# T HE U NIVERSITY O F Q UEENSLAND Foundation Year THERMOCHEMISTRY II - PowerPoint PPT Presentation

1 / 25

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.

I am the owner, or an agent authorized to act on behalf of the owner, of the copyrighted work described.

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

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.

- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -

#### Presentation Transcript

THE UNIVERSITY

OF QUEENSLAND

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

Standard Heats of Formation

### 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

• 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

• 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)

• 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)

• 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).

Vapour

 Hcond = -ve

 Hvap = +ve

Enthalpy

Liquid

 Hsolid = -ve

 Hfus = +ve

Solid

### 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.

Exothermic Solvation

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

Endothermic Solvation

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

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

### 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 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)+ 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)

• 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

• 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)

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

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

### 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/