T HE U NIVERSITY O F Q UEENSLAND Foundation Year THERMOCHEMISTRY II

Lesson Overview Thermochemistry Specific heat capacity Endothermic and Exothermic Equations Calorimetry Thermochemical Equations Heats of Changes of State Heat Capacity Hess’s Law Standard Heats of Formation

Thermochemical Equations 1 A thermochemical equation is one that includes energy changes. In exothermic reactions, heat is a product (it's being formed), so a reaction of this kind might look like this: A + B ---> C + D + heat And similarly, if a reaction is endo, then it acts like a reactant (goes on the left side): A + B + heat ---> C + D

Thermochemical Equations 2 There are two ways to write a thermochemical equation: 2 C 2 H 6 (g) + 7 O 2 (g)  4 CO 2 (g) + 6 H 2 O(g)  H = kJ or 2 C 2 H 6 (g) + 7 O 2 (g)  4 CO 2 (g) + 6 H 2 O(g) kJ Are these reactions exothermic or endothermic?

Thermochemical Equations 3 For endothermic reactions, energy must be added to the reactants to make it happen. Heat may be considered as a reactant. 2 NH 3 (g) + 92 kJ  N 2 (g) + 3 H 2 (g) or 2 NH 3 (g)  N 2 (g) + 3 H 2 (g)  H = +92 kJ When writing thermochemical equations, the state symbols must be included. This is because changing the state of a chemical involves energy changes.

Heat and Changes of State (1) All solids absorb heat in melting to liquids. The heat absorbed by one mole of a substance in melting from a solid to a liquid at a constant temperature is called the molar heat of fusion (  H fus.). The heat lost when one mole of a liquid changes to a solid at a constant temperature is the molar heat of solidification (  H solid. )

Heat and Changes of State (2) The amount of heat absorbed by one mole of a liquid that is undergoing evaporation is called the molar heat of vaporisation. (  H vap ) The condensation of 1 mole of vapour releases heat as the molar heat of condensation (  H cond ).

 H solid = -ve  H cond = -ve  H fus = +ve  H vap = +ve Vapour Liquid Solid Enthalpy

Heat of Solution Heat changes can occur when a substance is dissolved in a solvent. The heat change caused by dissolution of one mole of substance is the molar heat of solution,  H soln.

Exothermic Solvation

Endothermic Solvation

Thermochemical Equation Calculations Thermochemical equations obey several simple rules that make computation of the enthalpy change in a reaction easy. 1.The magnitude of  H is directly proportional to the amount of reactants used in the reaction  H for a reaction is equal in magnitude but opposite in sign to the reverse reaction. 3.Hess' Law: The value of  H for a reaction is the same no matter what path is used to get from reactants to products.

Hess’s Law of Summation Hess's Law of Heat Summation states: If you add two or more thermochemical equations to give a final equation, then you can also add the heat changes to give the final heat changes.

Hess’s Law of Summation(2) To find the enthalpy for: C(s, diamond)  C(s, graphite) C(s, graphite)+ O 2 (g)  CO 2 (g)  H = kJ C(s, diamond)+O 2 (g)  CO 2 (g)  H =-395.4kJ Write the reverse of equation (a) to give CO 2 (g)  C(s, graphite)+O 2 (g)  H = kJ

Hess’s Law of Summation(3) Adding the equations should give us our original equation. Now do the same thing with the enthalpy values :- i.e.:  H = kJ +  H = kJ  H = -1.9kJ

Heats of Formation The standard heat of formation (H f ) of a compound is the change in enthalpy that accompanies the formation of one mole of the compound from its elements with all substances in their standard states at 25 o C. The H for a reaction is the difference between the standard heats of formation of all the reactants and products. ie:H=   H f (products) -   H f (reactants)

References 1. tik.htmlhttp:// tik.html vies/5.2small.movhttp://chemed.chem.purdue.edu/demos/movies/small_mo vies/5.2small.mov oc/Hsolv1gif.gifhttp://wine1.sb.fsu.edu/chm1046/notes/SolnProp/SolnPr oc/Hsolv1gif.gif 6. io/text_images/ io/text_images/

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