1 Chapter 6 EnergyThermodynamics

2 Energy is... n The ability to do work. n Conserved. n made of heat and work. n a state function. n independent of the path, or how you get from point A to B. n Work is a force acting over a distance. n Heat is energy transferred between objects because of temperature difference.

3 The universe n is divided into two halves. n the system and the surroundings. n The system is the part you are concerned with. n The surroundings are the rest. n Exothermic reactions release energy to the surroundings. n Endo thermic reactions absorb energy from the surroundings.

4 Potential energy Heat

5 Potential energy Heat

6 Direction n Every energy measurement has three parts. 1. A unit ( Joules of calories). 2. A number how many. 3. and a sign to tell direction. n negative - exothermic n positive- endothermic

7 System Surroundings Energy  E <0

8 System Surroundings Energy  E >0

9 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 positive. n Work done on system by surroundings is negative. n Thermodynamics- The study of energy and the changes it undergoes.

10 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.

11 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

12 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 = J

13 Examples n What amount of work is done when 46 L of gas is expanded to 64 L at 15 atm pressure? w = - P  V -270 L atm n If 2.36 J of heat are absorbed by the gas above, what is the change in energy? J L atm x J/L atm

14 Examples A balloon is being inflated to its full extent by heating the air inside it. In the final stages of this process, the volume of the balloon changes from 4.00 x 106 L to 4.50 x 106 L by the addition of 1.3 x 108 J of energy as heat. Assuming tha the balloon expands against a constant pressure of 1.0 atm, calculate  E for this process. A balloon is being inflated to its full extent by heating the air inside it. In the final stages of this process, the volume of the balloon changes from 4.00 x 106 L to 4.50 x 106 L by the addition of 1.3 x 108 J of energy as heat. Assuming tha the balloon expands against a constant pressure of 1.0 atm, calculate  E for this process.

15 Enthalpy – background…. n n MOST experiments are carried out under constant pressure (i.e. in an open beaker or test tube). n n Since heat (q) at constant pressure is often being calculated in chemistry, a NEW symbol was used to describe this specific case of heat transfer.

16 Enthalpy n n The heat content of a substance at constant pressure is called enthalpy and has the symbol “H”. n H = E + PV n n Like internal energy (E), enthalpy (H) is impossible to measure. The change in enthalpy (∆H) is often used to symbolize the heat of a reaction.  H =  E + P  V  H =  E + P  V

17 Heat at constant pressure n the heat at constant pressure q p can be calculated from  E = q p + w = q p - P  V  E = q p + w = q p - P  V q p =  E + P  V =  H q p =  E + P  V =  H

18 Example When 1 mole of methane is burned at constant pressure, 890 kJ of energy is released as heat. Calculate the change in enthalpy for a process in which a 5.8 g sample of methane is burned at constant pressure. Examples – Book, # 35

19 Calorimetry n Measuring heat. n Use a calorimeter. n Two kinds n Constant pressure calorimeter (called a coffee cup calorimeter) n heat capacity for a material, C is calculated C= heat absorbed/  T =  H/  T C= heat absorbed/  T =  H/  T n specific heat capacity = C/mass

20 Calorimetry n molar heat capacity = C/moles heat = specific heat x m x  T heat = specific heat x m 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. A coffee cup calorimeter measures  H. A coffee cup calorimeter measures  H. n An insulated cup, full of water. n The specific heat of water is 1 cal/gºC Heat of reaction=  H = sh x mass x  T Heat of reaction=  H = sh x mass x  T

21 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. 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?

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

24 Properties n intensive properties not related to the amount of substance. n density, specific heat, temperature. n Extensive property - does depend on the amount of stuff. n Heat capacity, mass, heat from a reaction.

25 Hess’s Law n Enthalpy is a state function. n It is independent of the path. 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 reversed the sign of  H is changed If the reaction is multiplied, so is  H If the reaction is multiplied, so is  H

26 N2N2 2O 2 O2O2 NO 2 68 kJ NO kJ -112 kJ H (kJ)

27 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.

28  Hº= -394 kJ  Hº= -286 kJ Example Given calculate  Hº for this reaction Given calculate  Hº for this reaction  Hº= kJ

29 Example Given Calculate  Hº for this reaction  Hº= +77.9kJ  Hº= +495 kJ  Hº= kJ

30 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 n There is a table in Appendix 4 (pg A22)

31 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. n Write the equation for the formation of methanol CH 3 OH.

32 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. C 2 H 5 OH +3O 2 (g)  2CO 2 + 3H 2 O C 2 H 5 OH +3O 2 (g)  2CO 2 + 3H 2 O n which leads us to this rule.

33 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. C 2 H 5 OH +3O 2 (g)  2CO 2 + 3H 2 O C 2 H 5 OH +3O 2 (g)  2CO 2 + 3H 2 O n which leads us to this rule.