1. Hess’s Law of Additivity of Reaction Enthalpies 1

2.  I will be able to use proper scientific terminology to describe Hess’s Law  I will be able to use Hess’s law to calculate enthalpies in multiple step reactions 2

3.  describe Hess’s law  use Hess’s law to calculate enthalpies in multiple step reactions 3

4.  Calorimetry is an excellent technique for determining enthalpy changes, BUT is no longer useful when reactions are too slow and temperature changes would be too small to measure  Chemists utilize methods to deal with this problem based on the principle that net (or overall) changes in some properties of a system are independent of the way the system changes from initial state (reactants) to the final state (products) 4

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6. The value of the ∆H for any reaction that can be written in steps equals the sum of the values of ∆H for each of the individual steps. OR If two or more equations with known enthalpy changes can be added together to form a new “target” equation, then their enthalpy changes may be similarly added together to yield the enthalpy change of the target equation 6

7. Written as an equation ∆H target = ∆H 1 + ∆H 2 + ∆H 3 + … ∆H target = ∑ ∆H known 7

8. Example 1 Carbon can react with oxygen to form carbon monoxide. The carbon monoxide then reacts with oxygen to produce carbon dioxide. The two equations below represent this pathway. C (s) + ½ O 2(g)  CO (g) ∆Hº = -110.5 kJ CO (g) + ½ O 2(g)  CO 2(g) ∆Hº = -283.0 kJ Carbon can also react with oxygen to produce carbon dioxide directly. C (s) + O 2(g)  CO 2(g) ∆Hº = -393.5 kJ 8

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10.  Hess’s Law allows chemists to determine the enthalpy change of a reaction without direct calorimetry, using two familiar rules for chemical equations and enthalpy changes 1.If a chemical equation is reversed, then the sign of the ∆H changes. 2.If the coefficients of a chemical equation are altered by multiplying or dividing by a constant factor, then the ∆H is altered in the same way. 10

11. Practice 1 A typical automobile engine uses a lead-acid battery. During discharge, the following chemical reaction takes place. 2PbO (s) + 2H 2 SO 4(l)  2PbSO 4(aq) + 2H 2 O (l) Determine the enthalpy change of this reaction, given the following equations. (1) 2PbO (s) + 2SO 3(g)  2PbSO 4(s) ∆Hº = -775 kJ (2) SO 3(g) + H 2 0 (l)  H 2 SO 4(l) ∆Hº = -133 kJ Answer: -509 kJ 11

12. Practice 2 Mixing household cleansers can result in the production of hydrogen chloride gas, HC1(g). Not only is this gas dangerous in its own right, but it also reacts with oxygen to form chlorine gas and water vapour. 4HCl (g) + O 2(g)  2Cl 2(g) + 2H 2 O (g) Determine the enthalpy change of this reaction, given the following equations. (1) H 2(g) + Cl 2(g)  2HC1 (g) ∆Hº = -185 kJ (2) H 2(g) + ½ O 2(g)  H 2 O (l) ∆Hº = -285.8 kJ (3) H 2 O (g)  H 2 O (l) ∆Hº = -40.7 kJ Answer: -120.2 kJ 12

13. 13 Practice pp. 326-327 UC # 1, 2, 3 Multi-Step Enthalpy Practice p. 329 UC # 4, 5 Additional Practice p. 330 UC # 1, 2, 3