1. Unit 8: Thermochemistry

2. Kinetic Molecular Theory
Collective hypotheses about the particulate nature of matter and the surrounding space
Greeks - earliest written ideas on atoms
Current view
Matter comprised of microscopic particles - atoms
Atoms combine to form molecules
Many macroscopic phenomena can be traced to interactions on this level
The molecules of a gas are in constant and random motion
The temperature of a gas depends on its average kinetic energy

3. Temperature scales Measured with thermometer
Defined with reference to various reference points
Fahrenheit
Celsius
Kelvin
Conversion formulas
Fahrenheit to Celsius
Celsius to Fahrenheit
Celsius to Kelvin

4. Heat Vs. Temp
Heat is the total E of molecular motion in a substance while temp is a measure of the avg E of molecular motion in a substance
Heat E depends on the number of particles (the size or mass). Temperature does not
(Videos)

5. EX: the temp of 20 mL of water might be the same as the temp of 1 L of water, but the liter of water has more heat because it has more water and thus more total thermal energy.
Adding or removing heat (energy) will increase or decrease the temp. Higher temperatures mean that the molecules are moving, vibrating and rotating with more E

6. Heat (energy) moves from higher temperatures to lower temperatures.
Temperature is not energy, but a measure of it (measure of hotness or coldness)
Heat
50 degrees C
20 degrees C

7. Units of heat Metric units Joule (J) – SI unit
calorie (cal) - energy needed to raise temperature of 1 g of water 1 degree Celsius
kilocalorie (kcal, Calorie, Cal) - energy needed to raise temperature of 1 kg of water 1 degree Celsius
English system
British thermal unit (BTU) - energy needed to raise the temperature of 1 lb of water 1 degree Fahrenheit
Mechanical equivalence
4.184 J = 1 cal

8. Land heats up and cools down faster than water
Specific Heat
a. . Some things heat up or cool down faster than others.
Land heats up and cools down faster than water

9. b. Specific heat is the amount of heat required to raise the temperature of 1 g of a material by one degree (C or K).
1) C water = J / g C
2) C sand = J / g C
This is why land heats up quickly during the day and cools quickly at night and why water takes longer.

10. Why does water have such a high specific heat?
Water molecules have strong intermolecular forces called hydrogen bonds; therefore it takes more heat energy to break them. Metals have weak bonds and do not need as much energy to break them.

11. How to calculate changes in thermal energy
Q = m x T x C
Q Q = change in thermal energy
m m = mass of substance
 T T= change in temperature (Tf – Ti)
Ccc c = specific heat of substance

12. rature of water are measured
c. A calorimeter is used to help measure the specific heat of a substance.
First, mass and its Q value, its mass, and its T, its C can be calculated
rature of water are measured
Then heated sample is put inside and heat flows into water
This gives the heat lost by the substance
T is measured for water to help get its heat gain

13. 2H2 (g) + O2 (g) 2H2O (l) + energy
Exothermic process is any process that gives off heat – transfers thermal energy from the system to the surroundings.
2H2 (g) + O2 (g) H2O (l) + energy
H2O (g) H2O (l) + energy
Endothermic process is any process in which heat has to be supplied to the system from the surroundings.
energy + H2O (s) H2O (l)
energy + 2HgO (s) Hg (l) + O2 (g)
6.2

14. DH = H (products) – H (reactants)
Enthalpy (H) is used to quantify the heat flow into or out of a system in a process that occurs at constant pressure.
DH = H (products) – H (reactants)
DH = heat given off or absorbed during a reaction at constant pressure
Hproducts < Hreactants
Hproducts > Hreactants
DH < 0
DH > 0
6.4

15. Thermochemical Equations
Is DH negative or positive?
System gives off heat
Exothermic
DH < 0
890.4 kJ are released for every 1 mole of methane that is combusted at 250C and 1 atm.
CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l)
DH = kJ
6.4

16. Thermochemical Equations
Is DH negative or positive?
System absorbs heat
Endothermic
DH > 0
6.01 kJ are absorbed for every 1 mole of ice that melts at 00C and 1 atm.
H2O (s) H2O (l)
DH = 6.01 kJ
6.4

17. Thermochemical Equations
The stoichiometric coefficients always refer to the number of moles of a substance
H2O (s) H2O (l)
DH = 6.01 kJ/mol ΔH = 6.01 kJ
If you reverse a reaction, the sign of DH changes
H2O (l) H2O (s)
DH = kJ
If you multiply both sides of the equation by a factor n, then DH must change by the same factor n.
2H2O (s) H2O (l)
DH = 2 mol x 6.01 kJ/mol = 12.0 kJ
6.4

18. Thermochemical Equations
The physical states of all reactants and products must be specified in thermochemical equations.
H2O (s) H2O (l)
DH = 6.01 kJ
H2O (l) H2O (g)
DH = 44.0 kJ
How much heat is evolved when 266 g of white phosphorus (P4) burn in air?
P4 (s) + 5O2 (g) P4O10 (s) DHreaction = kJ
1 mol P4
123.9 g P4 x
3013 kJ
1 mol P4 x
266 g P4
= 6470 kJ
6.4

19. Standard enthalpy of formation (DH0) is the heat change that results when one mole of a compound is formed from its elements at a pressure of 1 atm. f
The standard enthalpy of formation of any element in its most stable form is zero.
DH0 (O2) = 0 f
DH0 (C, graphite) = 0 f
DH0 (O3) = 142 kJ/mol f
DH0 (C, diamond) = 1.90 kJ/mol f
6.6

20. Practice Problem
How much heat is produced by the combustion of 1.00 lb (454 g) of propane (C3H8)? The ΔH for the combustion of propane is kJ mol-1.
Convert the grams of propane to moles of propane and then find the amount of heat produced:

21. The standard enthalpy of reaction (DH0 ) is the enthalpy of a reaction carried out at 1 atm.
rxn
aA + bB cC + dD
DH0
rxn
dDH0 (D) f
cDH0 (C) = [ + ] -
bDH0 (B)
aDH0 (A)
DH0
rxn = S
DH0 (products) f - S
DH0 (reactants) f
Hess’s Law: When reactants are converted to products, the change in enthalpy is the same whether the reaction takes place in one step or in a series of steps.
(Enthalpy is a state function. It doesn’t matter how you get there, only where you start and end.)
6.6

22. 2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)
Benzene (C6H6) burns in air to produce carbon dioxide and liquid water. How much heat is released per mole of benzene combusted? The standard enthalpy of formation of benzene is kJ/mol.
2C6H6 (l) + 15O2 (g) CO2 (g) + 6H2O (l)
DH0
rxn
DH0 (products) f = S
DH0 (reactants) -
DH0
rxn
6DH0 (H2O) f
12DH0 (CO2) = [ + ] -
2DH0 (C6H6)
DH0
rxn =
[ 12 × × ] – [ 2 × ] = kJ
-6535 kJ
2 mol
= kJ/mol C6H6
6.6

23. Calculate the standard enthalpy of formation of CS2 (l) given that:
C(graphite) + O2 (g) CO2 (g) DH0 = kJ
rxn
S(rhombic) + O2 (g) SO2 (g) DH0 = kJ
rxn
CS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = kJ
rxn
1. Write the enthalpy of formation reaction for CS2
C(graphite) + 2S(rhombic) CS2 (l)
2. Add the given rxns so that the result is the desired rxn.
rxn
C(graphite) + O2 (g) CO2 (g) DH0 = kJ
2S(rhombic) + 2O2 (g) SO2 (g) DH0 = x2 kJ
rxn +
CO2(g) + 2SO2 (g) CS2 (l) + 3O2 (g) DH0 = kJ
rxn
C(graphite) + 2S(rhombic) CS2 (l)
DH0 = (2x-296.1) = 86.3 kJ
rxn
6.6

24. DHsoln = Hsoln - Hcomponents
The enthalpy of solution (DHsoln) is the heat generated or absorbed when a certain amount of solute dissolves in a certain amount of solvent.
DHsoln = Hsoln - Hcomponents
Which substance(s) could be used for melting ice?
Which substance(s) could be used for a cold pack?
6.7

25. Energy Diagrams Activation energy (Ea) for the forward reaction
Activation energy (Ea) for the reverse reaction
(c) Delta H
50 kJ/mol
300 kJ/mol
150 kJ/mol
100 kJ/mol
-100 kJ/mol
+200 kJ/mol

26. Phase Changes
The boiling point
The normal boiling point
11.8

27. The critical temperature (Tc)
The critical pressure (Pc)
11.8

28. Where vaporization occurs?
Can you find…
The Triple Point?
Critical pressure?
Critical temperature?
Where fusion occurs?
Where vaporization occurs?
Melting point (at 1 atm)?
Boiling point (at 6 atm)?
Carbon Dioxide

29. The melting point of a solid or the freezing point
H2O (s) H2O (l)
The melting point of a solid or the freezing point
11.8

30. H2O (s) H2O (g)
11.8

31. Molar heat of fusion (DHfus) is the energy required to melt 1 mole of a solid substance.
11.8

32. 11.8

33. Sample Problem
How much heat is required to change 36 g of H2O from -8 deg C to 120 deg C?
Step 1: Heat the ice Q=mcΔT
Q = 36 g x 2.06 J/g deg C x 8 deg C = J = 0.59 kJ
Step 2: Convert the solid to liquid ΔH fusion
Q = 2.0 mol x 6.01 kJ/mol = 12 kJ
Step 3: Heat the liquid Q=mcΔT
Q = 36g x J/g deg C x 100 deg C = J = 15 kJ

34. Now, add all the steps together
Sample Problem
How much heat is required to change 36 g of H2O from -8 deg C to 120 deg C?
Step 4: Convert the liquid to gas ΔH vaporization
Q = 2.0 mol x kJ/mol = kJ
Step 5: Heat the gas Q=mcΔT
Q = 36 g x 2.02 J/g deg C x 20 deg C = J = 1.5 kJ
Now, add all the steps together
0.59 kJ + 12 kJ + 15 kJ + 88 kJ kJ = 118 kJ