CHEMICAL THERMODYNAMICS 1 chem 16 4 slides per page

# If the problem asks for the amount of heat involved

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Unformatted text preview: of Dissolution in: NaCl(s) Na+(aq) + Cl-(aq) ∆Hdisso = (∆Hf, Na+(aq)o + ∆Hf, Cl-(aq)o ) – (∆Hf, NaCl(s)o ) 6 1/7/2014 Using Enthalpies Example 1: Calculate the enthalpy change for the reaction of one mole of H2(g) with one mole of F2(g) to form two moles of HF(g) at 25oC and one atmosphere. • If the problem asks for the amount of heat involved (either released or absorbed) in the process: ∆H rxn = (2∆H o HF(g) ) - (∆H fo F2 ( g ) + ∆H fo H 2 ( g) ) f heat ‘q’ of process = ∆Hprocess • moles of limiting reagent kJ q = ∆H process ( mole ) x n LR (mole) moles means stoichiometry yeah! :D • Example 2: Calculate the heat involved in which 15.0 g of aluminum reacts with oxygen to form Al2O3 at 25oC and one atmosphere. Calculate ∆Hrxn = 2∆H o HF(g) f From appendix of book, ∆H o HF(g) = −271 kJ/mole HF f 2 mole HF − 271 kJ ∴ ∆H rxn = mole of reaction mole HF = − 542 kJ/mole of reaction • Nitrogen oxide (NO) has been found to be a key component in many biological processes. It also can react with oxygen to give the brown gas NO2. When one mole of NO reacts with Oxygen, 57.0 kJ of heat is evolved. = (2 • -1676 kJ/mol) – (0 + 0) = -3352 kJ/mole of reaction Calculate # of moles = 15.0 g Al/ 26.98 g Al/mole Al = 0.555967383 moles Al Calculate Heat = 0.555967383 moles Al x (-3352 kJ/ 4 moles Al) = -465.9006672 kJ = -466 kJ or 466 kilojoules heat released a.) What is the ∆Hrxn? b.) How many grams of nitrogen oxide must react with an excess of oxygen to liberate ten kilojoules of heat? 7 1/7/2014 4FeO(s) + O 2(g) → 2Fe2O3(s) Indirect Determination of ∆H: H...
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## This document was uploaded on 03/26/2014.

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