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 or 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 by the system is negative. n Heat absorbed by the system is positive. n Work done by the system on surroundings is positive. n Work done on the 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. (“cannot be created or destroyed”) n AKA, Law of conservation of energy.  E = q + w, where  E = q + w, where  q = heat  w = work n Take the system’s point of view to decide signs.

11 What is work? n From physics, work is a force acting over a distance. 1. w= F x  d (physics) 2. P = F/ area 3. d = V/area 4. w= (P x area) x  (V/area)= P  V (chemistry) n Work can be calculated by multiplying pressure by the change in volume at constant pressure. n Units: 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. 1 L atm = J 1 L atm = J

13 Examples 1. What amount of work is done when 15 L of gas is expanded to 25 L at 2.4 atm pressure? 2. If 2.36 J of heat are absorbed by the gas above. what is the change in energy? 3. How much heat would it take to change the gas without changing the internal energy of the gas?

14 Enthalpy (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:   E = q p + w = q p – P  V (w = -PΔV)  q p =  E + P  V =  H, so  q p = ΔH

15 Calorimetry n Measures heat; uses a calorimeter. n Two kinds: 1.Constant pressure calorimeter (called a coffee cup calorimeter) 2.Constant volume calorimeter (called a bomb calorimeter)

16 Calorimetry Specific heat capacity: the energy required to raise the temperature of one gram of a substance by one degree Celsius Specific heat capacity: the energy required to raise the temperature of one gram of a substance by one degree Celsius n Heat capacity for a material (C) is calculated:  C = heat absorbed/  T =  H/  T  specific heat capacity = J/g °C  molar heat capacity = J/moles °C

17 Calorimetry Energy released by a reaction (heat) = specific heat capacity x mass x  T Energy released by a reaction (heat) = specific heat capacity x mass x  T heat = molar heat capacity x moles x  T heat = molar heat capacity 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. –An insulated cup, full of water. –The specific heat of water is 1 cal/gºC or 4.18 J/gºC –Heat of reaction=  H = q = c x m x  T

18 Examples 4. 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. 5. A 24.3 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?

19 ASIM Labs 1. Mixing Warm and Cold (Calorimetry) 2. Specific Heat of Metals

20 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

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

22 Properties n Intensive properties not related to (independent of) the amount of substance. n EX: density, specific heat, temperature. n Extensive property - does depend on the amount of stuff. n EX: heat capacity, mass, heat from a reaction.

23 Hess’s Law n Enthalpy is a state function; it is independent of the path. We can add reaction equations to come up with the desired final product, and add the  H’s. We can add reaction equations to come up with the desired final product, and add the  H’s. n Two rules: 1.If the reaction is reversed, the sign of  H is changed. 2.If the reaction is multiplied, so is  H.

24 N2N2 + O 2 2NO 68 kJ 2NO kJ -112 kJ H (kJ) N 2 + 2O 2 → 2NO 2

25 Standard Enthalpy n The enthalpy change for a reaction at standard conditions (25ºC, 1 atm, 1 M solutions) Symbol:  H° Symbol:  H° 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.

26  Hº= -394 kJ  Hº= -286 kJ Example 6. Given n calculate  Hº for this reaction  Hº= kJ

27 Example 7.Given Calculate  Hº for this reaction  Hº= +77.9kJ  Hº= +495 kJ  Hº= kJ

28 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 form 1 mole of a compound from its elements in their standard states. –Standard states are 1 atm, 1M and 25ºC –For an element it is 0 n There is a table in Appendix 4 (pg A21)

29 Standard Enthalpies of Formation n Need to be able to write the equations, starting with the individual elements. 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. –Start with the elemental forms!!!

30 Since we can manipulate the equations: n We can use heats of formation to figure out the heat of reaction. n This leads us to this rule: 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