Chap 6 Sec 5

# Chap 6 Sec 5 - ∆ H for the reaction Ca H 2 O 2> Ca(OH...

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Chap 6 Sec 5 notes F09.doc Page 1 Hess’s Law: Chap 6 Section 5 Hess’s Law You can add thermochemical equations up just like algebraic ones and the H values just follow along. Example 1 Use Rxn 1 H 2 + 1/2 O 2 -----> H 2 O(l) H = 285.8 kJ/mole and Rxn 2 H 2 + 1/2 O 2 -----> H 2 O( g ) H = 241.8 kJ/mole to find the H of the reaction H 2 O(g) -----> H 2 O(l) a) Notice we need gaseous water (steam) on the reactant side so Rxn 2 (and of H.) H 2 O(g) -----> H 2 + 1/2 O 2 H = + 241.8 kJ/mole b) The reaction with liquid water already has the water on the product side where it belongs so we can just H 2 + 1/2 O 2 -----> H 2 O(l) H = 285.8 kJ/mole H 2 O(g) -----> H 2 + 1/2 O 2 H = + 241.8 kJ/mole H 2 O(g) -----> H 2 O(l) H = –44.0 kJ/mole This is a very exothermic reaction which is why you can easily get scalded by hot steam if it condenses on your skin. Example 2: Calculate the

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Unformatted text preview: ∆ H for the reaction Ca + H 2 + O 2-----> Ca(OH) 2 (s) Using Rxn 1 Ca(OH) 2 (s) -----> CaO + H 2 O(l) ∆ H = + 65.4 kJ and Rxn 2 H 2 + 1/2 O 2-----> H 2 O (l) ∆ H = – 285.8 kJ and Rxn 3 2 Ca + O 2-----> 2CaO ∆ H = – 1270.2 kJ Chap 6 Sec 5 notes F09.doc Page 2 Get Ca(OH) 2 (s) on the product side of Rxn 1 ) CaO + H 2 O(l) -----> Ca(OH) 2 (s) ∆ H = – 65.4 kJ Rxn 2 H 2 O (l) will cancel out the H 2 O (l) on the reaction side so just put in without changes. H 2 + 1/2 O 2-----> H 2 O (l) ∆ H = – 285.8 kJ Ca + 1/2 O 2-----> CaO 1/2 x ∆ H = – 1270.2 kJ) = – 635.1 kJ Together CaO + H 2 O(l)-----> Ca(OH) 2 (s) ∆ H = – 65.4 kJ H 2 + 1/2 O 2-----> H 2 O (l) ∆ H = – 285.8 kJ Ca + 1/2 O 2-----> CaO ∆ H = – 635.1 kJ Ca + H 2 + O 2-----> Ca(OH) 2 (s) ∆ H = – 986.3 kJ Notice: Practice Problems...
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## This note was uploaded on 04/06/2010 for the course CHEM 1A taught by Professor Okamura during the Fall '08 term at UC Riverside.

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Chap 6 Sec 5 - ∆ H for the reaction Ca H 2 O 2> Ca(OH...

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