1. 15-2 Heat --Calorimetry
Calorimetry is the act or process of measuring heat.
See pg 523/524 for calorimeter.
We use the formula mw Cw ΔT w =moCoΔTo
W= water o=object…solve for Co
We can set the 2 heats equal because the
heat absorbed by the water = heat lost by the object.
If you have heat of an object, its mass, and its change in temperature only, then you would use the formula…

2. Thermochemistry
Thermochemistry is the study of heat changes that accompany reactions and physical changes.
To help define where heat is and when it moves out of a ‘system’, we define system as the container where a reaction occurs, the surroundings as the area outside of the container and …
Universe= system + surroundings
Enthalpy(H) is the heat content of a system at constant pressure.
The enthalpy (heat) of a reaction is found by…
ΔHrxn =ΔHprod - ΔHreact

3. Thermochemical Equations
The sign of ΔHrxn is – if heat is lost to surroundings, and + if heat is gained.
You may also see the energy on the product side if it is heat lost, and on the reactant side if it is heat gained.
4Fe (s) O2(g)  2Fe2O3(s) kJ (Exothermic)
ΔHrxn =-1625kJ
21 kJ + H2S (g)  H2(g) S(s) (Endothermic)
ΔHrxn = +21 kJ

4. Enthalpy Calculations
For enthalpy calculations you will have an enthalpy value H and multiply by the moles of that substance.
q=mH
These H values can come from…
heat of combustion = heat from burning 1 mole of a substance. **
heat of vaporization= heat from vaporizing 1 mole of a substance*
heat of condensation= heat from condensing 1 mole of a substance**
heat of fusion = heat from melting 1 mole of a substance*
heat of solidification= heat from freezing 1 mole of a substance*
* = heat absorbed
**= heat given off

5. Heats Values for Water Hvap =Hcond =40.7 kJ /mol = 2.26 kJ /g
Hfus=Hsolid =6.01 kJ /mol = .334 kJ /g
See pg 530 for other substances
Any time you have a state change, there is no temperature change and you use q=mH where m is mass with a mass value of H and m is moles with a mole value of H.

6. Calculating Enthalpy Changes
When you have a temperature change, you use…
q=moCoΔTo
When you have a state change or other H use…
q=mH
Hess’s Law states that the total heat given off or absorbed by a series of reactions is equal to the summation of their enthalpies for the individual reactions.

7. Applying Hess’s Law Calculate the energy change for the reaction…
2S(s) O 2(g)  2SO 3(g)
Look at your handout or pg 535 of the text.
This also applies to problems where you have temperature changes and state changes…
Example: Calculate the heat resulting from melting 10.0 g of ice at 0.0oC , then heating it to 100oC, then turning it to steam at 100oC.
Hvap =Hcond =40.7 kJ /mol = 2.26 kJ /g
Hfus=Hsolid =6.01 kJ /mol = .334 kJ /g

8. Solution First melt…. Use qm =mH (m is mass in grams)
H fus water = .334 kJ /g
q= (10.0g)(.334 kJ /g)=3.34 kJ
Second change temp. ….. qt =mtCtΔTt
q=(10.0 g)(4.184 J/goC)(100.0 oC)= J
Finally ….. Use qv =mH (m is mass in grams)
H vap water = = 2.26 kJ /g
q= (10.0g)(2.26 kJ /g)=22.6 kJ
Apply Hess’s Law and add the 3 q values.
q total =3.34 kJ kJ kJ=30.12 kJ

9. Standard Heats of Formation
Find the small chart of pg 538 and your extended chart on page 979.
These values are heats of formation for different substances.
Heats of formation for ground state elements is always 0kJ.
We can solve for the heats of formation by using
ΔHrxn =ΔHprod - ΔHreact

10. The Summation Equation
ΔHrxn = ∑ ΔHprod - ∑ ΔHreact
See Heats of formation chart R-11