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THERMOCHEMISTRY OR THERMODYNAMICS Chapter 6 Chemical Reactivity What drives chemical reactions? How do they occur? The first is answered by THERMODYNAMICS.

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Presentation on theme: "THERMOCHEMISTRY OR THERMODYNAMICS Chapter 6 Chemical Reactivity What drives chemical reactions? How do they occur? The first is answered by THERMODYNAMICS."— Presentation transcript:

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2 THERMOCHEMISTRY OR THERMODYNAMICS Chapter 6

3 Chemical Reactivity What drives chemical reactions? How do they occur? The first is answered by THERMODYNAMICS and the second by KINETICS. We have already seen a number of driving forces for reactions that are PRODUCT-FAVORED. formation of a precipitateformation of a precipitate gas formationgas formation H 2 O formation (acid-base reaction)H 2 O formation (acid-base reaction) electron transfer in a batteryelectron transfer in a battery

4 Energy and Chemistry ENERGY is the capacity to do work or transfer heat. HEAT is the form of energy that flows between 2 samples because of their difference in temperature. Other forms of energy Other forms of energy lightelectricalnuclear kinetic potential

5 Law of Conservation of Energy Energy can be converted from one form to another but can neither be created nor destroyed. (E universe is constant)

6 The Two Types of Energy Potential: due to position or composition - can be converted to work PE = mgh (m = mass, g = acceleration of gravity, and h = height) Kinetic: due to motion of the object KE = 1 /2 mv 2 (m = mass, v = velocity)

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

8 Kinetic and Potential Energy Kinetic energy energy of motion. Translation Translation Rotation Rotation Vibration Vibration

9 Units of Energy 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

10 Temperature v. Heat Temperature reflects random motions of particles, therefore related to kinetic energy of the system. Heat involves a transfer of energy between 2 objects due to a temperature difference

11 Extensive & Intensive Properties Extensive properties depends directly on the amount of substance present. massmass volumevolume heatheat heat capacity (C)heat capacity (C) Intensive properties is not related to the amount of substance. temperaturetemperature concentrationconcentration pressurepressure specific heat (s)specific heat (s)

12 State Function Depends only on the present state of the system - not how it arrived there. It is independent of pathway. Energy change is independent of the pathway (and, therefore, a state function), while work and heat are dependent on the pathway.

13 System and Surroundings System: That on which we focus attention Surroundings: Everything else in the universe Universe = System + Surroundings

14 Exo and Endothermic Heat exchange accompanies chemical reactions. Exothermic: Heat flows out of the system (to the surroundings). Endothermic: Heat flows into the system (from the surroundings).

15 ENERGY DIAGRAMS Exothermic Endothermic

16 Endo- and Exothermic Surroundings System q system > 0 w > 0 w > 0 heat ENDOTHERMIC E goes up

17 Endo- and Exothermic Surroundings System q system > 0 w > 0 w > 0 heat Surroundings System q system < 0 w < 0 w < 0 heat ENDOTHERMICEXOTHERMIC E(system) goes up E(system) goes down

18 First Law First Law of Thermodynamics: The energy of the universe is constant.

19 EnthalpyEnthalpy H = H final - H initial H = H final - H initial If H final > H initial then H is positive Process is ENDOTHERMIC If H final < H initial then H is negative Process is EXOTHERMIC

20 First Law E = q + w E = q + w E = change in systems internal energy E = change in systems internal energy q = heat w = work

21 Piston moving a distance against a pressure does work.

22 Work w & V must have opposite signs, since work is being done by the system to expand the gas. w & V must have opposite signs, since work is being done by the system to expand the gas. w system = -P V 1 Latm = J 1 J = kgm 2 /s 2 P = F/A F = PA w = F h w = PA h w = P V

23 Enthalpy Enthalpy = H = E + PV E = H P V E = H P V H = E + P V H = E + P V At constant pressure, At constant pressure, q P = E + P V, where q P = H at constant pressure H = energy flow as heat (at constant pressure) H = energy flow as heat (at constant pressure)

24 Some Heat Exchange Terms specific heat capacity (s) heat capacity per gram = J/°C g or J/K g molar heat capacity (s) heat capacity per mole = J/°C mol or J/K mol

25 Specific Heat Capacity SubstanceSpec. Heat (J/gK) H 2 O4.184 Al0.902 glass0.84 Aluminum

26 H o = - q p H o = - q p q p = ms t Simple Calorimeter q = heat (J) m = mass (g) s = specific heat (j/gC o ) t = change in temperature (C o ) t = change in temperature (C o ) H o = change in enthalpy (kJ) H o = change in enthalpy (kJ)

27 Specific Heat Capacity 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? 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? where T = T final - T initial where T = T final - T initial heat gain/lost = q = m s T

28 Specific Heat Capacity 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? where T = T final - T initial q = (0.902 J/gK)(25.0 g)( )K q = J heat gain/lost = q = m s T

29 Specific Heat Capacity 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? q = J Notice that the negative sign on q signals heat lost by or transferred out of Al. Notice that the negative sign on q signals heat lost by or transferred out of Al.

30 Bomb Calorimeter

31 Heat Capacity E = q v & q v = -(C t + ms t) E = q v & q v = -(C t + ms t) E = change in internal energy (J) E = change in internal energy (J) q v = heat at constant volume (J) C = heat capacity (J/C o ) t = changein temperature (C o ) t = changein temperature (C o )

32 REMEMBER!!! In regular calorimetry pressure is constant, but the volume will change so: q p = - H q p = E + p V In bomb calorimetry, volume is constant so: q v = E since = zero. since p V = zero.

33 Calculate heat of combustion of octane. C 8 H /2 O 2 --> 8 CO H 2 O Burn 1.00 g of octaneBurn 1.00 g of octane Temp rises from to o CTemp rises from to o C Calorimeter contains 1200 g waterCalorimeter contains 1200 g water Heat capacity of bomb = 837 J/KHeat capacity of bomb = 837 J/K H c m = -(C t + ms t) where H c is heat of combustion. H c m = -(C t + ms t) where H c is heat of combustion. Measuring Heats of Reaction CALORIMETRY

34 Step 1Calc. heat transferred from reaction to water. q = (4.184 J/gK)(1200 g)(8.20 K) = 41,170 J Step 2Calc. heat transferred from reaction to bomb. q = C t = (837 J/K)(8.20 K) = 6860 J = (837 J/K)(8.20 K) = 6860 J Step 3Total heat evolved 41,170 J J = 48,030 J 41,170 J J = 48,030 J Heat of combustion of 1.00 g of octane = kJ Measuring Heats of Reaction CALORIMETRY

35 Hesss Law Reactants Products The change in enthalpy is the same whether the reaction takes place in one step or a series of steps.

36 Standard States Compound -For a gas, pressure is exactly 1 atmosphere. -For a solution, concentration is exactly 1 molar. -Pure substance (liquid or solid), it is the pure liquid or solid. Element -The form [N 2 (g), K(s)] in which it exists at 1 atm and 25°C.

37 Calculations via Hesss Law 1.If a reaction is reversed, H is also reversed. N 2 (g) + O 2 (g) 2NO(g) H = 180 kJ 2NO(g) N 2 (g) + O 2 (g) H = 180 kJ 2.If the coefficients of a reaction are multiplied by an integer, H is multiplied by that same integer. 6NO(g) 3N 2 (g) + 3O 2 (g) H = 540 kJ

38 Using Enthalpy Consider the decomposition of water H 2 O(g) kJ ---> H 2 (g) + 1/2 O 2 (g) Endothermic reaction heat is a reactant H = kJ H = kJ

39 Making H 2 from H 2 O involves two steps. H 2 O(l) + 44 kJ ---> H 2 O(g) H 2 O(g) kJ ---> H 2 (g) + 1/2 O 2 (g) H 2 O(l) kJ --> H 2 (g) + 1/2 O 2 (g) Example of HESSS LAW If a rxn. is the sum of 2 or more others, the net H is the sum of the Hs of the other rxns. Using Enthalpy

40 Calc. H for S(s) + 3/2 O 2 (g) --> SO 3 (g) S(s) + O 2 (g) --> SO 2 (g) kJ SO 2 (g) + 1/2 O 2 (g) --> SO 3 (g) kJ _______________________________________ S(s) + 3/2 O 2 (g) --> SO 3 (g) kJ Using Enthalpy

41 S solid SO 3 gas SO 2 gas direct path + 3/2 O 2 H = kJ energy +O 2 H 1 = kJ + 1/2 O 2 H 2 = kJ H along one path = H along one path = H along another path H along another path H along one path = H along one path = H along another path H along another path

42 This equation is valid because H is a STATE FUNCTION These depend only on the state of the system and not how it got there. Unlike V, T, and P, one cannot measure absolute H. Can only measure H. H along one path = H along one path = H along another path H along another path H along one path = H along one path = H along another path H along another path

43 Standard Enthalpy Values NIST (Natl Institute for Standards and Technology) gives values of H o f = standard molar enthalpy of formation H o f = standard molar enthalpy of formation This is the enthalpy change when 1 mol of compound is formed from elements under standard conditions. H o f is always stated in terms of moles of product formed. See Appendix A21-A24.

44 H o f, standard molar enthalpy of formation H 2 (g) + 1/2 O 2 (g) --> H 2 O(g) H o f = kJ/mol H o f = kJ/mol By definition, H o f = 0 for elements in their standard states.

45 Using Standard Enthalpy Values Use H s to calculate enthalpy change for H 2 O(g) + C(graphite) --> H 2 (g) + CO(g) (product is called water gas)

46 Using Standard Enthalpy Values 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 H f of H 2 O vapor = kJ/mol C(s) + 1/2 O 2 (g) --> CO(g) H f of CO = kJ/mol H f of CO = kJ/mol

47 Using Standard Enthalpy Values 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) H o net = +131 kJ H o net = +131 kJ 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.

48 Change in Enthalpy Can be calculated from enthalpies of formation of reactants and products. H rxn ° = n p H f (products) n r H f (reactants) H rxn ° = n p H f (products) n r H f (reactants) H is an extensive property--kJ/mol H is an extensive property--kJ/mol For the reaction: 2H 2 (g) + O 2 (g) ---> 2H 2 O (g) Enthalpy would be twice the H value for the combustion of hydrogen. Enthalpy would be twice the H value for the combustion of hydrogen.

49 Using Standard Enthalpy Values 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 = H o f (prod) - H o f (react) H o rxn = H o f (prod) - H o f (react)

50 Using Standard Enthalpy Values CH 3 OH(g) + 3/2 O 2 (g) --> CO 2 (g) + 2 H 2 O(g) H o rxn = H o f (prod) - H o f (react) H o rxn = H o f (prod) - H o f (react) H o rxn = H o f (CO 2 ) + 2 H o f (H 2 O) 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)} - {3/2 H o f (O 2 ) + H o f (CH 3 OH)} = ( kJ) + 2 ( kJ) = ( kJ) + 2 ( kJ) - {0 + ( kJ)} - {0 + ( kJ)} H o rxn = kJ per mol of methanol H o rxn = kJ per mol of methanol H o rxn is always in terms of moles of reactant. H o rxn is always in terms of moles of reactant.

51 Pathway for the Combustion of Methane

52 Schematic diagram of the energy changes for the combustion of methane.

53 Greenhouse Effect Greenhouse Gases: CO 2 H 2 O CH 4 N 2 O -- a warming effect exerted by the earths atmosphere due to thermal energy retained by absorption of infrared radiation. thermal energy retained by absorption of infrared radiation.


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