LEC 03 CHAP 07 - Lecture 3 Using Thermochemical Equations....

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TOUPADAKIS Chapter 7 Lecture 3 Lecture 3 Using Thermochemical Equations. Hess’s Law. Standard Enthalpy of Formation of a Substance. Determination of Δ H o : A summary. Indirect determination of Δ H o : A summary.
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TOUPADAKIS Chapter 7 Lecture 3 Using Thermochemical Equations Enthalpy and therefore Δ H is an extensive property. Enthalpy is directly proportional to the amount of substance. N 2 (g) + O 2 (g) 2 NO(g) Δ H = +180.50 kJ ½ N 2 (g) + ½ O 2 (g) NO(g) Δ H = +90.25 kJ Δ H changes sign when the process is reversed NO(g) ½ N 2 (g) + ½ O 2 (g) Δ H ° = 90.25 kJ Δ H depends on the physical state of the reactants and products. ½ N 2 (g) + ½ O 2 (g) NO(g) Δ H ° = + 90.25 kJ CH 4 (g) + 2O 2 (g) CO 2 (g) + 2H 2 O (l) Δ H ° = 890 kJ CH 4 (g) + 2O 2 (g) CO 2 (g) + 2H 2 O (g) Δ H = 802 kJ
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TOUPADAKIS Chapter 7 Lecture 3 Hess’s Law Combining Reaction Enthalpies – If a process occurs in stages or steps (even hypothetically),
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This note was uploaded on 04/18/2010 for the course CHEM 2B 993029 taught by Professor Toupadakis during the Spring '10 term at UC Davis.

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LEC 03 CHAP 07 - Lecture 3 Using Thermochemical Equations....

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