease when we make a withdrawal (a “negative” event). Similarly, in thermodynamics, we need to keep track of whether our system is losing heat to the surroundings or gaining heat from the surroundings when a physical or chemical change takes place. Work done on or by a system must also be accounted for carefully.

14. Slide 14 of 57 Illustration of sign conventions used in thermodynamics FIGURE 7-10 Copyright © 2011 Pearson Canada Inc. General Chemistry: Chapter 7

15. Slide 15 of 57 Functions of State Copyright © 2011 Pearson Canada Inc. General Chemistry: Chapter 7 Any property that has a unique value for a specified state of a system is said to be a function of state or a state function. Water at 293.15 K and 1.00 atm is in a specified state. d = 0.99820 g/mL This density is a unique function of the state. It does not matter how the state was established.

16. Slide 16 of 57 Functions of State U is a function of state. – Not easily measured.  U has a unique value between two states. – Is easily measured. Copyright © 2011 Pearson Canada Inc. General Chemistry: Chapter 7

17. Slide 17 of 57 Path Dependent Functions Copyright © 2011 Pearson Canada Inc. General Chemistry: Chapter 7 Changes in heat and work are not functions of state. – Remember example 7-5, w = -1.24 x 10 2 J in a one step expansion of gas: let us consider a two step process from 2.40 atm to 1.80 atm, and finally to 1.20 atm.

18. Slide 18 of 57 Path Dependent Functions Copyright © 2011 Pearson Canada Inc. General Chemistry: Chapter 7 FIGURE 7-11 A two-step expansion for the gas shown in Figure 7-8 w = (-1.80 atm)(1.36-1.02)L – (1.30 atm)(2.04-1.36)L = -0.61 L atm – 0.82 L atm = -1.43 L atm = -1.44 x 10 2 J Compared -1.24  10 2 J for the one stage process

19. Copyright  2011 Pearson Canada Inc. 7 - 19

20. First Law Example Ex. A system undergoes a series of four changes in which (a) the system absorbs 4.88 kJ of heat from its surroundings, (b) the system does 1.25 kJ of electrical work on its surroundings, (c) the system gives up 210 J of heat to the surroundings and (d) the surroundings do 1.00 kJ of pressure volume do work on the system. Evaluate ΔU System for the series of four changes outlined.

21. Hydrocarbon Combustion Example Ex. The heat (enthalpy) of combustion of CH 4 (g) is -890 kJ∙mol -1 at 298 K. How much heat is released when 9.00 g of methane is burned at constant pressure at 298 K? “Asides”: This is a highly exothermic reaction. The balanced chemical reaction for the combustion is CH 4 (g) + 2 O 2 (g) → CO 2 (g) + 2 H 2 O(l) Combustion of hydrocarbons drives much of our industry and our transportation systems. The amount of CO 2 (g) being produced is enormous and is thought to contribute to climate change/global warming.

22. Heat Capacity Units Ex. The specific heat capacity of Hg(l) is 0.140 J∙K -1 ∙g -1. Find the molar heat capacity of Hg(l). Hint: The problem is easily solved once the units for molar heat capacity are written.

23. PV Work and Mole Counting For the reactions below no PV work is done in only one case (at constant P and T). Which one? (a) CaC 2 (s) + 2 H 2 O(l) → C 2 H 2 (g) + Ca(OH) 2 (aq) (b) K 2 CO 3 (s) + 2 HCl(aq) → 2 KCl(aq) + H 2 O(l) + CO 2 (g) (c) H 2 (g) + Cl 2 (g) → 2 HCl(g) (d)H 2 (g) + ½ O 2 (g) → H 2 O(l) (e)SO 2 Cl 2 (g) + 2 H 2 O(l) → H 2 SO 4 (aq) + 2 HCl(g)

24. ReactionSystem Expands ? w System Δn Gas (a) CaC 2 (s) + 2 H 2 O(l) → C 2 H 2 (g) + Ca(OH) 2 (aq)Yes -ve +1 (b) K 2 CO 3 (s) + 2 HCl(aq) → 2 KCl(aq) + H 2 O(l) +CO 2 (g)Yes -ve +1 (c)H 2 (g) + Cl 2 (g) → 2 HCl(g)No 0 0 Contracts +ve -1.5 (e) SO 2 Cl 2 (l) + 2 H 2 O(l) → H 2 SO 4 (aq) + 2 HCl(g)Yes -ve +2

25. Heats of Combustion Given time, we will practice writing thermochemical equations corresponding to combustion processes for common organic compounds.