1. CHAPTER 17
THERMOCHEMISTRY

2. ENERGY
Thermochemistry is the study of the heat changes that occur during chemical reactions and physical changes of state.
Heat (q) is a form of energy that always flows from a warmer object to a cooler object.
Energy is the capacity to do work or to supply heat.
Kinetic energy and Chemical Potential Energy
energy stored in chemicals because of their compositions

3. ENERGY Remember the law of conservation of energy?
In any chemical or physical process, energy is neither created nor destroyed.
All energy can be accounted for as work, stored energy, or heat.
System Surroundings = Universe
Any specific part of the universe that attention is focused on
Includes everything outside the system.
Constant

4. HEAT CHANGES Endothermic Process Endothermic Process
A process that absorbs heat from the surroundings
Surroundings cool down
System heats up
Heat change > 0 (positive)
Endothermic Process
A process that loses heat to the surroundings
Surroundings heat up
System cools down
Heat change < 0 (negative)

5. HEAT CAPACITY The SI unit of heat (q) and energy (E) is the joule (J).
The amount of heat it takes to change an object’s temperature by 1ºC is the heat capacity (J/ºC) of that object.
The amount of heat that it takes to change the temperature of 1g of a substance by 1ºC is the specific heat of that object.
One calorie is the quantity of heat that raises the temperature of 1 gram of water 1ºC.
1 Calorie = 1000 calories and 1 Joule = cal

6. HEAT CAPACITY VS. SPECIFIC HEAT
Specific Heat (J/g.oC)
The amount of heat it would take to raise the temperature of a 1g of a substance by 1oC
Independent of mass
The specific heat of steel is about 0.5 J/g.oC
Heat Capacity (J/ºC)
The amount of heat it would take to raise the temperature of a sample of a substance by 1oC
Dependent on Mass
It would take much more heat to raise the temperature of a steel girder by 1oC than it would to raise the temperature of a steel nail by the same amount
.002 kg = 1J
200 kg = 1 x 105J

7. HEAT CAPACITY VS. SPECIFIC HEAT PROBLEMS
Heat Capacity (J/oC)
How much heat is required to raise 150g of silver by 1oC?
(Specific Heat of Ag is 0.24J/g.oC)
What is the specific heat of silver if takes 36J to increase a 150g sample of silver from 25oC to 26oC?

8. CALORIMETRY
Calorimetry is the accurate and precise measurement of the heat change for chemical and physical processes.
The enthalpy change (△H), is the heat change(q) for a process at constant pressure.
q = △H = mC△T
Constant Volume Calorimetry

9. CALORIMETRY CALCULATIONS
To measure heat changes using calorimetry Measure initial temperature of water (SURROUNDINGS) 2. Perform reaction inside insulated container. 3. Record final water temperature. 4. Use specific heat of water to measure enthalpy change with the formula qsys=△H= - qsurr = - mC△T For endothermic reactions qsys is positive qsurr is negative For exothermic reactions qsys is negative qsurr is positive

10. EXAMPLE PROBLEM
The following acid-base reaction is performed in a coffee cup calorimeter:
H+(aq) + OH-(aq) --> H2O(l)
55mL of water containing 0.1mol of HCl and 55mL of water containing 0.1mol of NaOH are combined in a calorimeter.
The temperature of the water rises from 25.0°C to 26.2°C when 0.10 mol of each substance react.
Calculate qwater
Calculate ΔH for the reaction
Calculate ΔH if 1.00 mol OH- reacts with 1.00 mol H+

11. SOLUTION

12. THERMOCHEMICAL EQUATIONS
A chemical equation that includes the amount of heat produced or absorbed during the reaction is the thermochemical equation.
Exothermic Reaction
The heat of reaction explains the heat released or absorbed during a chemical reaction (the change in enthalpy, △H). kJ
Endothermic Reaction NH4NO3(s) + H2O(l) KJ --> NH4OH(aq) + HNO3(aq)

13. THERMOCHEMICAL EQUATIONS
The heat of reaction for the complete burning of one mole of a substance heat of combustion
The heat absorbed by one mole of a substance while melting molar heat of fusion (△Hfus)
The heat lost when one mole of a substance freezes molar heat of solidification (△Hsolid)
The heat absorbed when one mole of a substances vaporizes molar heat of vaporization (△Hvap)
The heat absorbed when one mole of a substances condenses molar heat of condensation (△Hcond)
The heat change caused by dissolution of one mole of a substance molar heat of solution (△Hsoln)

14. MOLAR HEAT CALCULATIONS
The heat of combustion of methane is -890KJ/mol. How much heat is released when 45g of methane are combusted?
How much heat would be required to melt 523g of osmium is its molar heat of fusion is KJ/mol?

15. Potentially Useful Information
For water…
ΔHfus = 6.01 KJ/mol
ΔHvap= 40.7 KJ/mol
What about…
ΔHsolid
ΔHcond

16. PHASE CHANGES AND ENTHALPY
To calculate the amount of heat absorbed or released as a substance changes state, q=mcDT for pure states and molar heats for phase changes.
FOR WATER
ΔHvap= 40.7 KJ/mol
solid
ΔHfus = 6.01 KJ/mol

17. Sample Problem
Calculate the total enthalpy change when 42.0 grams of ice at -10oC is warmed to water at 20oC.

18. HESS’S LAW
Hess’s law of heat summation states that 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 change.
EXAMPLE
Calculate the value of Ho/kJ for the following reaction using the listed thermochemical equations:
2 F2(g) + 2 H2O(l) HF(g) + O2(g)
H2(g) + F2(g) HF(g) Ho/kJ = kJ H2(g) + O2(g) H2O(l) Ho/kJ = kJ
H2(g) + F2(g) HF(g) Ho/kJ = kJ
2 H2O(l) H2(g) + O2(g) Ho/kJ = kJ
2 F2(g) + 2 H2O(l) HF(g) + O2(g) Ho/kJ = +30 kJ

19. STANDARD HEATS OF FORMATION
The standard heat of formation (△Hfo) 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 oC.
For free elements △Hf°= 0 in their standard states.
Standard heat of reaction...
△H°= S△Hfo(products) - S△Hfo(reactants)
*Page 530 – standard heats of formation table

20. EXAMPLE
Calculate the heat of reaction (kJ) using the listed heats of formation:
CH4(g) + 4 F2(g) CF4(g) + 4 HF(g)
Hof CH4 = kJ/mole Hof CF4 = kJ/mole Hof HF = kJ/mole
DHfo = ((-925kJ/mol) + (4moles x kJ/mol)) - ((1 mole x kJ/mol) + (1 x 0)) = 1935kJ