T he u niversity o f q ueensland foundation year thermochemistry ii
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T HE U NIVERSITY O F Q UEENSLAND Foundation Year THERMOCHEMISTRY II. Lesson Overview. Thermochemistry. Heat Capacity. Endothermic and Exothermic Equations. Specific heat capacity. Calorimetry. Thermochemical Equations. Heats of Changes of State. Hess’s Law.

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

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T he u niversity o f q ueensland foundation year thermochemistry ii

THE UNIVERSITY

OF QUEENSLAND

Foundation Year

THERMOCHEMISTRY II


Lesson overview

Lesson Overview

Thermochemistry

Heat Capacity

Endothermic and Exothermic Equations

Specific heat

capacity

Calorimetry

Thermochemical

Equations

Heats of Changes

of State

Hess’s Law

Standard Heats of Formation


Thermochemical equations 1

Thermochemical Equations 1

  • A thermochemical equationis 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

Thermochemical Equations 2

  • There are two ways to write a thermochemical equation:

  • 2 C2H6(g) + 7 O2(g)4 CO2(g) + 6 H2O(g)

    H = -2855 kJ

    or

  • 2 C2H6(g) + 7 O2(g) 4 CO2(g) + 6 H2O(g) + 2855 kJ

  • Are these reactions exothermic or endothermic?


Thermochemical equations 3

Thermochemical Equations 3

  • For endothermic reactions, energy must be added to thereactants to make it happen.

  • Heat may be considered as areactant.

  • 2 NH3(g) + 92 kJ N2(g) + 3 H2(g)

    or

  • 2 NH3(g) N2(g) + 3 H2(g)H = +92 kJ

    • When writing thermochemical equations, the state symbolsmust be included.This is because changing the state of achemical involves energy changes.


Heat and changes of state 1

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 (DHfus.).

  • The heat lost when one mole of a liquid changes to a solid at a constant temperature is themolar heat of solidification (DHsolid. )


Heat and changes of state 2

Heat and Changes of State(2)

  • The amount of heat absorbed by one mole of a liquid that is undergoing evaporation is called themolar heat of vaporisation.

    (D Hvap)

  • The condensation of 1 mole of vapour releases heat as the molar heat of condensation (DHcond).


T he u niversity o f q ueensland foundation year thermochemistry ii

Vapour

 Hcond = -ve

 Hvap = +ve

Enthalpy

Liquid

 Hsolid = -ve

 Hfus = +ve

Solid


Heat of solution

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, DHsoln.


T he u niversity o f q ueensland foundation year thermochemistry ii

Exothermic Solvation

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


T he u niversity o f q ueensland foundation year thermochemistry ii

Endothermic Solvation

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


T he u niversity o f q ueensland foundation year thermochemistry ii

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


Thermochemical equation calculations

Thermochemical Equation Calculations

  • Thermochemical equations obey several simple rules that make computation of the enthalpy change in a reaction easy.

  • The magnitude of DH is directly proportional to the amount of reactants used in the reaction

  • DH for a reaction is equal in magnitude but opposite in sign to the reverse reaction.

  • Hess' Law: The value of DH 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 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

Hess’s Law of Summation(2)

To find the enthalpy for:

  • C(s, diamond)  C(s, graphite)

  • C(s, graphite)+ O2(g) CO2(g)  H = -393.5kJ

  • C(s, diamond)+O2(g) CO2(g) H =-395.4kJ

    Write the reverse of equation (a) to give

  • CO2(g)  C(s, graphite)+O2(g)  H = +393.5kJ


Hess s law of summation 3

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.: DH = -395.4kJ

    + DH = +393.5kJ

    DH = -1.9kJ


Heats of formation

Heats of Formation

  • The standard heat of formation (Hf) 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 25oC.

  • The H for a reaction is the difference between the standard heats of formation of all the reactants and products.

  • ie:H=  Hf(products) - Hf(reactants)


T he u niversity o f q ueensland foundation year thermochemistry ii

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


T he u niversity o f q ueensland foundation year thermochemistry ii

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


References

References

  • http://www.chem.vt.edu/RVGS/ACT/notes/Chap_8_Triptik.html

  • http://apchem.virtualave.net/concepts/thermochem.html

  • http://chemed.chem.purdue.edu/demos/movies/small_movies/5.2small.mov

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

  • http://wine1.sb.fsu.edu/chm1046/notes/SolnProp/SolnProc/Hsolv1gif.gif

  • http://cwx.prenhall.com/petrucci/medialib/media_portfolio/text_images/


End of lecture

End of Lecture


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