Thermochemistry. Energy Flow in Thermochemistry

Transcription

1 Thermochemistry! part of thermodynamics (study of heat)! the study of energy transfer during reactions! defined through system and surroundings Energy Flow in Thermochemistry!E = E final - E initial = E products - E reactants system loses energy!e < 0 system gains energy!e > 0

2 eat and Work: Two Forms of Energy eat (q) Work (w) as thermal energy as all other (mechanical, electrical etc)!e = q + w eat no work, w = 0 hot!e = q cold In this case, q and!e are negative

3 Work no heat, q = 0 force applied by surroundings!e = w Work no heat, q = 0 force applied by surroundings!e = w In this case, w and!e are positive

4 Units of Energy The SI unit 1 Joule = 1 kg m 2 /s 2 calorie: energy to heat 1 g of 2 by 1 o C 1 cal = J 1 kj = 1000 J 1 kcal = 1000 cal The First Law of Thermodynamics!E universe = 0 energy is exchanged, not created or destroyed

5 Energy is a State Function!E is independent of how the change takes place 1. C + 2 C + 2 C 2 2. overall!e is the same! C + 2 C2 Enthalpy (): eat at Constant P pressure is equal in the initial and final states w = -P!V initial state final state Enthalpy: = E + PV

6 Enthalpy (): eat at Constant P!E = q P + w = q P - P!V! =!E + P!V initial final w = -P!V q P =!E + P!V =! the change in enthalpy equals heat lost or gained at constant pressure! Enthalpy is a State Function! is independent of how the change takes place 1. C + 2 C + 2 C 2 2. overall! is the same! C + 2 C2

7 ! and!e! =!E + P!V 1. Reactions that do not involve gases: K + Cl " KCl + 2 ; P!V! 0;!!!E 2. Reactions where the amount (mol) of gases does not change. N " 2N; P!V = 0;! =!E 3. Reactions in which the amount of gases does change: " 2 2 ;!» P!V;!!!E Exothermic and Endothermic Processes new definitions for processes - exothermic: heat is released endothermic: heat is consumed

8 Exothermic and Endothermic Processes new definitions for processes - exothermic: heat is released endothermic: heat is consumed energy diagram for exothermic process: A B A! < 0 B Time! in Reactions and ther Processes negative C " C exothermic reaction! = -891 kj/mol positive 2 (s) " 2 (l)! = 6.02 kj/mol endothermic process

9 C " C Practice Problem Assuming complete heat transfer, combustion of how many L of methane (at STP) is required to melt 2.2 lb (1 kg) of ice?! = -891 kj/mol 2 (s) " 2 (l)! = 6.02 kj/mol Types of Enthalpy eat of combustion (! comb ) C " C ! comb = -891 kj/mol eat of formation (! f ) 2K + Br 2 " 2KBr! =! f < 0 eat of fusion (! fus ) NaCl(s) " NaCl(l)! =! fus eat of vaporization (! vap ) C 6 6 (l) " C 6 6 (g)! =! vap

10 Types of Enthalpy eat of combustion (! comb ) C " C ! comb = -891 kj/mol eat of formation (! f ) 2K + Br 2 " 2KBr! =! f < 0 eat of fusion (! fus ) NaCl(s) " NaCl(l)! =! fus > 0 eat of vaporization (! vap ) C 6 6 (l) " C 6 6 (g)! =! vap Types of Enthalpy eat of combustion (! comb ) C " C ! comb = -891 kj/mol eat of formation (! f ) 2K + Br 2 " 2KBr! =! f < 0 eat of fusion (! fus ) NaCl(s) " NaCl(l)! =! fus > 0 eat of vaporization (! vap ) C 6 6 (l) " C 6 6 (g)! =! vap > 0

11 Key Use of!: Bond Strength ! f kj/mol!!!e 2 2 Key Use of!: Bond Strength bonds break! f kj/mol!!!e 2 2 new bonds form

12 Key Use of!: Bond Strength bonds break! f kj/mol!!!e 2 2 new bonds form!e (and!) is mostly due to differences in bond strength in reactants and products. Rabbit Eating Butter! kj/g butter burn C g! comb and nutritional value

13 Rabbit Eating Butter! kj/g butter 1 g burn or feed to rabbit C 2 + 2! comb and nutritional value! comb of C 4 and C 3 C Methane (natural gas)! comb = 890 kj/mol + 2 C! comb = 55.5 kj/g + C + 2! comb = 727kJ/mol! comb = 22.7 kj/g C + Methanol

14 Calorimetry measures! rxn, which is q p, by measuring heat transfer Calorimetry measures! rxn, which is q p, by measuring heat transfer

15 Specific eat Capacity, c c (J/g K) = q mass #!T specific heat capacity: the quantity of heat required to change the temperature of 1 gram of substance by 1 K Calorimetry: Constant-Pressure q P =! commom use: to determine specific heat capacity of a solid

16 Calorimetry: Constant-Pressure q P =! commom use: to determine specific heat capacity of a solid heat exchange q A + q B = 0 q A = -q B Calorimetry: Constant-Pressure Sample Problem 6.4 (follow-up). As a purity check for industrial diamonds, a carat (1 carat = g) diamond is heated to o C and immersed in g of water at o C in a constant pressure calorimeter. After equilibration, the final recorded temperature is o C. What is the specific heat capacity of the diamond? (c water = J/g. K) c = J/g. K

17 Calorimetry: Constant-Volume!E = q + w!e = q V + P!V q V =!E!! the difference is < 1% Stoichiometry of Thermochemical Equations sign magnitude 2 2 " moles " moles 1 mole 2 moles 2 moles 1 mole! = 572 kj! = -572 kj " 2! = -286 kj 2 1 mole 1 mole 0.5 moles

18 Stoichiometry of Thermochemical Equations sign magnitude 2 2 " moles " moles 1 mole 2 moles 2 moles 1 mole! = 572 kj! = -572 kj " 2 2! f = -286 kj/mol Stoichiometry of Thermochemical Equations kj Fundamental Strategy Input (g, ml, kg, etc) $ moles (starting)% $ moles (final) $ utput (g, ml, kg, etc)` kj " 2 2! f = -286 kj/mol

19 Stoichiometry of Thermochemical Equations Sample Problem 6.57(b) A mercury mirror forms inside a test tube by the thermal decomposition of mercury (II) oxide: 2g " 2g + 2! rxn = kj If 275 kj of heat is absorbed, how many grams of mercury form? ess s Law ess s law of heat summation: the enthalpy change of an overall process is the sum of the enthalpy changes of its individual steps A + B " C + D C + D " E + F A + B " E + F! = a! = b! overall = a+b

20 ess s Law ess s law of heat summation: the enthalpy change of an overall process is the sum of the enthalpy changes of its individual steps 4N " 2N 2 5! =? 2N 2 5 " 4N + 3 2! = kj/mol (N 2 5 ) 2N + 2 " 2N 2! = kj/mol (N) (sample problem 6.7) Standard eats of Reaction (! rxn ) o! at standard states is! o standard states: gases: 1 atm and the gas behaving ideally aqueous solution: 1 M concentration pure substance: most stable form at 1 atm, 25 o C usually

21 o Standard eats of Formation (! f ) C(graphite) " C 4 2Na + Cl 2 " 2NaCl! f o = kj/mol! f o = kj/mol 4C(graphite) " 2C 2 5! f o = kj/mol! for elements at standard state! f o = 0! o rxn and! f o! o rxn = &m!o f(products) - &n!o f(reactants) m,n - stoichiometric coefficients

22 ! o rxn and! f o Use the following information to find! o f of methanol C 3 + 3/2 2 " C ! o comb = kj/mol! o f of C 2 = kj/mol! o f of 2 = kj/mol