1. 1 Chapter 6 EnergyThermodynamics

2. 2 Energy is... n Conserved n Made of heat and work. –Work is a force acting over a distance –Heat is energy transferred between objects because of temperature difference. n A state function which means that the result is independent of the path, or how you get from point A to B.

3. 3 The universe n Is divided into two halves: the system and the surroundings. n The system is the part you are concerned with. n The surroundings are the rest. n Unfortunately, it is easier to measure the effect on the surroundings than the system directly

4. 4 The universe n Exothermic reactions release energy to the surroundings. n Endothermic reactions absorb energy from the surroundings.

5. 5 Potential energy Heat

6. 6 Potential energy Heat

7. 7 Direction n Every energy measurement has three parts. 1. A unit ( Joules or calories). 2. A number how many. 3. A sign to tell direction. n negative – exothermic –System to the surrounding n positive- endothermic –Surroundings to the system

8. 8 System Surroundings Energy  E <0

9. 9 System Surroundings Energy  E >0

10. 10 Same rules for heat and work n Heat given off is negative. n Heat absorbed is positive. n Work done by system on surroundings is negative. n Work done on system by surroundings is positive. n Thermodynamics- The study of energy and the changes it undergoes.

11. 11 First Law of Thermodynamics n The energy of the universe is constant. n Law of conservation of energy. n q = heat n w = work  E = q + w  E = q + w n Take the systems point of view to decide signs. Punch Line:  E = q (At constant P) Punch Line:  E = q (At constant P)

12. 12 What is work? n Work is a force acting over a distance. w= F x  d w= F x  d n P = F/ area n d = V/area w= (P x area) x  (V/area)= P  V w= (P x area) x  (V/area)= P  V n Work can be calculated by multiplying pressure by the change in volume at constant pressure. n units of liter - atm L-atm

13. 13 Work needs a sign n If the volume of a gas increases, the system has done work on the surroundings. n work is negative w = - P  V w = - P  V n Expanding work is negative. n Contracting, surroundings do work on the system w is positive. n 1 L atm = 101.3 J

14. 14 Example #1 n What amount of work is done when 15 L of gas is expanded to 25 L at 2.4 atm pressure? –volume increase – work is being done by the system on the surroundings (-) w = - P  V w = -(2.4atm)(25L-15L) w = -24L atm -24 L atm x (101.3J/Latm) = -2431 J = - 2.4kJ

15. 15 Example #2 If 2.36 J of heat are absorbed by the gas above. what is the change in energy?  E = q + w  E = 2.36 J + -2431 J  E = -2429 J

16. 16 Enthalpy (H) n H = E + PV (that’s the definition) –at constant pressure.  H =  E + P  V  H =  E + P  V –the heat at constant pressure q p can be calculated from  E = q p + w = q p – P  V (w=-P  V)  E = q p + w = q p – P  V (w=-P  V) q p =  E + P  V =  H q p =  E + P  V =  H –PUNCH LINE q p =  H

17. 17 Calorimetry n Measuring heat using a calorimeter. n Two kinds – the first kind being a –Constant pressure calorimeter (called a coffee cup calorimeter) n heat capacity for a material, C is calculated (heat required to change a substances temperature) C= heat absorbed/  T =  H/  T C= heat absorbed/  T =  H/  T

18. 18 Calorimetry n Specific heat capacity = C/mass n Molar heat capacity = C/moles heat = specific heat x mass x  T heat = specific heat x mass x  T heat = molar heat x moles x  T heat = molar heat x moles x  T n Make the units work and you’ve done the problem right.

19. 19 Calorimetry A coffee cup calorimeter measures  H. A coffee cup calorimeter measures  H. n The specific heat of water is 1 cal/gºC or 4.184 J/gC Heat of reaction=  H = SH x mass x  T Heat of reaction=  H = SH x mass x  T

20. 20 Examples n The specific heat of graphite is 0.71 J/gºC. Calculate the energy needed to raise the temperature of 75 kg of graphite from 294 K to 348 K.  H = SH x mass x  T  H = SH x mass x  T  H = 0.71J/gC (75000g)(54C)  H = 0.71J/gC (75000g)(54C)  H = 2876 kJ  H = 2876 kJ

21. 21 Examples n A 46.2 g sample of copper is heated to 95.4ºC and then placed in a calorimeter containing 75.0 g of water at 19.6ºC. The final temperature of both the water and the copper is 21.8ºC. What is the specific heat of copper?  H surr = -  H sys  H surr = -  H sys 75.0g (4.184J/g C)(2.2C) = -[46.2g(SH)(-73.6C)] 75.0g (4.184J/g C)(2.2C) = -[46.2g(SH)(-73.6C)] 690J = 3400g C (SH).203 J/g C = SH

22. 22 Calorimetry n Constant volume calorimeter is called a bomb calorimeter. n Material is put in a container with pure oxygen. Wires are used to start the combustion. The container is put into a container of water. n The heat capacity of the calorimeter is known and tested. Since  V = 0, P  V = 0,  E = q Since  V = 0, P  V = 0,  E = q

23. 23 Bomb Calorimeter n thermometer n stirrer n full of water n ignition wire n Steel bomb n sample

24. 24 Bomb Calorimeter n n A bomb calorimeter works in a similar manner as the coffee cup calorimeter, but there is one significant difference. n n In a coffee cup calorimeter, the reaction takes place in the water. In a bomb calorimeter, the reaction takes place in a sealed metal can, which is then placed in the water (contained in an insulated container).

25. 25 Bomb Calorimeter n n Analysis of the heat flow is a bit more complex than it was for the coffee cup calorimeter, because the heat flow absorbed by the metal parts of the calorimeter (the “bomb” part) must be taken into account: n n qrxn = - (qwater + qbomb) –where qwater = 4.18 J/(g·°C) x mwater x ΔT –The heat flow of the bomb is: –qbomb = Ccal x ΔT

26. 26 One More Twist – Phase Changes Molar Heat of Fusion – Energy required to change 1 mol from a solid to a liquid (visa versa) 6kJ/mol for water Molar Heat of Vaporization – Energy required to change 1 mol from a liquid to a gas (vise versa) 40.7kJ/mol for water

27. 27 Phase Changes If 20g of ice at 0C is placed in a calorimeter containing 100ml of water at 60C, what is the final temperature of the system? q phasechange + q heating = -q water 20g (1mol/18.02g) = 1.10mol 1.10mol(6000J/mol) + 20g(4.184J/gC)(T f -0) = -100g(4.184J/gC)(T f -60) 6600 + 83.68T f = -418.4T f + 25104 6600 + 83.68T f = -418.4T f + 25104 502.88 T f = 18504 502.88 T f = 18504 T f = 36.8 C T f = 36.8 C

28. 28 Problem Solving Strategies n n The 1st step is to ascertain whether the process is constant pressure (open to the atmosphere) or constant volume. –If it’s constant pressure, use ΔH = −ΔEsurr; –for constant volume it’s ΔErxn = −ΔEsurr. n n In many problems –ΔEsurr = (mass)(specific heat)ΔT. –To use this equation, you must determine the part of the overall system that is changing temperature.

29. 29 Problem Solving Strategies n n For example, if we add 1.0 g of Mg to 100.0 mL of 1.0 M HCl, it is not the Mg that is changing temperature, but rather the 100.0 mL of acidic solution in which the Mg is reacting. n n In this example, then, –ΔEsurr = (100.0 g)(4.184 J/[g oC])ΔT, –where we have used the usual assumptions stated above. We would not use the mass of Mg (1.0 g) and the specific heat of Mg (1.02 J/[g oC]) in the ΔEsurr equation because it’s not the Mg that is changing temperature.

30. 30 Problem Solving Strategies n n Always ask yourself this question: “What is actually changing temperature in this process?” n n Wherever the thermometer goes, that’s what is changing temperature. n n * Note: in some problems – the heat capacity is given, so you’ll use Δ Esurr = C Δ T, where C is heat capacity)

31. 31 Problem Solving Strategies n n In the equations Δ H = −Δ Esurr or Δ Erxn = −Δ Esurr, the units on both sides are joules (J)! n n Therefore, if the problem asks you to find an answer in J or kJ per mole (or g), you need to find J first and then divide by the appropriate moles (or g) n n Likewise, if the problem gives you ΔH (or ΔErxn) and asks you to find one of the terms in ΔEsurr, you first need to make sure that the units on ΔH (or ΔErxn) are J and not J per mole or J per gram

32. 32 Hess’s Law n Enthalpy is a state function. We can add equations to to come up with the desired final product, and add the  H We can add equations to to come up with the desired final product, and add the  H n Two rules –If the reaction is reversed the sign of  H is changed –If the reaction is multiplied, so is  H

33. 33 N2N2 2O 2 O2O2 NO 2 68 kJ NO 2 180 kJ -112 kJ H (kJ)

34. 34 Standard Enthalpy n The enthalpy change for a reaction at standard conditions (25ºC, 1 atm, 1 M solutions) Symbol  Hº Symbol  Hº n When using Hess’s Law, work by adding the equations up to make it look like the answer. n The other parts will cancel out.

35. 35  Hº= -394 kJ  Hº= -286 kJ Example Given calculate  Hº for this reaction Given calculate  Hº for this reaction  Hº= -1300. kJ

36. 36 Example Given Calculate  Hº for this reaction  Hº= +77.9kJ  Hº= +495 kJ  Hº= +435.9kJ

37. 37 Standard Enthalpies of Formation n Hess’s Law is much more useful if you know lots of reactions. n Made a table of standard heats of formation. The amount of heat needed to for 1 mole of a compound from its elements in their standard states. n Standard states are 1 atm, 1M and 25ºC n For an element it is 0

38. 38 Standard Enthalpies of Formation n Need to be able to write the equations. n What is the equation for the formation of NO 2 ? ½N 2 (g) + O 2 (g)  NO 2 (g) ½N 2 (g) + O 2 (g)  NO 2 (g) n Have to make one mole to meet the definition.

39. 39 Since we can manipulate the equations n We can use heats of formation to figure out the heat of reaction. n Lets do it with this equation: Δ H = ∑ nΔ H f (products) – ∑ nΔ H f (reactants) Δ H = ∑ nΔ H f (products) – ∑ nΔ H f (reactants)

40. 40 Example n Use the table of standard enthalpies of formation at 25°C to calculate Δ H for the reaction 4NH 3 ( g ) + 5O 2 ( g ) → 6H 2 O( g ) + 4NO( g ) 4NH 3 ( g ) + 5O 2 ( g ) → 6H 2 O( g ) + 4NO( g ) Δ H = ∑ nΔ H f (products) – ∑ nΔ H f (reactants) Δ H = ∑ nΔ H f (products) – ∑ nΔ H f (reactants) = [6 Δ H f (H 2 O) + 4 Δ H f (NO)] – [4 Δ H f (NH 3 ) + 5 Δ H f (O 2 )] = [6 Δ H f (H 2 O) + 4 Δ H f (NO)] – [4 Δ H f (NH 3 ) + 5 Δ H f (O 2 )] = 6(–241.8) kJ mol –1 + 4(90.3) kJ mol –1 – 4(–46.1 kJ mol –1 ) – 5 × 0 = 6(–241.8) kJ mol –1 + 4(90.3) kJ mol –1 – 4(–46.1 kJ mol –1 ) – 5 × 0 = –1450.8 kJ mol –1 + 361.2 kJ mol –1 + 184.4 kJ mol –1 = –1450.8 kJ mol –1 + 361.2 kJ mol –1 + 184.4 kJ mol –1 = –905.2 kJ mol –1 = –905.2 kJ mol –1