Chemistry 30 – Unit 1B Thermochemical Changes To accompany Inquiry into Chemistry PowerPoint Presentation prepared by Robert Schultz

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

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

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

Chapter 10, Section 10.1 Practice Problem 1, page 374 – find ∆H for the reaction: C 2 H 4 (g) + H 2 O(l) C 2 H 5 OH(l), using the following equations: (1)C 2 H 5 OH(l) + 3 O 2 (g) 3 H 2 O(l) + 2 CO 2 (g) ∆H= kJ (2)C 2 H 4 (g) + 3 O 2 (g) 2 H 2 O(l) + 2 CO 2 (g) ∆H= kJ The desired equation needs C 2 H 5 OH(l) on the right, so flip equation 1 It needs C 2 H 4 (g) on the left, so leave equation 2 as is Strategy:

Chapter 10, Section H 2 O(l) + 2 CO 2 (g) C 2 H 5 OH(l) + 3 O 2 (g) ∆H= kJ C 2 H 4 (g) + 3 O 2 (g) 2 H 2 O(l) + 2 CO 2 (g) ∆H= kJ C 2 H 4 (g) + H 2 O(l) C 2 H 5 OH(l) ∆H= kJ + Add the equations and their ∆H’s Note the sign change

Chapter 10, Section 10.1 Potential Energy Diagram for previous example: Reaction Coordinate E p (kJ) C 2 H 4 (g) + 3 O 2 (g) 2 H 2 O(l) + 2 CO 2 (g) + H 2 O(l) C 2 H 5 OH(l) + 3 O 2 (g) ∆H = kJ ∆H = kJ ∆H=-44.2 kJ

Chapter 10, Section 10.1 Example: Practice Problem 2, page 374 Find ∆H for: Pb(s) + PbO 2 (s) + 2 H 2 SO 4 (l) 2 PbSO 4 (aq) + 2 H 2 O(l) using the equations: (1) Pb(s) + PbO 2 (s) + 2 SO 3 (g) 2 PbSO 4 (aq) ∆H o = -775 kJ (2) SO 3 (g) + H 2 O(l) H 2 SO 4 (l) ∆H o = -133 kJ Strategy: Use equation (1) as is, double equation (2) and flip it

Chapter 10, Section 10.1 Pb(s) + PbO 2 (s) + 2 SO 3 (g) 2 PbSO 4 (aq) ∆H o = -775 kJ 2 H 2 SO 4 (l) 2 SO 3 (g) + 2 H 2 O(l) ∆H o = +266 kJ Pb(s) + PbO 2 (s) + 2 H 2 SO 4 (l) 2 PbSO 4 (aq) + 2 H 2 O(l) ∆H o = -509 kJ + Handout BLM Do Nelson Chemistry WS 48 Lab 10.A page 375

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

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: S 8 (s) + O 2 (g) SO 2 (g) 2 C(s) + 3 H 2 (g) + ½ O 2 (g) C 2 H 5 OH(l)

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

Chapter 10, Section 10.1 Substances with very large negative 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 H 2 (g) CH 4 (g) kJ

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

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

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

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

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

Chapter 10, Section 10.1 Practice Problem 10a, page 383 CH 3 OH(l) + 3/2 O 2 (g) CO 2 (g) + 2 H 2 O(g)

Chapter 10, Section b Since the equation is balanced with 1 mol of CH 3 OH(l), therefore, and,

Chapter 10, Section 10.1 Example: Practice Problem 12, page 383 C 7 H 16 (l) + 11 O 2 (g) 7 CO 2 (g) + 8 H 2 O(l) Good vocabulary question!

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

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

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

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

Chapter 10, Section 10.2 How do the relative amounts of energy from phase, chemical, and nuclear changes compare? phase changes: 10 0 to 10 2 kJ/mol source: intermolecular forces (bonds) chemical changes: 10 2 to 10 5 kJ/mol source: chemical bonds (also called intramolecular bonds) nuclear changes: 10 6 to kJ/mol source: nuclear bonds or intranuclear bonds, the bonds that hold particles of the nucleus together

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

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

Chapter 10, Section 10.2 Since equation is balanced with 1 mol of C 4 H 10, C 4 H 10 (g) + 13/2 O 2 (g) 4 CO 2 (g) + 5 H 2 O(g)

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:

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:

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

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

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

Chapter 10, Section 10.3