Chapter 5: Thermochemistry

1. Chapter 5: Thermochemistry Law of Conservation of Energy: Energy is neither created nor destroyed during a chemical or physical change. However... It can be transformed into a different type. Kinetic energy: energy associated with motion. EK = ½ mv2 Potential energy: stored energy that will be released if the object falls. EP = mgh Electrostatic potential energy: energy stored as a result of interactions between charged particles. k = 8.99 × 109 J-m/C2 Qe = -1.60 x 10-19 C

2. Energy is usually defined as the ability to do work (w) or to transfer heat (q). If we heat the system (+q), the expansion of the gas will lift the piston and do work on the surroundings (-w). The surroundings (w = F•d) As a result of the addition of heat and the loss of energy from the system, the internal energy (sum of all types of energy within the system) has changed. DE = Ef - Ei The system Note: It is very difficult to know the absolute internal energy of a system. Fortunately, DE can be found from q and w  DE = q + w

3. DE = -1150 J + 480 J = -670 J Two gases, A(g) and B(g), are confined in a cylinder-and-piston arrangement like that in Figure 5.3. Substances A and B react to form a solid product: A(g) + B(g)→ C(s). As the reactions occurs, the system loses 1150 J of heat to the surroundings. The piston moves downward as the gases react to form a solid. As the volume of the gas decreases under the constant pressure of the atmosphere, the surroundings do 480 J of work on the system. What is the change in the internal energy of the system? DE = q + w +480 J q = -1150 J w = gains heat. Recall: In an endothermic process, the system (Heat enters the system.) In an exothermic process, the system loses heat. (Heat exits the system.)

4. State functions depend only on the initial and final state, and are independent of the path taken. Initial DEP = Ef - Ei = -10 J This system has lost 10 J of potential energy to the surroundings. Examples of state functions Final DE, DH, DS, DG, DT EP = 0 J EP = 10 J Q: Are q and w state functions? No, both are dependent upon the path taken. (Think about trying to move a box across the floor.)

5. Thus, DH = qp +DH endothermic -DH exothermic Enthalpy: the amount of heat energy transferred when the system is under constant pressure. DE = qp + w DE = DH + w The only time w is nonzero is when gases are involved. w = F•d = -PDV DE = DH - PDV P DH = DE + PDV Recall:

6. Enthalpies of reaction (DHrxn): If 2 mol of H2 and 1 mol of O2 are combusted, DHrxn = -483.6 kJ 2 H2(g) + O2(g) 2 H2O (g) DH = -483.6 kJ Q: What would DH be if 4 mol of H2 and 2 mol of O2 were combusted? A: -967.2 kJ Note: While DH is a state function, it is NOT an intensive property. The enthalpy change depends upon the quantity of matter involved in the reaction. Q: What would DH be if 6 g of H2 was combusted in an excess of O2? Excess = the amount of O2 doesn’t matter 6 g H2/(2 g/mol) = 3 mol H2 x = -725.4 kJ

7. Hess’s Law states that if a reaction is carried out in a series of steps, ΔH for the overall reaction will equal the sum of the enthalpy changes for the individual steps. EX: The enthalpies of reaction for the combustion of C to CO2 and for CO to CO2 are shown below: DH = -393.5 kJ C + O2 CO2 CO + ½ O2 CO2 DH = -283.0 kJ flip What is the DHrxn for C + ½O2 CO ? DH 0.5 C + O2 CO2 -393.5 kJ + + CO2 CO + ½ O2 +283.0 kJ C + ½ O2 CO -110.5 kJ

8. C + O2 CO + ½ O2 CO2 An Enthalpy Diagram DH = —110.5 kJ DH = -393.5 kJ DH = —283.0 kJ

9. 2 AlBr3 + 3 Cl2 2 AlCl3 + 3 Br2 2 AlBr3 + 3 Cl2 2 AlCl3 + 3 Br2 Energy But how do we determine the heat content in the first place? DHrxn = Heat content of products – heat content reactants DHrxn < 0 Reaction is exothermic

10. 2 AlBr3 2 Al + 3 Br2 DHrxn DHf(AlBr3) = 2 Heat of formation, DHf • The DHf of all elements in their standard state equals zero. • The DHf of all compounds is the molar heat of reaction for synthesis of the compound from its elements DHrxn (AlBr3): Are forming 2 moles of AlBr3 so the DHrxn is two times larger than the DHf for one mole of AlBr3 Therefore, • Since the DHrxn can be used to find DHf, this means that DHf can be used to find DHrxnWITHOUT having to do all of the calorimetric measurements ourselves!! The Law of Conservation of Energy strikes again!!

11. Hess’s Law: DHrxn = S DHf(products) – S DHf(reactants) 6 CO2 (g) + 6 H2O (l) C6H12O6(s) + 6 O2(g) From DHf tables: DHf(C6H12O6) = -1250 kJ/mol DHf(CO2) = -393.5 kJ/mol DHf(H2O) = -285.8 kJ/mol DHrxn = [DHf(C6H12O6) + 6 DHf(O2)] – [6 DHf(CO2) + 6 DHf(H2O)] DHrxn = [-1250 kJ/mol] – [6(-393.5 kJ/mol) + 6(-285.8 kJ/mol)] DHrxn = +2825.8 kJ/mol

12. Calorimetry: The measurement of heat flow Calorimeter: A device to measure heat flow. Heat capacity: the amount of heat needed to make the temperature an object increase by 1C. Q: Is heat capacity an intensive or extensive property? A: Extensive. Think about how long it takes a large amount of water to heat up as compared to a much smaller amount with the same starting temperature. Specific heat: the amount of heat needed to increase the temperature of 1 gram of a substance by 1C. Q: Is specific heat an intensive or extensive property? A: Intensive because it is now on a per gram basis.

13. Q: What do you think the molar heat capacity is? A: The amount of heat needed to increase the temperature of one mole of a substance by 1C. q = mCpDT q = heat, m = mass, Cp = specific heat, DT = Tf - Ti Ex. How much heat is needed to warm 250 g of water (about 1 cup) from 22°C to near its boiling point, 98°C? The specific heat of water is 4.18 J/g−K. (b) What is the molar heat capacity of water? q = mCpDT = (250 g)(4.18 J g-1K-1)(98C – 22C) q = 7.9 x 104 J b)

14. If the same amount of heat was used to heat 100 g of water [Cp(liquid water) = 4.184 J/g°C], what would be the DT of the water? How much heat is required by a 100 g paraffin candle to increase the temperature by 5 °C? Cp(paraffin) = 2.1 J/g°C q = Cp(mass)(DT) q = (2.1 J/g°C)(100 g)(5 °C) q = 1050 J q = Cp(mass)(DT) 1050 J = (4.184 J/g°C)(100g)(DT) For the same amount of heat and mass, DT decreases as the specific heat of the substance increases DT = 2.5 °C

15. Constant Pressure Calorimetry q = mCpT T = Tf - Ti -(1x105 g)(0.13 J/g°C)(Tf – 327°C) = (1x104 g)(4.18 J/g°C)(Tf – 20°C) 10 kg H2O If the temperature of the lead is 327°C before it hits the water, what is the final temperature of the lead and the water? 100 kg Pb 200 ft -qPb = qH2O Ti = 20 °C Tf = 93 °C Cp(Pb) = 0.13 J/g°C Cp (H2O) = 4.18 J/g°C

16. Constant Volume Calorimetry: Bomb Calorimetry • The heat from the reaction (usually a combustion) causes the temperature of the Bomb and the surrounding water to increase. • The heat capacity of the bomb calorimeter, Ccal, must be determined using a reaction for which the DHcomb is known. qrxn = -CcalDT

17. Ex: A 0.5865-g sample of lactic acid (HC3H5O3) is burned in a calorimeter whose heat capacity is 4.812 kJ/°C. The temperature increases from 23.10°C to 24.95°C. Calculate the heat of combustion of lactic acid (a) per gram and (b) per mole. Ccal = 4.812 kJ/C DT = Tf – Ti = 24.95C – 23.10C = 1.85C qrxn = -CcalDT = (-4.812 kJ/C)(1.85C) = -8.90 kJ (a) Per gram: -8.90 kJ/0.5865 g = -15.2 kJ/g (-15.2 kJ/g)(90.079 g/mol) = -1370 kJ/mol (b) Per mole: FW(HC3H5O3) = 90.079 g/mol