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. ( dependant only on present state- independent of the path, or how you get from point A to B. ( difference between altitude differences and distance differences) n Work is a force acting over a distance. ( not a state function) n Heat is energy transferred between objects because of temperature difference. ( not a state function)

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 Endothermic 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 negative. n Work done on system by surroundings is positive. 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 n What amount of work is done when 15 L of gas is expanded to 25 L at 2.4 atm pressure? n If 2.36 J of heat are absorbed by the gas above. what is the change in energy? n How much heat would it take to change the gas without changing the internal energy of the gas? Examples

14 Enthalpy n abbreviated H n H = E + PV (that’s the definition) n at constant pressure.  H =  E + P  V  H =  E + P  V n the heat at constant pressure q p can be calculated from q p =  E + P  V =  H q p =  E + P  V =  H

15 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 n C= heat absorbed/ change in temperature Or C=  H/  T Or C=  H/  T n specific heat capacity = C/mass

16 Calorimetry n molar heat capacity = C/moles heat = specific heat x m x  T heat = specific heat x m x  T n Specific heat is given or look it up, Mass is mass of the water or substance thermometer is in  T is the change of the temp. you will measure Mass is mass of the water or substance thermometer is in  T is the change of the temp. you will measure n Use this if in moles 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

17 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?

18 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

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

20 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.

21 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

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

23 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.

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

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

26 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. Standard states are 1 atm, 1M and 25ºC n For an element it is 0 n There is a table in Appendix 4 (pg A20)

27 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.

28 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.

29 The equation to use…

30 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. 4NH 3 +7O 2 (g)  4NO 2 + 6H 2 O 4NH 3 +7O 2 (g)  4NO 2 + 6H 2 O n which leads us to this rule.

31 which leads us to this rule. 4NH3+7O2(g) ® 4NO2 + 6H2O First figure how to break down each compound reactant, and how to create each compound product. 2 nd – multiply those formation numbers by the moles in your problem. 3 rd - all elements have 0 formation heat.

32 4NH 3 +7O 2 (g)  4NO 2 + 6H 2 O n So NH 3 N 2 NO 2 n H 2 n O 2 H 2 O n See pg 247 or back of book for # n Heat of formation for NH3 is -46, this is decomp so 46, and need 4 of them n Heat NO 2 is 34 x4 and for water -28 x6 n Add all that up, make sure the equ. balances

33 CH 4 + O 2 = CO 2 + H 2 0 n You try!!!

34 Present sourses of energy n Petroleum, natural gas, coal n New energy= coal conversion, Hydrogen fuel ethanol n Read