Thermochemistry

1. Thermochemistry Chapter 6 HW: Vocab and #2-62 every other even

2. Basics Thermochemistry – Law of Conservation of Energy –

3. Nature of Energy E is known and recognized as – E is defined as –

4. Types of Energy KE – PE – Radiant – Thermal – Chemical –

5. Units of Energy SI – Non SI – 1kcal = 1lb burned approx. =

6. Work Work – Chemical Work – System – Surroundings –

7. Types of Systems Open – Closed – Isolated –

8. Heat vs Temp Heat – Temp –

9. Types of Rxns Exothermic – Endothermic –

10. Enthalpy (ΔH) Enthalpy (H) * * * *

11. Enthalpy cont. * * * *

12. Thermochemical Equations An equation that shows the ΔH as well as the mass relationships H2O (s)  H2O (l) ΔH = 6.01kJ * *

13. Thermochemical Equations cont. * *

14. Thermochemical Equations cont. Problem: Calculate the heat evolved when 266g of white phosphorus burns in air according to the equation P4(s) + 5O2(g)  P4O10(s) ΔH = -301.3 kJ

15. Calorimetry • Measurement • Depends on

16. Specific Heat • Symbol is • Defined as • Is an • Has units of • Constant values that are in Table 6.1 or in the chemical handbook • Constant for water is –*know*

17. Heat Capacity • Symbolized by a • Defined as • Is an • Has units of • Not as commonly used at c • Equation to convert

18. Specific Heat Equation • q=mcΔT • q= • Negaive amt = • Positive amt = • m= • c= • ΔT= • ΔT=

19. Example Problem • A 110g sample of copper is heated to 82.4oC and then placed in a container of water at 22.3oC. The final T of the water and the copper is 24.9oC. What was the mass of the water in the original container assuming that all the heat lost by the copper is gained by the water?

20. Types of Calorimetry 1. Bomb calorimeter • Measures heats of • System is • qsystem = • qrxn =

21. Types of Calorimeter • Regular Calorimeter • Measures heats of • Simple device • qsystem= • qrxn=

22. Calorimetry • We will keep all calculations basic with the initial equation of

23. Heating Curve

24. Heating Curve The T of the system usually increases when E is applied. However, when the E supplied is used for phase transition, a change in physical state, the

25. Standard Enthalpy • We can find H of the rxn when we know the individual H of each reactant and product • ΔHorxn=∑ΔHof products – ∑ΔHof reactants

26. Standard Enthalpy • ΔHorxn=∑ΔHof products – ∑ΔHof reactants • ∑ΔHof is the standard heat of formation • Is the heat change that • T is at a standard • Constant values can be found in Table 6.3 and Appendix 3 • Elemental states are • Must consider the

27. Standard Enthalpy • Calculations can be made 2 ways: 1. Direct Method 2. Indirect Method or Hess’s Law

28. Direct Method • Look up all values for reactants and products in and put into equation multiplying by the mole values in the balanced reaction

29. Practice Problems • Calculate the standard heat of formation for the reaction below: 2H202(l)  2H20(l) + O2(g) • Calculate the standard heat of formation of gaseous carbon monoxide based on the following reaction: 2CO(g) + C(s)  C3O2(g) ΔHo= 127.3kJ

30. Indirect Method a.k.a. Hess’s Law • Based on • When reactions are converted to products, the change in H is the same if it takes place in • Arrange a series of steps in such a way that when added,

31. Hess’s Law • _______________of the individual steps is sometimes necessary to achieve the overall reaction

32. Practice Problems Calculate the ΔH if given the following information: Indiv. Rxns ΔHo Si(s) + 2H2(g)  SiH4(g) 34 kJ/mol Si(s) + O2(g)  SiO2(s) -911kJ/mol H2(g) + ½O2(g)  H2O(g) -242kJ/mol Overall SiH4(g) + 2O2(g)  SiO2(s) + 2H2O(g)

33. Practice Problems • p. 221 Practice Exercise

34. Some interesting quantities of energy.

35. C4H10(l) + 13/2O2(g) 4CO2(g) + 5H2O(g) K(s) + 1/2Br2(l) KBr(s) NaCl(s) NaCl(l) C6H6(l) C6H6(g) Some Important Types of Enthalpy Change heat of combustion (DHcomb) heat of formation (DHf) heat of fusion (DHfus) heat of vaporization (DHvap)

36. DP = Pfinal - Pinitial DV = Vfinal - Vinitial DT = Tfinal - Tinitial Thermodynamics is the scientific study of the interconversion of heat and other kinds of energy. State functions are properties that are determined by the state of the system, regardless of how that condition was achieved. energy , pressure, volume, temperature DE = Efinal - Einitial Potential energy of hiker 1 and hiker 2 is the same even though they took different paths.

37. C3H8 + 5O2 3CO2 + 4H2O Chemical energy lost by combustion = Energy gained by the surroundings system surroundings First law of thermodynamics – energy can be converted from one form to another, but cannot be created or destroyed. DEsystem + DEsurroundings = 0 or DEsystem = -DEsurroundings Exothermic chemical reaction!

38. First Law of Thermodynamics = Law of conservation of energy DEuniverse = DEsystem + DEsurroundings = 0 Units of Energy Joule (J) 1 J = 1 kg*m2/s2 Calorie (cal) 1 cal = 4.184 J British Thermal Unit 1 Btu = 1055 J

39. Figure 6.2 Energy diagrams for the transfer of internal energy (E) between a system and its surroundings. DE = Efinal - Einitial = Eproducts - Ereactants

40. Energy, E Work (w) done on surroundings (w < 0) DE<0 A system losing energy as work only. Figure 6.4 Zn(s) + 2H+(aq) + 2Cl-(aq) H2(g) + Zn2+(aq) + 2Cl-(aq)

41. Another form of the first law for DEsystem DE = q + w DE is the change in internal energy of a system q is the heat exchange between the system and the surroundings w is the work done on (or by) the system w = -PDVwhen a gas expands against a constant external pressure

42. Table 6.1 The Sign Conventions* for q, w and DE + = DE q w + + + + - depends on sizes of q and w - + depends on sizes of q and w - - - * For q: + means system gains heat; - means system loses heat. * For w: + means work done on system; - means work done by system.

43. PROBLEM: When gasoline burns in a car engine, the heat released causes the products CO2 and H2O to expand, which pushes the pistons outward. Excess heat is removed by the car’s cooling system. If the expanding gases do 451 J of work on the pistons and the system loses 325 J to the surroundings as heat, calculate the change in energy (DE) in J, kJ, and kcal. kcal kJ -776 J -0.776 kJ 4.184 kJ 103J Determining the Change in Internal Energy of a System PLAN: Define system and surroundings, assign signs to q and w and calculate DE. The answer should be converted from J to kJ and then to kcal. SOLUTION: q = - 325 J w = - 451 J DE = q + w = -325 J + (-451 J) = -776 J = -0.776 kJ = -0.185 kcal

44. DV > 0 -PDV < 0 P x V = x d3 = F x d = w wsys < 0 F d2 Dw = wfinal - winitial Work Done On the System w = F x d w = -P DV Work is not a state function. initial final

45. Work done on the system Pressure-volume work.

46. Practice Problem A sample of nitrogen gas expands in volume from 1.6L to 5.4L at constant temperature. What is the work done in joules if the gas expands: a) against a vacuum b) against a constant pressure of 3.7atm

48. Chemistry in Action: Making Snow DE = q + w q = 0 w < 0, DE < 0 DE = CDT DT < 0, SNOW!

49. Enthalpy and the First Law of Thermodynamics DE = q + w At constant pressure: q = DH and w = -PDV DE = DH - PDV DH = DE + PDV

50. The Meaning of Enthalpy w = - PDV DH ≈ DE in H = E + PV 1. Reactions that do not involve gases. where H is enthalpy 2. Reactions in which the number of moles of gas does not change. DH = DE + PDV qp = DE + PDV = DH ΔE = ΔH – Δ(PV) ΔE = ΔH – Δ(nRT) ΔE = ΔH – RTΔn 3. Reactions in which the number of moles of gas does change but q is >>> PDV.