Geothermal power Wairakei North Island, New Zealand

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2 Geothermal power Wairakei North Island, New Zealand 2

3 3 Burning peanuts supply sufficient energy to boil a cup of water. Burning sugar (sugar reacts with KClO 3, a strong oxidizing agent)

4 4 These reactions are PRODUCT FAVORED They proceed almost completely from reactants to products, perhaps with some outside assistance.

5 5 2 H 2 (g) + O 2 (g) --> 2 H 2 O(g) + heat and light This can be set up to provide ELECTRIC ENERGY in a Oxidation: in a. 2 H 2 ---> > 4 H e - Reduction: 4 e - + O H 2 O ---> > 4 OH - CCR, page 845

6 6 ENERGY is the capacity to do work or transfer heat. HEAT is the form of energy that flows between 2 objects because of their difference in temperature. Other forms of energy light electrical kinetic and potential

7 7 Potential energy energy a motionless body has by virtue of its position.

8 8 Positive and negative particles (ions) attract one another. Two atoms can bond As the particles attract they have a lower potential energy NaCl composed of Na + and Cl - ions.

9 9 Positive and negative particles (ions) attract one another. Two atoms can bond As the particles attract they have a lower potential energy

10 10 Kinetic energy energy of motion Translation

11 11 Kinetic energy energy of motion. rotate vibrate translate

12 12!" PE + KE = Internal energy (E or U) Int. E of a chemical system depends on number of particles type of particles temperature

13 13!" PE + KE = Internal energy (E or U) QuickTime and a Graphics decompressor are needed to see this picture.

14 14!" The higher the T the higher the internal energy So, use changes in T (T) to monitor changes in E (E).

15 15 Thermodynamics is the science of heat (energy) transfer. Heat energy is associated with molecular motions. Heat transfers until #$ is established.

16 % 16 Heat always transfer from hotter object to cooler one. &thermic: heat transfers from to ''( %( ). T(system) ) goes down T(surr) ) goes up

17 % 17 Heat always transfer from hotter object to cooler one. ( %thermic: heat transfers from ''( %( ). T(system) ) goes up T (surr( surr) ) goes down

18 18 All of thermodynamics depends on the law of ( * ( +( ), The total energy is unchanged in a chemical reaction. If PE of products is less than reactants, the difference must be released as KE.

19 19 PE Reactants Kinetic Energy Products PE of system dropped. KE increased. Therefore, you often feel a T increase.

20 '( +( ) 20 1 calorie = heat required to raise temp. of 1.00 g of H 2 O by 1.0 o C cal = 1 kilocalorie = 1 kcal 1 kcal = 1 Calorie (a food calorie ) But we use the unit called the JOULE 1 cal = joules James Joule

21 21 The heat required to raise an object s T by 1 C. Which has the larger heat capacity?

22 -- 22 How much energy is transferred due to T difference? The heat!#" lost or gained is related to a) sample mass b) change in T and c) specific heat capacity Specific heat capacity = heat lost or gained by substance (J) (mass, g)(t change, K)

23 -- 23 Substance Spec. Heat (J/g K g K) H 2 O Ethylene glycol 2.39 Al glass 0.84 Aluminum

24 24 -- If 25.0 g of Al cool from 310 o C to 37 o C, how many joules of heat energy are lost by the Al? Specific heat capacity = heat lost or gained by substance (J) (mass, g)(t change, K)

25 -- 25 If 25.0 g of Al cool from 310 o C to 37 o C,, how many joules of heat energy are lost by the Al? heat gain/lose = q = (sp. ht.)(mass)(t) where T = T final - T initial q = (0.897 J/g K)(25.0 g)(37-310)k q = J Notice that the negative sign on q signals heat lost by or transferred OUT of Al.

26 ( 26 #.!-,,"! "!/"

27 0 Changes of state involve energy!" Ice J/g (heat of fusion) -----> > Liquid water #.!"! "! " 27

28 28 Liquid ---> > Vapor Requires energy (heat). This is the reason a) you cool down after swimming b) you use water to put out a fire. + energy

29 Evaporate water Heat water Melt ice Note that T is constant as ice melts

30 30 What quantity of heat is required to melt 500. g of ice and heat the water to steam at 100 o C? Heat of fusion of ice = 333 J/g Specific heat of water = 4.2 J/g K Heat of vaporization = 2260 J/g +333 J/g J/g

31 31 What quantity of heat is required to melt 500. g of ice and heat the water to steam at 100 o C? 1. To melt ice q = (500. g)(333 J/g) = 1.67 x 10 5 J 2. To raise water from 0 o C to 100 o C q = (500. g)(4.2 J/g K)(100-0)K = 2.1 x 10 5 J 3. To evaporate water at 100 o C q = (500. g)(2260 J/g) = 1.13 x 10 6 J 4. Total heat energy = 1.51 x 10 6 J = 1510 kj

32 2 32 What drives chemical reactions? How do they occur? The first is answered by %( and the second by (. Have already seen a number of driving forces for reactions that are PRODUCT-FAVORED FAVORED. formation of a precipitate gas formation H 2 O formation (acid-base reaction) electron transfer in a battery

33 2 But energy transfer also allows us to predict reactivity , 33 So, let us consider heat transfer in chemical processes.

34 34 CO 2 (s, -78 o C) ---> > CO 2 (g, -78 o C) Heat transfers from surroundings to system in endothermic process.

35 35 CO 2 (s, -78 o C) ---> > CO 2 (g, -78 o C) A regular array of molecules in a solid -----> > gas phase molecules. Gas molecules have higher kinetic energy.

36 62% 36 CO 2 gas E = E(final) - E(initial) = E(gas) - E(solid) CO 2 solid

37 37 CO 2 (s, -78 o C) ---> > CO 2 (g, -78 o C) Two things have happened! Gas molecules have higher kinetic energy. Also, WORK is done by the system in pushing aside the atmosphere.

38 %( heat energy transferred E = q + w energy change work done by the system 27

39 heat transfer in (endothermic), +q heat transfer out (exothermic), -q 39 SYSTEM E = q + w w transfer in (+w) w transfer out (-w)

40 40 ( 6 Most chemical reactions occur at constant P, so Heat transferred at constant P = q p q p = H where H = enthalpy and so E = H + w (and w is usually small) H = heat transferred at constant P E H = change in heat content of the system H = H final - H initial

41 41 H = H final - H initial If H final final > H initial then H is is positive Process is ENDOTHERMIC If H final ( 6 Process is is ENDOTHERMIC If H final final < H initial then H is is negative Process is EXOTHERMIC If H final Process is is EXOTHERMIC

42 42 ' ( )( 6 Consider the formation of water H 2 (g) + 1/2 O 2 (g) --> > H 2 O(g) + O(g) kj Exothermic reaction heat is a product and H = kj

43 ' ( )( 6 43 H 2 + O 2 gas Making liquid H 2 O from H 2 + O 2 involves two exothermic steps. H 2 O vapor Liquid H 2 O

44 ' ( )( 6 44 Making H 2 O from H 2 involves two steps. H 2 (g) + 1/2 O 2 (g) ---> > H 2 O(g) kj H 2 O(g) ---> > H 2 O(liq) + 44 kj H 2 (g) + 1/2 O 2 (g) --> > H 2 O(liq) kj Example of , : 4 / /8 9 9,

45 860 62% 45 Forming H 2 O can occur in a single step or in a two steps. H total is the same no matter which path is followed. Figure 6.18, page 227

46 860 62% 46 Forming CO 2 can occur in a single step or in a two steps. H total is the same no matter which path is followed. Figure 6.18, page 227

47 Σ H along one path = Σ H along another path 47 This equation is valid because H is a +'( ( These depend only on the state of the system and not how it got there. V, T, P, energy and your bank account! Unlike V, T, and P, one cannot measure absolute H. Can only measure H.

48 48 -* Most H values are labeled H o Measured under P = 1 bar Concentration = 1 mol/l T = usually 25 o C with all species in standard states e.g., C = graphite and O 2 = gas

49 -* 49 Depend on how the reaction is written and on phases of reactants and products H 2 (g) + 1/2 O 2 (g) --> > H 2 O(g) 2 H 2 (g) + O 2 (g) --> > 2 H 2 O(g) H 2 O(g) ---> > H 2 (g) + 1/2 O 2 (g) H 2 (g) + 1/2 O 2 (g) --> > H 2 O(liquid) H = -242 kj H = -484 kj H = +242 kj H = -286 kj

50 50 -* NIST (Nat l Institute for Standards and Technology) gives values of /. - the enthalpy change when 1 mol of compound is formed from elements under standard conditions. See Table 6.2 and Appendix L

51 / 4-51 H 2 (g) + 1/2 O 2 (g) --> > H 2 O(g) H o f (H 2 O, g)= kj/mol By definition, H o f = 0 for elements in their standard states.

52 ' -* 52 Use H s to calculate enthalpy change for H 2 O(g) + C(graphite) --> > H 2 (g) + CO(g) (product is called water gas )

53 ' -* 53 H 2 O(g) + C(graphite) --> > H 2 (g) + CO(g) From reference books we find H 2 (g) + 1/2 O 2 (g) --> > H 2 O(g) H f of H 2 O vapor = kj/mol C(s) ) + 1/2 O 2 (g) --> CO(g) H f of CO = kj/mol

54 ' -* 54 H 2 O(g) --> > H 2 (g) + 1/2 O 2 (g) H o = +242 kj C(s) ) + 1/2 O 2 (g) --> CO(g) H o = -111 kj H 2 O(g) + C(graphite) --> > H 2 (g) + CO(g) net = +131 kj H o net To convert 1 mol of water to 1 mol each of H 2 and CO requires 131 kj of energy. The water gas reaction is ENDOthermic thermic.

55 ' -* 55 Calculate H of reaction? ALL In general, when ALL enthalpies of formation are known, H o rxn = H o rxn = ΣH o f (products) fo - ΣH o f (reactants) fo Remember that always = final initial

56 ' -* 56 Calculate the heat of combustion of methanol, i.e., H o rxn for CH 3 OH(g) + 3/2 O 2 (g) --> > CO 2 (g) + 2 H 2 O(g) H o rxn = ΣH o f (prod) - Σ H o f (react)

57 ' -* 57 CH 3 OH(g) + 3/2 O 2 (g) --> > CO 2 (g) + 2 H 2 O(g) H o rxn = ΣH o f (prod) - Σ H o f (react) H o rxn = H o f (CO 2 ) + 2 H o f (H 2 O) - {3/2 H o f (O 2 ) + H o f (CH 3 OH)} = (-393.5( kj) + 2 (-241.8( kj) - {0 + (-201.5( kj)} H o rxn = kj per mol of methanol

58 6 58 Constant Volume Bomb Calorimeter Burn combustible sample. Measure heat evolved in a reaction. Derive E for reaction.

59 59 Some heat from reaction warms water q water = (sp. ht.)(water mass)(t) Some heat from reaction warms bomb q bomb = (heat capacity, J/K)(T) Total heat evolved = q total = q water + q bomb

60 6 60 Calculate heat of combustion of octane. C 8 H /2 O 2 --> 8 CO H 2 O Burn 1.00 g of octane Temp rises from to o C Calorimeter contains 1200 g water Heat capacity of bomb = 837 J/K

61 6 Step 1 Calc. heat transferred from reaction to water. q = (4.184 J/g K)(1200 g)(8.20 K) = 41,170 J Step 2 Calc. heat transferred from reaction to bomb. q = (bomb heat capacity)(t) = (837 J/K)(8.20 K) = 6860 J Step 3 Total heat evolved 41,170 J J = 48,030 J Heat of combustion of 1.00 g of octane = kj 61