# Physical Chemistry for the Chemical and Biological Sciences

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Unformatted text preview: Section 3.0. The 1 st Law of Thermodynamics (CHANG text – Chapter 4) 3.1. Revisiting Heat Capacities 3.2. Definitions and Concepts 3.3. The First Law of THERMODYNAMICS 3.4. Enthalpy 3.5. Adiabatic Expansion of an Ideal Gas 3.6. Thermochemistry 3.7. Bond Energies and Bond Enthalpies 3.6. Thermochemistry Goals…. (1) To define enthalpy of reaction ( Δ R H ° ) (2) To understand how to calculate Δ R H from Δ f H ° for compounds (1) To use Hess’s Law to calculate Δ f H ° 3.6. Thermochemistry ……is the study of energy changes in chemical reactions…. • For a constant pressure process the heat of reaction (q p ) is equal to the enthalpy of reaction ( Δ R H ) • Exothermic reaction (gives off heat) Δ R H = negative value Endothermic (absorbs heat) Δ R H = positive value • at a pressure of 1 bar and T = 298 K we say Δ R H = Δ R H ° the standard enthalpy of reaction • Δ R H ° has units of kJ • Δ R H ° is the enthalpy change when reactants in standard state are converted to products in standard state Exothermic Exothermic Endothermic Endothermic The standard enthalpy change for a chemical rxn. is defined as: Δ R H ° = v H ° ( products ) - v H ° ( reactants ) H ° is the standard molar enthalpy v is the stoichiometric coefficient (number of moles) Example : a A + b B c C + d D The standard enthalpy of reaction is …………… Δ R H ° = + (c mol) H ° (C) + (d mol) H ° (D) – (a mol) H ° (A) – (b mol) H ° (B) Unfortunately we can’t measure the absolute values of H ° for substances so we use standard molar enthalpy of formation instead……… Δ f H ° so Δ R H ° = v Δ f H ° ( products ) - v Δ f H ° ( reactants ) E n t h a l p y Elements Reactants Products R H ° Δ f H ° ( r e a c t a n t s ) Δ f H ° ( p r o d u c t s ) We can calculate H ° simply by measuring the difference between the heat of formation ( Δ f H ° ) of the products and the reactants. How can we calculate Δ R H ° ? Example: C (graphite) + O 2 (g) CO 2 (g) R H ° = – 393.5 kJ Δ R H ° = v Δ f H ° ( products ) - v Δ f H ° ( reactants ) = (1 mol) Δ f H ° (CO 2 ) - (1 mol) Δ f H ° ( graphite) – (1 mol) Δ f H ° (O 2 ) = – 393.5¡kJ Note: we assign Δ f H ° = 0 for elements in their “most stable allotropic forms” at a particular temperature example at 298 K Δ f H ° (O 2 ) = 0 kJ mol-1 Δ f H ° ( graphite) = 0 kJ mol-1 so Δ R H ° for combustion of graphite may be expressed as: Δ R H ° = (1 mol) Δ f H ° (CO 2 ) = – 393.5 kJ “Allotropy is the ability of a chemical to exhibit in a number...
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3b - Section 3.0 The 1 st Law of Thermodynamics(CHANG text...

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