Thermochemistry is the study of the energy changes that occur during chemical reactions.

1. Thermochemistry Thermochemistryis the study of the energy changes that occur during chemical reactions. Thermochemistry is part of thermodynamics, the study of heat, work, and energy in general and its transformations. Chemical reactions and physical processes like melting or dissolving can either produce heat (e.g. combustion reactions) or require heat (e.g. melting ice or dissolving NaCl in water).

2. Thermochemistry Chemical reactions that produce heat (feel hot) are calledexothermic. Chemical reactions that absorb heat (feel cold) are calledendothermic.

3. Energy Energy is the capacity to do work or transfer heat. Energy can either be kinetic or potential. Kinetic energy is the energy of motion. KE = ½ mv2

4. Energy Potential energyis “stored” energy related to an object’s position or to the composition of the substance (stored chemical energy due to the arrangement of atoms).

5. Work and Heat work(w) - energy used to change the motion of an object. (w = force x distance) heat (q) - energy transferred from a hotter object to a cooler one. (Temperature determines the direction of heat flow - it’s always hot to cold.) Temperature is related to the velocity of atoms and molecules (T  KE). So, temperature is proportional to heat energy, but it is not the same as heat.

6. Heat vs. Temperature A bucket of warm water can contain more heat energy than a burning match because the bucket of water has more mass (i.e. mass figures into how much heat is present - not just temperature).

7. Energy Units KE = ½ mv2 What is the kinetic energy of a 2 kg object moving at 1 m/s (2.2 mi/hr)? ½ (2 kg)(1 m/s)2 = 1 kg·m2/s2 = (1 kg·m/s2) · m Newton - a unit of force 1 Nm = 1 Joule (J) (w = F x d)

8. Energy Units A calorie (cal) was originally defined as the heat energy needed to increase the temperature of 1 gram of water from 14.5 C to 15.5 C. Now 1 calorie = 4.184 J (exactly). Joules and calories are both small units of energy. Heat and work are therefore often measured in kilojoules (kJ) or kilocalories (kcal) instead.

9. Energy Units Food calories are actually kilocalories (sometimes called “big calories”). A candy bar containing 260 Calories can produce enough heat to increase the temperature of 260 kg (almost 70 gallons) of water by 1 C or 1 gallon of water by almost 70 C.

10. Energy Units ABritish thermal unit (BTU or Btu) is the amount of heat needed to raise the temperature of one pound of water by 1 F. 1 Btu = 1054.35 J = 1.05435 kJ (Note that the Btu is only about 5% bigger than a kilojoule.)

11. Energy Units Given appropriate conversion factors, you should be able to convert between any two heat units. Convert 6.50 kcal to Btu. (1 cal = 4.184 J ; 1 Btu = 1054.35 J)

12. System and Surroundings In thermodynamics, it is important to be able to distinguish the system from the surroundings. The system is the object, reaction, or process that you are studying. The surroundings is everything else - literally. Surroundings System   

13. System and Surroundings The First Law of Thermodynamics says that energy is conserved. Energy cannot be created or destroyed - it can only move from one place to another (be transferred) or change into some other form of energy (be transformed). Surroundings System   

14. System and Surroundings Heat (q) and work (w) can go from the system to the surroundings or from the surrounding to the system. Surroundings System   q w  (Remember that the direction of heat flow is determined by which has the higher temperature.)

15. System and Surroundings Surroundings System   q w  Heat flows into the system when the system is colder than the surroundings - i.e. when the system is absorbing heat (endothermic). Heat flows to the surroundings from the system when the system is producing heat and is hotter than the surroundings (exothermic).

16. Internal Energy Surroundings System   q w  Heat and work going into or out of the system change theinternal energy(E) of the system. The change in a system’s internal energy (E) is the difference between the energy present after the heat and work transfer is complete and how much internal energy was present initially. E = Efinal - Einitial

17. Internal Energy Change (E) It’s hard to calculate Efinal and Einitial because you have to know all of the kinetic energy and potential energy components of the system (i.e. kinetic energy from motion of Earth - rotational and orbital, plus vibrations of atoms, electronic energy, etc.). However, you can calculate E by keeping track of the heat and work going into or out of the system. Know this. E = q + w

18. Internal Energy Change (E) Heat and work going into the system increase internal energy and can therefore be considered positive quantities with respect to the system. Heat and work coming out of the system represent an energy loss by the system and can therefore be considered negative quantities with respect to the system. Surroundings System   + + q w   

19. Internal Energy Change (E) Calculate the internal energy change for a system when it absorbs 125 J of heat while performing 95 J of work. Calculate the internal energy change in kJ for a system that releases 48.5 kcal of heat while 54.3 kJ of work are done on the system. (1 cal = 4.184 J)

20. PV Work When a gas in a cylinder expands, it has to push against the pressure applied to the piston - i.e. it has to exert a force. Since the piston moves as the gas expands, the force is exerted over a distance. Therefore work is being done. w = F x d (or h) h

21. PV Work Pressure is force per unit area (F/A). The volume change (V) in the cylinder = A x h. Combing these we get: P x V = F x A x h A Areas cancel, leaving us with: PV = F x h Since a volume increase (positive h) means the system is doing work, the internal energy change for the system is negative. Therefore, in the end we say: w = PV

22. State Functions E is a state function. In thermodynamics, a state function is any quantity whose value is independent of the pathway taken to get that quantity. Example: When you climb a mountain, your change in altitude is a state function. altitude

23. State Functions Temperature is also a state function. It doesn’t matter whether your room temperature water was once ice water or boiling water. It is now room temperature - that’s all that matters.

24. State Functions Volume and pressure are also state functions. However, heat and work are not state functions. Their values depend upon how the system loses heat or does work.

25. Enthalpy State functions are easier to work with, so the fact that heat is not a state function is a problem. Solution: Define a new quantity - enthalpy (H) - as the internal energy plus the PV work. H = E + PV Since E, P, and V are all state functions, H is a state function.

26. Enthalpy If H = E + PV we can then say that H = E + (PV) If pressure is constant (as in a system open to the atmosphere or in a cylinder with a moveable piston), then H = E + PV H = qP + w  w = qP where qP stands for heat flow measured under constant pressure conditions.

27. Enthalpy Change (H) Therefore enthalpy change (H) = the heat flow under constant pressure conditions (i.e. when the only work done is PV work). Therefore heat flow under constant pressure conditions (H or qP) is a state function. Even though we went to the trouble of defining this new state function, it turns out PV work is usually very small. Therefore, in most cases H  E.

28. Enthalpy Change (H) H = qP ; H = Hfinal - Hinitial We often put subscripts after the H to specify what process the heat flow is associated with. Hrxn = enthalpy of reaction or heat of reaction Hcomb = enthalpy or heat of combustion Hsoln = enthalpyor heatof solution (dissolution) Hvap = enthalpyor heatof vaporization Hfus = enthalpy or heat of fusion

29. Sign of H If the total enthalpy (or heat) of the products (Hfinal) is lower than the total enthalpy of the reactants (Hinitial), then heat is released during the reaction (H < 0). Thus an exothermic reaction is associated with a negative H.

30. Sign of H When the products have a higher enthalpy than the reactants, His positive and the reaction is endothermic. When you reverse a reaction, the size of the H is the same but the sign is opposite.

31. Thermochemical Equations A balanced chemical equation plus the H is called a thermochemical equation. 2 H2(g) + O2 (g)  2 H2O(g) H = -483.6 kJ Enthalpy changes depend upon the states of the substances involved, so the states should be specified. 2 H2(g) + O2 (g)  2 H2O(l) H = -571.7 kJ Enthalpy changes are extensive properties. (The more stuff you burn, the more heat you get.)

32. Using Thermochemical Equations You can incorporate H’s into stoichiometric relationships. 2 H2(g) + O2 (g)  2 H2O(g) H = -483.6 kJ Calculate the enthalpy change when 60.0 g of hydrogen burns to make water vapor.

33. Using Thermochemical Equations Calculate the enthalpy change needed to produce 80.0 g of H2O2. 2 H2O2 (l)  2 H2O(g) + O2(g) H = -196 kJ

34. Using Thermochemical Equations How many grams of liquid butane are needed to produce 1.50 x 104 kJ of heat in the following reaction? 2 C4H10(l) + 13 O2(g)  8 CO2(g) + 10 H2O(l) H = -5711 kJ

35. Learning Check How many grams of oxygen are needed to produce 655 kJ of heat in the following reaction? 2 C4H10(l) + 13 O2(g)  8 CO2(g) + 10 H2O(l) H = -5711 kJ

36. Learning Check How many liters of oxygen at STP are needed to produce 655 kJ of heat in the following reaction? 2 C4H10(l) + 13 O2(g)  8 CO2(g) + 10 H2O(l) H = -5711 kJ

37. Thermochemical Equations Until now, I’ve insisted on whole number coefficients in balanced equations. However, coefficients of thermochemical equations are always read as moles. While you can’t have half of an oxygen molecule, you can have half a mole of oxygen molecules. Therefore, coefficients in thermochemical equations can be fractions: H2(g) + ½ O2(g)  H2O(g) H = -242 kJ

38. Calorimetry • Measuring heat flows for processes is called calorimetry. A device used to measure heat flows is called a calorimeter. Calorimeters can be open (“coffee cup”) or closed (“bomb calorimeters”).

39. Calorimetry • Adding heat to a substance or an object increases its temperature. How much the temperature rises depends upon how much heat is added, the mass of the substance, and its heat capacity. • heat capacity (C) - the amount of heat needed to raise the temperature of a substance or object by 1 C or 1 K. (Note: T in K = T in C.) • Units: kJ/K kcal/C cal/C (Btu/F)

40. Calorimetry • molar heat capacity(Cm) - the amount of heat needed to raise the temperature of one mole of substance by 1 C or 1 K. (Heat capacity per mole.) • Units: J/molK or J/mol-K (Jmol-1K-1) J/molC • specific heat capacity (Cs) - the amount of heat needed to raise the temperature of one gram of substance by 1 C or 1 K. (Heat capacity per gram.) • Units: J/gK (Jg-1K-1) J/gC (Jg-1C-1)

41. Specific Heat Capacities In units of cal/gC, the Cs of H2O(l) = 1.00 cal/gC.

42. Calorimetry • The calorimetry equation: • q = m x Cs x T • Calculate the amount of heat required to raise the temperature of a 60.0 g block of Al by 125 C. • (Cs of Al = 0.90 J/gK) Know this.

43. Calorimetry • Calculate the temperature change when 4.59 kJ of heat is added to a 150.0 g block of Fe (Cs = 0.45 J/gK). • If the initial temperature of the block was 23 C, what is the final temperature of the block?

44. Calorimetry • Calculate the mass of an ingot of Ag (Cs = 0.237 J/gK) that increased its temperature from 25 C to 300. C when 20.0 kcal of heat were added. (1 cal = 4.184 J) • Given that silver has a density of 10.5 g/cm3, what is the volume of the silver ingot?

45. Calorimetry • What is the specific heat of a metal if a 60.0 gram block increases its temperature from 22.0 C to 48.5 C when 4.60 x 102 J of heat are added.

46. Learning Check • Calculate the temperature change when 5.50 kJ of heat is added to 150.0 g of water (Cs = 4.18 J/gK). • If the initial temperature of the water was 22.5 C, what is the final temperature of the block?