1. Things are “heating up” now!
Thermochemistry
Things are “heating up” now!

2. Nature of Energy Energy – the capacity to do work (or to produce heat)
Work is a force acting over a distance
Moving an object
Heat is a form of energy
Chemicals may store potential energy in their bonds
Can be released as heat energy

3. First Law of Thermochemistry
Law of conservation of energy – energy is neither created nor destroyed
Can only be converted from one form to another
Potential energy – energy due to position
Kinetic energy – energy due to the motion of an object
KE = ½ mv2
m = mass
v = velocity
Units are Joules

4. Heat & Temperature Heat (J) Temperature (°C) Measure of energy content
What is transferred during a temperature change
Temperature (°C)
Reflects random motion of particles in a substance
Indicates the direction in which heat energy will flow
Higher  lower

5. State Functions
A property of a system that depends only on its present state
Do not depend on what has happened in the system, or what might happen in the future
Independent of the pathway taken to get to that state
Eg: 1.0 L of water behind a dam has the same PE for work regardless of whether it flowed downhill to the dam or was taken uphill to the dam in a bucket.
PE is a state function dependent only on the current position of the water

6. Chemical Energy Exothermic Reactions Endothermic Reactions
Reactions that give off energy as they progress
Some of the PE stored in the chemical bonds is converted to thermal energy (random KE) through heat
Products are generally more stable (stronger bonds) than reactants
Endothermic Reactions
Reactions in which energy is absorbed from surroundings
Energy flows into the system to increase the PE of the system
Products are generally less stable (weaker bonds) than the reactants
Energy is needed to break bonds; released when formed

7. Thermodynamics System Energy ΔE = q + w q = heat w = work
Positive in endothermic reactions
Negative in exothermic reactions
w = work
Negative if the system does work
Positive if work is done on the system

8. Example
Calculate the ΔE for a system undergoing an endothermic process in which 18.7 kJ of heat flows and where 4.3 kJ of work is done on the system.
ΔE = q + w
q = 18.7 kJ (endothermic – gaining heat)
w = 4.3 kJ (positive – work is done on the system)
ΔE = 18.7 kJ kJ = 23.0 kJ
System has gained 23.0 kJ of energy

9. Thermodynamics - Gases
Work done by gases
w = -PΔV
By a gas (through expansion)
ΔV is positive
w is negative
To a gas (by compression)
ΔV is negative
w is positive
P is the P of surroundings

10. Example
Calculate the work associated with the expansion of a gas from 32 L to 75 L at a constant external pressure of 15 atm.
w = -PΔV
P = 15 atm
ΔV = 75 L – 32 L = 43 L
w = -(15 atm)(43 L)
w = -645 L · atm
-w because the gas expands so work is done to the surroundings
Energy flows out of the system

11. Enthalpy & Calorimetry
Enthalpy – measure of the energy released/absorbed by a substance when bonds are broken/formed during a reaction
Way to calculate heat flow in or out of a system
H = E + PV
In systems at constant pressure, where the only work is PV, the change in enthalpy is due only to energy flow as heat (ΔH = heat of reaction)
ΔH = Hproducts – Hreactants
Exothermic reactions: -ΔH
Endothermic reactions: +ΔH

12. Calorimetry Calorimetry – the science of measuring heat
Heat capacity (c)
Ratio of heat absorbed to increase in temperature
c = heat absorbed
temperature increase
Specific heat capacity – energy required to raise the temperature of 1 gram of a substance by 1°C
Molar heat capacity – energy required to raise the temperature of 1 mole of a substance by 1°C

13. q = mcΔT
q = heat
m = mass
c = specific heat
ΔT = change in T

14. Constant Pressure Calorimetry (soln)
Calculating Heat of Reaction (ΔH)
ΔH = (s)(m) (ΔT)
ΔH = (specific heat capacity) x (mass of soln) x (T change)
Heat of reaction is an extensive property
Dependent on the amount of substance
ΔH α moles of reactant

15. Constant Volume Calorimetry
Volume of bomb calorimeter cannot change
No work is done
Bomb calorimetry = constant volume
The heat capacity of the calorimeter must be known
kJ/°C
Bomb calorimetry – weighed reactants are placed inside a steel container and ignited
Used by industry to determine number of food calories that we consume
“Calories” on food labels are really kilocalories

16. ΔE = q + w, where w = 0  ΔE = q

17. Example – Bomb Calorimetry
Heat capacity of calorimeter = 11.3 kJ/°C
Compare heat of combustion for:
1.50 g methane T increase = 7.3°C
1.15 g hydrogen °C
g octane °C
Hcombustion CH4 = 54.0 kJ/g
Hcombustion H2 = kJ/g
Hcombustion C8H18 = 48.3 kJ/g

18. Hess’s Law

19. Hess’s Law
Hess’s Law - In going from a particular set of reactants to a particular set of products, the change in enthalpy (ΔH) is the same whether the reaction takes place in one step or a series of steps

20. Hess’s Law One step: Two step:
N2(g) + 2 O2(g)  2 NO2(g) ΔH = 68 kJ
Two step:
N2(g) + O2(g) 2 NO(g) ΔH1 = 180 kJ
2 NO(g) + O2(g)  2 NO2(g) ΔH2 = -112 kJ
N2(g) + 2 O2(g)  2 NO2(g) ΔH1 + ΔH2 = 68 kJ

21. Characteristics of Enthalpy Changes
If a reaction is reversed, the sign on ΔH is reversed
N2(g) + 2 O2(g)  2 NO2(g) ΔH = 68 kJ
2 NO2(g)  N2(g) + 2 O2(g) ΔH = -68 kJ
Magnitude of ΔH is directly proportional to the quantities of reactants and products in a reaction.
If the coefficients in a balanced reaction are multiplied by an integer, the value of ΔH is multiplied by the same integer

22. Using Hess’s Law Work backward from the final reaction
Reverse reactions as needed
Also reverse ΔH
Identical substances found on both sides of the summed equation cancel each other

23. Example Given the following data Calculate ΔH for the reaction
NH3(g)  ½ N2(g) + 3/2 H2(g) ΔH = 46 kJ
2 H2O(g)  2 H2(g) + O2(g) ΔH = 484 kJ
Calculate ΔH for the reaction
2 N2(g) + 6 H2O(g)  3 O2(g) + 4 NH3(g)

24. Standard Enthalpies of Formation
ΔHf°

25. Standard Enthalpy of Formation
The change in enthalpy that accompanies the formation of one mole of a compound from its elements with all elements in their standard state.
ΔHf° - Degree sign indicates that the process was carried out under standard conditions
Can measure enthalpy changes only by performing heat-flow experiments

26. Standard State For a compound For an element Gaseous state
Pressure of 1 atm
Pure liquid or solid
Standard state is the pure liquid or solid
Substance in solution
Concentration of 1 M
For an element
The form in which the element exists at 1 atm and 25°C

27. ΔHf° always given per mole of product with the product in its standard state
½ N2(g) + O2(g)  NO2(g)
ΔHf° = 34 kJ/mol
C(s) + 2 H2(g) + ½ O2(g)  CH3OH(l)
ΔHf° = -239 kJ/mol

28. Calculating Enthalpy Change
Elements in their standard states are not included
ΔHf° = 0
The change in enthalpy for a reaction can be calculated from the enthalpies of formation of the reactants and products
ΔH°reaction = npΔH°f(products) nrΔH°f(reactants)

29. Calculating ΔHf° CH4(g) + O2(g)  CO2(g) + 2 H2O(l)
Using Hess’s Law – break apart the reactants into their respective elements
CH4(g)  C(s) + 2 H2(g)
C(s) + 2 H2(g)  CH4(g) ΔHf° = -75 kJ/mol
O2(g)  O2(g) ΔHf° = 0 kJ/mol
C(s) + O2(g)  CO2(g) ΔHf° = -394 kJ/mol
H2(g) + ½ O2(g)  H2O(l) ΔHf° = -286 kJ/mol
2 H2O required: (-286 kJ/mol)(2) = ΔHf° = -572 kJ

30. Calculating ΔHf°
ΔHf° = (-ΔHf° CH4(g)) + (ΔHf° O2(g)) + (ΔHf° CO2(g)) (ΔHf° H2O(l))
ΔHf° = -(-75 kJ) + (0) + (-394 kJ) + (-572 kJ)
ΔHf° = -891 kJ

31. ΔHf°
If ΔHf° for a compound is negative, energy is released when the compound is formed from pure elements and the product is more stable that its constituent elements
Exothermic process
If ΔHf° for a compound is positive, energy is absorbed when the compound is formed from pure elements and the product is less stable than its constituent elements
Endothermic process