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Chemistry 30 – Unit 1B Thermochemical Changes

To accompany Inquiry into Chemistry. PowerPoint Presentation prepared by Robert Schultz robert.schultz@ei.educ.ab.ca. Chemistry 30 – Unit 1B Thermochemical Changes. Section 10.1. Hess’s Law: The enthalpy change of a chemical or physical process depends only on the initial and final systems

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Chemistry 30 – Unit 1B Thermochemical Changes

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  1. To accompany Inquiry into Chemistry PowerPoint Presentation prepared by Robert Schultz robert.schultz@ei.educ.ab.ca Chemistry 30 – Unit 1B Thermochemical Changes

  2. Section 10.1 • Hess’s Law: The enthalpy change of a chemical or physical process depends only on the initial and final systems • It is the sum of the of the enthalpy changes of the individual processes

  3. Chapter 10, Section 10.1 • The picture illustrates this – the gravitational Ep change for the two riders is the same • It doesn’t depend on the pathway

  4. Chapter 10, Section 10.1 • This principle, Hess’s Law, can be used to determine ∆H for reactions when ∆H can’t be measured by calorimetry • When using Hess’s Law, don’t worry about pure elements – get compounds with their appropriate coefficients on the proper side of the equation and cancel out any new compounds you add

  5. Chapter 10, Section 10.1 • Practice Problem 1, page 374 – find ∆H for the reaction: C2H4(g) + H2O(l) C2H5OH(l) , using the following equations: • C2H5OH(l) + 3 O2(g) 3 H2O(l) + 2 CO2(g) ∆H=-1366.8 kJ • C2H4(g) + 3 O2(g) 2 H2O(l) + 2 CO2(g) ∆H=-1411.0 kJ • The desired equation needs C2H5OH(l) on the right, so flip equation 1 • It needs C2H4(g) on the left, so leave equation 2 as is Strategy:

  6. Note the sign change Add the equations and their∆H’s Chapter 10, Section 10.1 + • 3 H2O(l) + 2 CO2(g) C2H5OH(l) + 3 O2(g)∆H=+1366.8 kJ • C2H4(g) + 3 O2(g) 2 H2O(l) + 2 CO2(g) ∆H=-1411.0 kJ C2H4(g) + H2O(l) C2H5OH(l) ∆H= -44.2 kJ

  7. C2H4(g) + 3 O2(g) ∆H=-44.2 kJ C2H5OH(l) + 3 O2(g) Ep (kJ) ∆H =-1411.0 kJ ∆H =+1366.8 kJ 2 H2O(l) + 2 CO2(g) + H2O(l) Reaction Coordinate Chapter 10, Section 10.1 • Potential Energy Diagram for previous example:

  8. Pb(s) + PbO2(s) + 2 H2SO4(l) 2 PbSO4(aq) + 2 H2O(l) • using the equations: • (1) Pb(s) + PbO2(s) + 2 SO3(g) 2 PbSO4(aq) ∆Ho = -775 kJ • SO3(g) + H2O(l) H2SO4(l) ∆Ho = -133 kJ • Strategy:Use equation (1) as is, double equation (2) and flip it Chapter 10, Section 10.1 Example: Practice Problem 2, page 374 Find ∆H for:

  9. + Chapter 10, Section 10.1 Pb(s) + PbO2(s) + 2 SO3(g) 2 PbSO4(aq) ∆Ho = -775 kJ 2 H2SO4(l) 2 SO3(g) + 2 H2O(l) ∆Ho = +266 kJ Pb(s) + PbO2(s) + 2 H2SO4(l) 2 PbSO4(aq) + 2 H2O(l) ∆Ho = -509 kJ Handout BLM 10.1.3 Do Nelson Chemistry WS 48 Lab 10.A page 375

  10. Chapter 10, Section 10.1 • With the right set of reaction enthalpy data and reaction equations, using Hess’s Law, you could calculate the enthalpy (change) for any chemical reaction • Chemists have tabulated data and prepared charts of standard molar enthalpies of formation

  11. Chapter 10, Section 10.1 • A formation reaction is a reaction that forms one mole of product from its elements in their standard states at SATP • Examples: • S8(s) + O2(g) SO2(g) • 2 C(s) + 3 H2(g) + ½ O2(g) C2H5OH(l)

  12. Chapter 10, Section 10.1 • Your Chemistry 30 Data Booklet, on pages 4 and 5, has a chart of Standard Molar Enthalpies of Formation • Standard molar enthalpies of formation can be used to: •predict thermal stability • write formation equations including energy terms •predict ∆H for other reactions

  13. Chapter 10, Section 10.1 • Substances with very largenegative enthalpies of formation are very thermally stable since it takes a lot of heat to decompose them to their elements • Which compound on your chart is most stable? Least stable? • Example: question 4a, page 379 • C(s) + 2 H2(g) CH4(g) + 74.6 kJ

  14. Chapter 10, Section 10.1 • Using enthalpies of formation, Hess’s Law becomes a shorter process • ∆H’s can be found using the following formula: Σ (sigma) is a Greek letter meaning “sum of” stands for standard molar enthalpy of formation n is moles of substance from the balanced equation

  15. Chapter 10, Section 10.1 • Why does this work? Because standard molar enthalpies of formation refer to reactions with only elements on the left side, and you don’t need to worry about elements when using Hess’s Law

  16. Chapter 10, Section 10.1 • Calculate the enthalpy change for: 2 ZnS(s) + 3 O2(g) 2 ZnO(s) + 2 SO2(g) Data: p. 4-5 of Data Booklet: Now use these numbers directly in the formula. No need to change signs. Why?!

  17. Chapter 10, Section 10.1 The subtraction changes the sign of the reactants Products are unchanged since they are already on the proper side Discuss significant digits in these calcs

  18. Chapter 10, Section 10.1 • Try Practice Problem 10, page 383 10b is an extension of the question, but you have already done this part in Chapter 9

  19. Chapter 10, Section 10.1 • Practice Problem 10a, page 383 CH3OH(l) + 3/2 O2(g) CO2(g) + 2 H2O(g)

  20. 10b Since the equation is balanced with 1 mol of CH3OH(l), • therefore, Chapter 10, Section 10.1 and,

  21. Chapter 10, Section 10.1 • Example: Practice Problem 12, page 383 C7H16(l) + 11 O2(g) 7 CO2(g) + 8 H2O(l) Good vocabulary question!

  22. Chapter 10, Section 10.1 • Do Nelson Chemistry Worksheet 49 • Do Nelson Chemistry Worksheet 50 • Do Worksheet 50A and Diploma Exam written response questions • Discuss Review Questions 2, 4, and 5 page 383

  23. Chapter 10, Section 10.2 • Canada, as your text says, has a very large per capita energy requirement (2nd only to the USA) • Why? • cold winter • large physical size leading to high transport costs • others?

  24. Chapter 10, Section 10.2 • In Canada, most of our energy is obtained from the exothermic combustion of fuels • A significant percentage comes from nuclear fission reactors

  25. Chapter 10, Section 10.2 • In Alberta, even our electricity comes mostly from combustion of fuels: • In 2008, 48.8 % came from combustion of coal • 38.4% came from combustion of natural gas • 7.1% came from hydroelectricity

  26. Chapter 10, Section 10.2 • How do the relative amounts of energy from phase, chemical, and nuclear changes compare? • phase changes: 100 to 102 kJ/molsource: intermolecular forces (bonds) • chemical changes: 102 to 105 kJ/molsource: chemical bonds (also called intramolecular bonds) • nuclear changes: 106 to 1010 kJ/molsource: nuclear bonds or intranuclear bonds, the bonds that hold particles of the nucleus together Not required for diploma exam

  27. Chapter 10, Section 10.2 • Efficiency – the ratio of useful energy produced (energy output) to energy used in its production (energy input), in % form

  28. Chapter 10, Section 10.2 • Example: Practice Problem 13, page 387 • Note that the molar enthalpy of combustion chart referred to in the question is not present in your Data Booklet, so we will do this question without the aid of the table

  29. Chapter 10, Section 10.2 • C4H10(g) + 13/2 O2(g) 4 CO2(g) + 5 H2O(g) Since equation is balanced with 1 mol of C4H10 ,

  30. Chapter 10, Section 10.2 • If the butane lighter was 100% efficient in heating the spoon, 0.70 g of butane would release: • Actual amount of useful energy absorbed:

  31. Chapter 10, Section 10.2 • Try Practice Problem 14, page 387 • Aluminium’s specific heat capacity, which you’ll need for the question is given on page 3 of the Data Booklet with a few others:

  32. Chapter 10, Section 10.2 • Practice Problem 14, page 387 • Energy input: • Useful energy output:

  33. Chapter 10, Section 10.2 • The previous problem, Practice Problem 14, page 387 is a more difficult question than you could expect to see on the Diploma Exam or on my test

  34. Chapter 10, Section 10.2 – 10.3 • Read pages 388 to 398 • Why does efficiency decrease dramatically as the number of stages in a process increases? • Do question 8 page 392 and questions 3, 4, 5, and 6 page 398

  35. Chapter 10, Section 10.3

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