1. 1 Thermochemistry Brown, LeMay Ch 5 AP Chemistry Adapted from: http:// matador.mvhs.fuhsd.org/kavita_gupta/ch5ppt.ppt http:// matador.mvhs.fuhsd.org/kavita_gupta/ch5ppt.ppt Monta Vista High School

2. 2 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 = 4.184 J 1.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 (1818-1889)

3. 3 2. 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. 4 Vocabulary System: “isolated” portion of study (typically just the chemicals in a reaction) Surroundings: everything else (container, room, Earth, etc.) Closed system: easiest to study because exchanges energy with surroundings but matter is not exchanged. 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

5. 5 5.2: Laws of Thermodynamics “0 th 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.) 1 st Law: Energy can be neither created nor destroyed, or, energy is conserved. 2 nd and 3 rd Laws: discussed in Ch. 19

6. 6 Internal energy, E Includes:  Translational motion  Rotational motion of particles through space  Internal vibrations of particles. It is difficult to measure all E, so the change in internal energy (  E) is typically measured:  E = E final - E initial  E > 0Increase in energy of system (gained fromsurroundings)  E < 0Decrease in energy of system (lost to surroundings)

7. 7 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:  E = q + w

8. 8 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

9. 9 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,  E = q + w = (-1150 J) + (-480 J) = - 1630 J  1630 J has been liberated from the system (C 8 H 18 and O 2 ) added to the surroundings (engine, atmosphere, etc.) Example:

10. 10 Heat & reactions Endothermic: energy added to system, +  E Exothermic: energy exits system, -  E E E

11. 11 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: E H2O 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. Consider the combustion of gasoline in a car engine vs. burning in the open.

12. 12 5.3: Enthalpy, H 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. Enthalpy (H): change in the heat content (q p ) of a reaction at constant pressure H = E + PV  H =  E + P  V(at constant P)  H = (q p + w) + (-w)  H = q p

13. 13 Sign conventions  H > 0 Heat is gained from surroundings +  H in endothermic reaction  H < 0 Heat is released to surroundings -  H in exothermic reaction

14. 14 5.4: Enthalpy of Reaction (  H rxn ) Also called heat of reaction: 1.Enthalpy is an extensive property (depends on amounts of reactants involved). Ex: CH 4 (g) + 2 O 2 (g)  CO 2 (g) + 2 H 2 O (l)  H rxn = - 890. kJ  Combustion of 1 mol CH 4 produces 890. kJ … of 2 mol CH 4 → (2)(-890. kJ) = -1780 kJ What is the  H of the combustion of 100. g CH 4 ?

15. 15 2.  H reaction = -  H reverse reaction CH 4 (g) + 2 O 2 (g)  CO 2 (g) + 2 H 2 O (l)  H = - 890. kJ CO 2 (g) + 2 H 2 O (l)  CH 4 (g) + 2 O 2 (g)  H = +890. kJ 9 minute review of thermochemistry concepts on You Tube

16. 88 kJ 802 kJ CO 2 (g) + 2 H 2 O (g) 3.  H rxn depends on states of reactants and products. CO 2 (g) + 2 H 2 O (g)  CH 4 (g) + 2 O 2 (g)  H = 802 kJ 2 H 2 O (l)  2 H 2 O (g)  H = 88 kJ So: CO 2 (g) + 2 H 2 O (l)  CH 4 (g) + 2 O 2 (g)  H = 890. kJ CH 4 (g) + 2 O 2 (g) 2 H 2 O (l) 2 H 2 O (g) CO 2 (g) + 2 H 2 O(l) CH 4 (g) + 2 O 2 (g) 890. kJ

17. 17 5.5: Calorimetry Measurement of heat flow Heat capacity, C: amount of heat required to raise T of an object by 1 K q = C  T Specific heat (or specific heat capacity, c ): heat capacity of 1 g of a substance q = m c  T 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  T = (40.0 g)(0.45 J/(g K))(100.0 – 0.0 ºC) = 1800 J

18. 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. 18

19. During the rxn: q rxn + q solution = 0, where q rxn is the heat gained or lost in the chemical reaction and q solution is the heat gained or lost by solution. Heat exchange in this system (q rxn ), is equal to enthalpy change. Assuming no heat transfer takes place between the system and surroundings, q rxn + q solution = 0 chemlab.truman.edu 19

20. 20 chemlab.truman.edu

21. 21 Sample Problems: 1. 0.500g of magnesium chips are placed in a coffee-cup calorimeter and 100.0 ml of 1.00 M HCl is added to it. The reaction that occurs is: Mg (s) + 2HCl (aq)  H 2 (g) + MgCl 2 (aq) The temperature of the solution increases from 22.2 o C (295.4 K) to 44.8 o C (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. -4.64 x 10 5 J/mol Mg

22. 22 Sample Problem #2:. 200. ml of 0.400 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 25.10 o C, and 27.78 o C 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).

23. Constant Volume Calorimeter 23 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 (q v ) is equal to internal energy change (d E). For a bomb calorimeter: Q rxn + q bomb + q water = 0 Bomb Calorimeter 2 minute video YT

24. Constant Volume Calorimeter: Bomb Calorimeter 24 www.chem.ufl.edu/~itl/2045/matter/FG05_016.GIF

25. Sample Problem 25 1. Octane (C 8 H 18 ) is a primary constituent of gasoline, and it burns in air as follows: C 8 H 18 (l) + 25/2 O 2 (g)  8CO 2 (g) + 9H 2 O (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 25.00 o C (298.15 K) to 33.20 o C (306.35 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. -48.1 kJ/g; -5.49 x 10 3 kJ/mol

26. Sample Problem: 26 2. A sample of table sugar (sucrose, C 12 H 22 O 11 ) weighing 1.00g is burned in a bomb calorimeter. The temperature of the 1.50 x 10 3 g of water in the calorimeter rises from 25.00 o C to 27.32 o C. 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

27. Sample Problem 27 When 1.000g of olive oil (glyceryl trioleate, C 57 H 104 O 6 ) is burned in pure oxygen in a bomb calorimeter, the temperature of the water bath increases from 22.000 o C to 22.241 o C. The calorimeter’s heat capacity is 9.032 kJ/ o C and glyceryl trioleate combusts with oxygen as follows: C 57 H 104 O 6 + 80O 2  57CO 2 + 52H 2 O. (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, -1.93 x 10 3 kJ/mol

28. 28 5.6: Hess’ Law If a rxn is carried out in a series of steps,  H rxn =  (  H steps ) =  H 1 +  H 2 +  H 3 + … Ex. What is  H rxn of the combustion of propane? C 3 H 8 (g) + 5 O 2 (g)  3 CO 2 (g) + 4 H 2 O (l) 3 C (s) + 4 H 2 (g)  C 3 H 8 (g)  H 1 = -103.85 kJ C (s) + O 2 (g)  CO 2 (g)  H 2 = -393.5 kJ H 2 (g) + ½ O 2 (g)  H 2 O (l)  H 3 = -285.8 kJ  H  rxn = 103.85 + 3(- 393.5) + 4(- 285.8) = - 2219.8 kJ 3[ ] 3( ) 4[ ] 4( ) Germain Hess (1802-1850) C 3 H 8 (g)  3 C (s) + 4 H 2 (g)  H 1 = +103.85 kJ

29. 29 5.7: Enthalpy of Formation (  H f ) Formation: a reaction that describes a substance formed from its elements NH 4 NO 3 (s) Standard enthalpy of formation (  H f  ): forms 1 mole of compound from its elements in their standard state (at 298 K) C 2 H 5 OH (l)  H f  = - 277.7 kJ   H f  of the most stable form of any element equals zero. H 2, N 2, O 2, F 2, Cl 2 (g) Br 2 (l), Hg (l) C (graphite), P 4 (s, white), S 8 (s), I 2 (s) Ex: 2 N 2 (g) + 4 H 2 (g) + 3 O 2 (g)  2 2 C (graphite) + 3 H 2 (g) + ½ O 2 (g) 

30. 30 Hess’ Law (again) Ex. Combustion of propane: C 3 H 8 (g) + 5 O 2 (g)  3 CO 2 (g) + 4 H 2 O (l) Given:Compound  H  rxn (kJ/mol) C 3 H 8 (g)-103.85 CO 2 (g)-393.5 H 2 O (l)-285.8 H 2 O (g)-241.82  H  rxn = [3(- 393.5) + 4(- 285.8)] – [1(-103.85) + 5(0)] = - 2219.8 kJ