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Chapter 5: Thermochemistry Law of Conservation of Energy:Energy is neither created nor destroyed during a chemical or physical change. It can be transformed.

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Presentation on theme: "Chapter 5: Thermochemistry Law of Conservation of Energy:Energy is neither created nor destroyed during a chemical or physical change. It can be transformed."— Presentation transcript:

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2 Chapter 5: Thermochemistry Law of Conservation of Energy:Energy is neither created nor destroyed during a chemical or physical change. It can be transformed into a different type. Kinetic energy: energy associated with motion.E K = ½ mv 2 Potential energy: stored energy that will be released if the object falls. E P = mgh Electrostatic potential energy: energy stored as a result of interactions between charged particles.  = 8.99 × 10 9 J-m/C 2 Q e = -1.60 x 10 -19 C

3 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 system 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.  E = E f - E i Note: It is very difficult to know the absolute internal energy of a system. Fortunately,  E can be found from q and w   E = q + w

4 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?  E = q + w q =-1150 Jw = +480 J  E = -1150 J + 480 J = -670 J Recall:In an endothermic process, the system gains heat. In an exothermic process, the system (Heat enters the system.) loses heat. (Heat exits the system.)

5 State functions depend only on the initial and final state, and are independent of the path taken. Final Initial  E P = E f - E i = -10 J E P = 10 J E P =0 J This system has lost 10 J of potential energy to the surroundings. Examples of state functions  E,  H,  S,  G,  T 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.)

6 Enthalpy: the amount of heat energy transferred when the system is under constant pressure. Thus,  H = q p  E = q p + w  E =  H + w The only time w is nonzero is when gases are involved. w = F d = -P  V P  E =  H - P  V  H =  E + P  V Recall: -H-H exothermic +H+H endothermic

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

8 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 CO 2 and for CO to CO 2 are shown below: C + O 2  CO 2  H = -393.5 kJ CO + ½ O 2  CO 2  H = -283.0 kJ What is the  H rxn for C + ½O 2  CO ? C + O 2  CO 2 -393.5 kJ HH CO 2  CO + ½ O 2 flip +283.0 kJ ++ 0.5 C + ½ O 2  CO-110.5 kJ

9 An Enthalpy Diagram CO + ½ O 2 C + O 2 CO 2  H = -393.5 kJ  H = —283.0 kJ  H = —110.5 kJ

10 2 AlBr 3 + 3 Cl 2 2 AlCl 3 + 3 Br 2 Energy 2 AlBr 3 + 3 Cl 2 2 AlCl 3 + 3 Br 2  H rxn = Heat content of products – heat content reactants  H rxn < 0 Reaction is exothermic But how do we determine the heat content in the first place?

11 Heat of formation,  H f The  H f of all elements in their standard state equals zero. The  H f of all compounds is the molar heat of reaction for synthesis of the compound from its elements  H rxn (AlBr 3 ): 2 Al + 3 Br 2 2 AlBr 3  H rxn 2  H f (AlBr 3 ) = Since the  H rxn can be used to find  H f, this means that  H f can be used to find  H rxn WITHOUT having to do all of the calorimetric measurements ourselves!! The Law of Conservation of Energy strikes again!! Are forming 2 moles of AlBr 3 so the  H rxn is two times larger than the  H f for one mole of AlBr 3 Therefore,

12 Hess’s Law:  H rxn =  H f (products) –  H f (reactants) 6 CO 2 (g) + 6 H 2 O (l)C 6 H 12 O 6 (s) + 6 O 2 (g)  H rxn = [  H f (C 6 H 12 O 6 ) + 6  H f (O 2 )] – [6  H f (CO 2 ) + 6  H f (H 2 O)] From  H f tables:  H f (C 6 H 12 O 6 ) = -1250 kJ/mol  H f (CO 2 ) = -393.5 kJ/mol  H f (H 2 O) = -285.8 kJ/mol  H rxn = [-1250 kJ/mol] – [6(-393.5 kJ/mol) + 6(-285.8 kJ/mol)]  H rxn = +2825.8 kJ/mol

13 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.

14 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 = mC p  T q = heat, m = mass, C p = specific heat,  T = T f - T i 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 = mC p  T= (250 g)(4.18 J g -1 K -1 )(98  C – 22  C) q = 7.9 x 10 4 J b)

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

16 200 ft 100 kg Pb 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? C p (Pb) = 0.13 J/g°CC p (H 2 O) = 4.18 J/g°C q = mC p  T  T = T f - T i T f = 93 °C -q Pb = q H2O -(1x10 5 g)(0.13 J/g°C)(T f – 327°C) = (1x10 4 g)(4.18 J/g°C)(T f – 20°C) 10 kg H 2 O T i = 20 °C Constant Pressure Calorimetry

17 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, C cal, must be determined using a reaction for which the  H comb is known. q rxn = -C cal  T

18 Ex: A 0.5865-g sample of lactic acid (HC 3 H 5 O 3 ) 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. C cal = 4.812 kJ/  C  T = T f – T i = 24.95  C – 23.10  C = 1.85  C q rxn = -C cal  T = (-4.812 kJ/  C)(1.85  C) = -8.90 kJ (a) Per gram:-8.90 kJ/0.5865 g = -15.2 kJ/g (b) Per mole: FW(HC 3 H 5 O 3 ) = 90.079 g/mol (-15.2 kJ/g)(90.079 g/mol) = -1370 kJ/mol


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