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

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Presentation on theme: "T HE U NIVERSITY O F Q UEENSLAND Foundation Year THERMOCHEMISTRY II."— Presentation transcript:

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

2 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

3 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

4 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 = -2855 kJ or 2 C 2 H 6 (g) + 7 O 2 (g)  4 CO 2 (g) + 6 H 2 O(g) + 2855 kJ Are these reactions exothermic or endothermic?

5 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.

6 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. )

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8 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 ).

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

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12 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.

13 http://wine1.sb.fsu.edu/chm1046/notes/SolnProp/SolnProc/Hsolv1gif.gif Exothermic Solvation

14 Endothermic Solvation http://wine1.sb.fsu.edu/chm1046/notes/SolnProp/SolnProc/SolnProc.htm

15 http://www.wou.edu/las/physci/ch412/heatsoln.gif

16 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.

17 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.

18 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 = -393.5kJ 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 = +393.5kJ

19 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 = -395.4kJ +  H = +393.5kJ  H = -1.9kJ

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21 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)

22 http://cwx.prenhall.com/petrucci/medialib/media_portfolio/text_images/FG07_16.JPG

23 http://cwx.prenhall.com/petrucci/medialib/media_portfolio/text_images/TB07_02.JPG

24 References 1.http://www.chem.vt.edu/RVGS/ACT/notes/Chap_8_Trip tik.htmlhttp://www.chem.vt.edu/RVGS/ACT/notes/Chap_8_Trip tik.html 2.http://apchem.virtualave.net/concepts/thermochem.htmlhttp://apchem.virtualave.net/concepts/thermochem.html 3.http://chemed.chem.purdue.edu/demos/movies/small_mo vies/5.2small.movhttp://chemed.chem.purdue.edu/demos/movies/small_mo vies/5.2small.mov 4.http://www.wou.edu/las/physci/ch412/heatsoln.gifhttp://www.wou.edu/las/physci/ch412/heatsoln.gif 5.http://wine1.sb.fsu.edu/chm1046/notes/SolnProp/SolnPr oc/Hsolv1gif.gifhttp://wine1.sb.fsu.edu/chm1046/notes/SolnProp/SolnPr oc/Hsolv1gif.gif 6.http://cwx.prenhall.com/petrucci/medialib/media_portfol io/text_images/http://cwx.prenhall.com/petrucci/medialib/media_portfol io/text_images/

25 End of Lecture


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