Slide 1 of Indirect Determination of  H: Hess’s Law  H is an extensive property. – Enthalpy change is directly proportional to the amount of substance in a system. Copyright © 2011 Pearson Canada Inc. General Chemistry: Chapter 7 N 2 (g) + O 2 (g) → 2 NO(g)  H° = kJ ½N 2 (g) + ½O 2 (g) → NO(g)  H° = kJ  H changes sign when a process is reversed NO(g) → ½N 2 (g) + ½O 2 (g)  H° = kJ

Hess’s Law and Enthalpies The ideas on the previous slide are familiar. ∆H is an intensive quantity. → The bigger the steak we want to barbecue the more moles (or grams) of propane we’ll have to burn.  H changes sign when a process is reversed. An analogy here would be climbing up or down a ladder. Climbing up our gravitational potential energy increases – climbing down our potential energy decreases (mgh rule!).

All ∆H’s Need Not Be Measured Elemental nitrogen and oxygen react to form a variety of oxides. We need not measure every ∆H experimentally – some can be calculated using data for a subset of all possible reactions. Reactions forming N X O Y molecules can be exothermic or endothermic. N 2 (g) + O 2 (g) → 2 NO(g) Endothermic In the atmosphere lightning supplies the energy needed to form NO(g).

Nature’s Synthesis of NO(g)! The reaction of N 2 (g) and O 2 (g) to give NO(g) is an endothermic process. In nature, lightning storms supply the needed energy!

All ∆H’s Need Not Be Measured The reaction of NO(g) with additional oxygen is exothermic. NO(g) + ½ O 2 (g) → NO 2 (g) Exothermic Knowing the ∆H for this rxn and the one on the previous slide (by experiment) we can calculate ∆H for the rxn ½ N 2 (g) + O 2 (g) → NO 2 (g) (red brown, LAX)

Smog in Southern California The formation of NO 2 (g) is one factor in smog formation.

Slide 7 of 57 Hess’s Law Schematically Copyright © 2011 Pearson Canada Inc. General Chemistry: Chapter 7

Slide 8 of 57 Hess’s Law of Constant Heat Summation Copyright © 2011 Pearson Canada Inc. General Chemistry: Chapter 7 If a process occurs in stages or steps (even hypothetically), the enthalpy change for the overall process is the sum of the enthalpy changes for the individual steps. ½N 2 (g) + O 2 (g) → NO 2 (g)  H° = kJ ½N 2 (g) + O 2 (g) → NO(g) + ½ O 2 (g)  H° = kJ NO(g) + ½O 2 (g) → NO 2 (g)  H°= kJ

Slide 9 of Standard Enthalpies of Formation, The enthalpy change that occurs in the formation of one mole of a substance in the standard state from the reference forms of the elements in their standard states. Copyright © 2011 Pearson Canada Inc. General Chemistry: Chapter 7 HofHof The standard enthalpy of formation of a pure element in its reference state is 0.

Slide 10 of 57 Liquid bromine vaporizing Copyright © 2011 Pearson Canada Inc. General Chemistry: Chapter 7 Br 2 (l) Br 2 (g)  H f ° = kJ

Slide 11 of 57 Diamond and graphite Copyright © 2011 Pearson Canada Inc. General Chemistry: Chapter 7

Element 79! Gold (Au), like most metals, is a shiny metal at room temperature (standard state!).

You can Mix a Metal and a Nonmetal If the “chemistry” is right! (One standard state?)

Enthalpies of Formation Heats of combustion of hydrocarbons are always –ve (exothermic reactions). Heats of formation of compounds can be +ve or –ve (endothermic and exothermic reactions). ½ H 2 (g) + ½ Cl 2 (g) → HCl(g) ∆H o f = kJ ½ H 2 (g) + ½ I 2 (g) → HI(g) ∆H o f = kJ If we knew the ∆H o f for every substance we could calculate the ∆H o value for every reaction.

Slide 15 of 57 Copyright © 2011 Pearson Canada Inc. General Chemistry: Chapter 7

Slide 16 of 57 Some standard enthalpies of formation at K FIGURE 7-18 Copyright © 2011 Pearson Canada Inc. General Chemistry: Chapter 7 number of carbons

Combining Heats of Formation The process for calculating a general enthalpy change from heat of formation relies on the fact that enthalpy, H, is a state function. Thus, we can imagine proceeding from the reactants to the products by a one step (direct process) or by a two step process: Step 1: Reactants → Constituent Elements Step 2: Constituent Elements → Products

REACTANTS PRODUCTS CONSTITUENT ELEMENTS

Heats of Formation The previous slide outlines a simple process for calculating Heats of Reactions (ΔH o’ s) if the relevant Heats of Formation for all reactants and all products are known. ΔH o Rxn = ΣΔH o f (Products) – ΣΔH o f (Reactants) In practice a particular temperature is specified for the Heats of Formation (usually 298K). Care must be taken to ensure that the correct phase for reactants and products has been specified.

Slide 20 of 57 Standard Enthalpies of Reaction Copyright © 2011 Pearson Canada Inc. General Chemistry: Chapter 7 FIGURE 7-20 Computing heats of reaction from standard enthalpies of formation  H overall = -2  H f ° NaHCO 3 + H f ° Na 2 CO 3 +  H f ° CO 2 +  H f ° H 2 O

Slide 21 of 57 Diagramatic representation of equation (7.21) FIGURE 7-21 Copyright © 2011 Pearson Canada Inc. General Chemistry: Chapter 7 ∆H° = ∑ p ∆H f °(products) - ∑ r ∆H f °(reactants) (7.21)

Examples: Use of Heats of Formation Class examples: Show how to calculate the standard enthalpy change for each of the following reactions using Heats of Formation data. (a)CaCO 3 (s) → CaO(s) + CO 2 (g) (b)2 Al(s) + 3 Cl 2 (g) → 2 AlCl 3 (s) (c) C 3 H 8 (g) + 5 O 2 (g) → 3 CO 2 (g) + 4 H 2 O(l)

Combustion Reactions – History Combustion reactions for hydrocarbons are important sources of energy for many processes. One could easily put together a spreadsheet to calculate the enthalpies of combustion for all hydrocarbons – if the heats of formation of the hydrocarbons were known. Historically this is backwards as enthalpies of combustion are used in most cases to determine heats of formation for hydrocarbons. n-octane will be used as an example.

8CO 2 (g) + 9H 2 O(l) H (Enthalpy) (kJ∙mol -1 ) “Elements” Products Combustion of Liquid n-Octane – C 8 H 18 (l) Reactants

Combustion of n-octane Class example: The standard enthalpy of combustion of n-octane was determined by experiment to be kJ∙mol -1 at 298K. Use heat of formation data for the products of combustion to determine the heat of formation of n-octane.

Combining Thermochemical Equations Unknown ΔH (and ΔU) values can be calculated by combining known thermochemical equations which are often more complex than those written for ΔH o f ’ s. It is often necessary to combine several thermochemical equations to obtain a ΔH o value not known from experiments. This requires that we compare the coefficients of substances appearing only once in the given and desired thermochemical equations.

Combining Themochemical Eqtns - Hydrazine Example: Determine ∆H o for the reaction N 2 H 4 (l) + 2 H 2 O 2 (l) → N 2 (g) + 4 H 2 O(l) Given: N 2 H 4 (l) + O 2 (g) → N 2 (g) + 2 H 2 O(l) ∆H o = kJ H 2 (g) + ½ O 2 (g) → H 2 O(l) ∆H o = kJ H 2 (g) + ½ O 2 (g) → H 2 O 2 (l) ∆H o = kJ

Combining Thermochemical Equations EquationThermochemical EquationMultiply Eq. 1→ 3 by: “Desired”N 2 H 4 (l) + 2H 2 O 2 (l) → N 2 (g) +4 H 2 O(l) ΔH o = ? Equation 1N 2 H 4 (l) + O 2 (g) → N 2 (g) +2H 2 O(l) ΔH o = kJ +1 (look at N 2 H 4 (l)) Equation 2+2 (look at H 2 O(l)) (Use Equation 2 last) Equation 3H 2 (g) + O 2 (g) → H 2 O 2 (l) ΔH o = kJ -2 (look at H 2 O 2 (l))

Results of Combining Equations Equation New Balanced Thermochemical Equations -1 x Eq (1)N 2 H 4 (l) + O 2 (g) → N 2 (g) +2 H 2 O(l) ΔH o = kJ -2 x Eq (3)2 H 2 O 2 (l) → 2 H 2 (g) + 2 O 2 (g) ) ΔH o = kJ +2 x Eq (2) 2 H 2 (g) + O 2 (g) ) → 2 H 2 O(l) ΔH o = kJ SUMN 2 H 4 (l) + 2 H 2 O 2 (l) → N 2 (g) +4 H 2 O(l) ΔH o = kJ