1. Brown, LeMay Ch 5 AP Chemistry
Thermochemistry
Brown, LeMay Ch 5
AP Chemistry

2. James Prescott Joule (1818-1889)
5.1: Thermochemistry
From Greek therme (heat); study of energy changes in chemical reactions
Energy: capacity do work or transfer heat
Joules (J) or calories (cal); 1 cal = J
Kinetic: energy of motion; dependent on mass & velocity
Applies to motion of large objects & molecules
Linked to thermal energy (object’s T above 0 K)
James Prescott Joule ( )
English, studied under John Dalton, took over his father’s job as brewer where he got interested in steam power.

3. Potential: stored in “fields” (gravitational and electrical/magnetic); dependant on position relative to another object
Applies to large objects where gravity is overriding force, but not significantly to molecules where gravity is negligible and electrostatic forces dominate
Associated with chemical energy; stored in arrangement of atoms or subatomic particles (electrostatic & nuclear forces, bonding between atoms)

4. Vocabulary
System: “isolated” portion of study (typically just the chemicals in a reaction)
Surroundings: everything else (container, room, Earth, etc.)
Three types of system
OPEN SYSTEM: can exchange mass and energy in the form of heat with its surroundings.
CLOSED SYSTEM: allows the transfer of energy in the form of heat but not mass.
ISOLATED SYTEM: does not allow the transfer of ether mass or energy.

5. Vocabulary Force: a push or pull on an object
Work: energy transferred to move an object a certain distance against a force: W = (F)(d)
Heat: energy transferred from a hotter object to a colder one

6. (So your 1 Calorie soft drink is really 1000 “real” calories!)
Energy Units
SI Unit for energy is the joule, J:
1 Joule=1 kg m2/s2
We sometimes use the calorie instead of the joule:
1 cal = J (exactly)
A nutritional Calorie: 1 Cal = 1000 cal = 1 kcal
(So your 1 Calorie soft drink is really 1000 “real” calories!)

7. Transferring Energy
There are 2 ways to transfer energy in or out of a system…
1) Work…applying a force to an object and moving it a distance (W=Fxd)
2) Heat…the energy transferred from hotter objects to colder objects
Energy is therefore capacity to do work OR transfer heat.
We will concern ourselves with the transfer of heat!

8. 5.2: Laws of Thermodynamics
“0th Law”: 2 systems are in thermal equilibrium when they are at the same T.
Thermal equilibrium is achieved when the random molecular motion of two substances has the same intensity (and therefore the same T.)
1st Law: Energy can be neither created nor destroyed, or, energy is conserved.
2nd and 3rd Laws: discussed in Ch. 19

9. The 1st Law of Thermodynamics
Energy cannot be created or destroyed.
The first law of thermodynamics is really just the law of conservation of energy.
The energy of (system + surroundings) is constant.
Thus any energy transferred from a system must be transferred to the surroundings (and vice versa).
Internal Energy: total energy of a system.
You cannot measure “absolute internal energy” so we measure the change in energy.
Change in internal energy…∆E= Ef - Ei

10. Internal energy, E DE = Efinal - Einitial
DE > 0 Increase in energy of system (gained from surroundings)
DE < 0 Decrease in energy of system (lost to surroundings)

11. When a system undergoes a chemical or physical change, the change in internal energy (E) is equal to the heat (q) added or liberated from the system plus the work (w) done on or by the system:
DE = q + w `

12. Sign Conventions (Table 5.1)
q > 0 Heat is added to system
q < 0 Heat is removed from system (into surroundings)
w > 0 Work done to system
w < 0 System does work on surroundings

13. Example:
Octane and oxygen gases combust within a closed cylinder in an engine. The cylinder absorbs 1150 J of heat and a piston is pushed down by 480 J during the reaction. What is the change in internal energy of the system?
q is (-) since heat leaves system; w is (-) since work is done by system. Therefore,
DE = q + w = (-1150 J) + (-480 J) = J
1630 J has been liberated from the system (C8H18 and O2) added to the surroundings (engine, atmosphere, etc.)

14. Endothermic and Exothermic Processes
• An endothermic process is one that absorbs heat from the surroundings. (+q)
An endothermic reaction feels cold.
Example--an “instant” ice pack
• An exothermic process is one that transfers heat to the surroundings. (-q)
An exothermic reaction feels hot.
Example--burning paper

15. Heat & reactions Endothermic: energy added to system, + DE
Exothermic: energy exits system, - DE E E

16. State Functions
A state function depends only on the initial and final states of a system.
Example: The altitude difference between Denver and Chicago does not depend on whether you fly or drive, only on the elevation of the two cities above sea level.
Similarly, the internal energy of 50 g of H2O(l) at 25 ºC does not depend on whether we cool 50 g of H2O(l) from 100 ºC to 25 ºC or heat 50 g of H2O(l) at 0 ºC to 25 ºC.

17. State functions
Property of a system that is determined by specifying its condition or its “state”
The value of a state function depends only on its present state and not on the history of the sample.
T & E are state functions.
Consider 50 g of water at 25°C: EH2O does not depend on how the water got to be 25°C (whether it was ice that melted or steam that condensed or…)
Work (w) and heat (q) are not state functions because the ratio of q and w are dependent on the scenario.

18. 5.3: Enthalpy, H ∆E is a state function.
Since most reactions occur in containers open to the air, w is often negligible. If a reaction produces a gas, the gas must do work to expand against the atmosphere. This mechanical work of expansion is called PV (pressure-volume) work.
∆E is a state function.

19. Enthalpy, H:
Enthalpy is heat transferred between the system and surroundings carried out under constant pressure. (Open containers are under constant pressure!)
Enthalpy is also a state function.
∆H is (+) for endothermic reactions…just like q!
∆H is (−) for exothermic reactions…just like q!

20. Enthalpy (H): change in the heat content (qp) of a reaction at constant pressure
H = E + PV
H = E + PV(at constant P)
H = (qp + w) + (-w)
H = qp

21. H < 0 Heat is released to surroundings - H in exothermic reaction
Sign conventions
H > 0 Heat is gained from surroundings + H in endothermic reaction
H < 0 Heat is released to surroundings - H in exothermic reaction
The enthalpy change that accompanies a reaction is called the enthalpy of reaction or heat of reaction (∆Hrxn).

22. Enthalpies of Reactions
Consider the equation for the production of water:
2H2(g) + O2(g) -----> 2H2O(g) ∆Hrxn = –483.6 kJ
-The equation tells us that kJ of energy are released to the surroundings when water is formed & the reaction would feel hot.
-These equations are called thermochemical equations.

23. Enthalpies of Reactions
Enthalpy is an “extensive property” which means it depends on the amount of reactant you start with. (“Intensive properties” do not depend on the quantity of the substance, i.e. – density.)
- Example: CH4(g) + 2O2(g)  CO2(g) + 2H2O(g) ∆H = -802 kJ
2CH4(g) + 4O2(g)  2CO2(g) + 4H2O(g) ∆H = kJ
If you reverse the reaction, change the sign on ∆H.
-Example: CO2(g) + 2H2O(g)  CH4(g) + 2O2(g) ∆H = +802 kJ
Enthalpy change for a reaction depends on the state of the reactants and products.
-Example: CH4(g) + 2O2(g) ----> CO2(g) + 2H2O(l) ∆H = -890 kJ
If water vapor is the product, ∆H = -802 kJ because…
2H2O(l) > 2H2O(g) ∆H = +88 kJ

24. 5.4: Enthalpy of Reaction (Hrxn)
Another example: heat of reaction:
Enthalpy is an extensive property (depends on amounts of reactants involved).
Ex: CH4 (g) + 2 O2 (g)  CO2 (g) + 2 H2O (l)
Hrxn = kJ
Combustion of 1 mol CH4 produces 890. kJ
… of 2 mol CH4 → (2)(-890. kJ) = kJ
What is the H of the combustion of 100. g CH4?

25. Hreaction = - Hreverse reaction
CH4 (g) + 2 O2 (g)  CO2 (g) + 2 H2O (l)
H = kJ
CO2 (g) + 2 H2O (l)  CH4 (g) + 2 O2 (g)
H = kJ
9 minute review of thermochemistry concepts on You Tube

26. Hrxn depends on states of reactants and products.
CO2 (g) + 2 H2O (g) CH4 (g) + 2 O2 (g) H = 802 kJ
2 H2O (l)  2 H2O (g) H = 88 kJ
So:
CO2 (g) + 2 H2O (l)  CH4 (g) + 2 O2 (g) H = 890. kJ
CH4 (g) + 2 O2 (g)
CH4(g) + 2 O2(g)
802 kJ
890. kJ
CO2 (g) + 2 H2O (g)
2 H2O (g)
88 kJ
2 H2O (l)
CO2(g) + 2 H2O(l)

27. 5.5: Calorimetry q = C DT q = m c DT Measurement of heat flow
Heat capacity, C: amount of heat required to raise T of an object by 1 K
q = C DT
Specific heat (or specific heat capacity, c): heat capacity of 1 g of a substance
q = m c DT
Ex: How much energy is required to heat 40.0 g of iron (c = 0.45 J/(g K) from 0.0ºC to 100.0ºC?
q = m c DT = (40.0 g)(0.45 J/(g K))(100.0 – 0.0 ºC)
= 1800 J

28. Calorimetry is an experimental technique used to measure the heat transferred in a physical or chemical process.
The apparatus used in this procedure is called as a “Calorimeter”. There are two types of calorimeters:
Constant Pressure Calorimeter: The coffee cup calorimeter is an example of this type of calorimeter. The system in this case is the “contents” of the calorimeter and the surroundings are cup and the immediate surroundings.

29. During the rxn:
qrxn + q solution = 0, where qrxn is the heat gained or lost in the chemical reaction and qsolutionis the heat gained or lost by solution. Heat exchange in this system (qrxn), is equal to enthalpy change. Assuming no heat transfer takes place between the system and surroundings, qrxn + q solution = chemlab.truman.edu

30. chemlab.truman.edu

31. Mg (s) + 2HCl (aq)  H2 (g) + MgCl2 (aq)
Sample Problems:
g of magnesium chips are placed in a coffee-cup calorimeter and ml of 1.00 M HCl is added to it. The reaction that occurs is:
Mg (s) + 2HCl (aq)  H2 (g) + MgCl2 (aq)
The temperature of the solution increases from 22.2oC (295.4 K) to 44.8 oC (318.0 K). What’s the enthalpy change for the reaction, per mole of Mg? (Assume specific heat capacity of solution is 4.20 J/(g * K) and the density of the HCl solution is 1.00 g/ml.)
Ans x 105 J/mol Mg

32. Sample Problem #2: ml of M HCl is mixed with the same amount and molarity of NaOH solution, inside a coffee-cup calorimeter. The temperature of the solutions before mixing was oC, and oC after mixing and letting the reaction occur. Find the molar enthalpy of the neutralization of the acid, assuming the densities of all solutions are 1.00 g/ml and their specific heat capacities are 4.20 J/(g * K).

33. Constant Volume Calorimeter
Constant Volume Calorimeter: The bomb calorimeter is an example of this type of calorimeter. The bomb, its contents and water are defined as the system. The heat generated by the combustion reaction warms the bomb and the water around it.
The heat measured at a constant volume (qv) is equal to internal energy change (d E). For a bomb calorimeter:
Qrxn + qbomb+ qwater = 0
Bomb Calorimeter 2 minute video YT 

34. Constant Volume Calorimeter: Bomb Calorimeter
Constant Volume Calorimeter: Bomb Calorimeter

35. Sample Problem
1. Octane (C8H18) is a primary constituent of gasoline, and it burns in air as follows:
C8H18 (l) + 25/2 O2 (g)  8CO2 (g) + 9H2O (g)
A sample of octane weighing 1.00g is burned in a constant volume calorimeter, which is an insulated container with 1.20 kg of water inside. The temperature of the water and the bomb rises from oC ( K) to oC ( K). The bomb’s heat capacity is 837 J/K. What is the heat of combustion per gram of octane? Per mole of octane?
Ans kJ/g; x 103 kJ/mol

36. Sample Problem:
2. A sample of table sugar (sucrose, C12H22O11) weighing 1.00g is burned in a bomb calorimeter. The temperature of the 1.50 x 103g of water in the calorimeter rises from oC to oC. The heat capacity of the bomb is 837 J/K and 4.20 J/(g * K) for water. Calculate the heat evolved per gram of sucrose and per mole of sucrose

37. Sample Problem
When 1.000g of olive oil (glyceryl trioleate, C57H104O6) is burned in pure oxygen in a bomb calorimeter, the temperature of the water bath increases from oC to oC. The calorimeter’s heat capacity is kJ/ oC and glyceryl trioleate combusts with oxygen as follows: C57H104O6 + 80O2  57CO2 + 52H2O.
(a) How many dietary Calories are in olive oil, per gram?
(b) What is the change in internal energy, E, in kJ for the combustion of 1 mol of glyceryl trioleate?
Ans. –0.521 Cal/g oil, x 103 kJ/mol

38. Laws of Thermochemistry
The magnitude of ΔH is directly proportional to the amount of reactant or product.
This rule allows us to relate ΔH for a thermochemical equation to the amount of product or reactant.
ΔH for a reaction is equal in magnitude but opposite in sign for the reverse reaction
The value of ΔH for a reaction is the same whether it occurs directly or in a series of steps. Called Hess’s Law

39. 5.6: Hess’ Law If a rxn is carried out in a series of steps,
Hrxn =  (Hsteps) = H1 + H2 + H3 + …
Germain Hess ( )
Ex. What is DHrxn of the combustion of propane?
C3H8 (g) + 5 O2 (g)  3 CO2 (g) + 4 H2O (l)
3 C (s) + 4 H2 (g)  C3 H8 (g) H1 = kJ
C (s) + O2 (g)  CO2 (g) H2 = kJ
H2 (g) + ½ O2 (g)  H2O (l) H3 = kJ
C3H8 (g)  3 C (s) + 4 H2 (g) H1 = kJ
Born in Switzerland, moved to St Petersburg at age 3 with his family. His father was an artist. Hess was also interested in geology.
3[ ] ( )
4[ ] ( )
Hrxn = ( ) + 4( ) = kJ

40. 5.7: Enthalpy of Formation (Hf)
Formation: a reaction that describes a substance formed from its elements
NH4NO3 (s)
Standard enthalpy of formation (Hf): forms 1 mole of compound from its elements in their standard state (at 298 K)
C2H5OH (l)
Hf = kJ
Hf of the most stable form of any element equals zero. H2, N2 , O2 , F2 , Cl2 (g)
Br2 (l), Hg (l)
C (graphite), P4 (s, white), S8 (s), I2 (s)
Ex: 2 N2 (g) + 4 H2 (g) + 3 O2 (g)  2
2 C (graphite) + 3 H2 (g) + ½ O2 (g) 

41. Hess’ Law (again) Ex. Combustion of propane:
C3H8 (g) + 5 O2 (g)  3 CO2 (g) + 4 H2O (l)
Given: Compound Hrxn (kJ/mol)
C3H8 (g)
CO2 (g)
H2O (l)
H2O (g)
Hrxn = [3( ) + 4( )] – [1( ) + 5(0)] = kJ