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Thermochemistry Chapters 6 and 18 TWO Trends in Nature Order Disorder High energy Low energy.

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Presentation on theme: "Thermochemistry Chapters 6 and 18 TWO Trends in Nature Order Disorder High energy Low energy."— Presentation transcript:

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2 Thermochemistry Chapters 6 and 18

3 TWO Trends in Nature Order Disorder High energy Low energy

4 Exothermic process is any process that gives off heat – transfers thermal energy from the system to the surroundings. Endothermic process is any process in which heat has to be supplied to the system from the surroundings. 2H 2 (g) + O 2 (g) 2H 2 O (l) + energy H 2 O (g) H 2 O (l) + energy energy + 2HgO (s) 2Hg (l) + O 2 (g) 6.2 energy + H 2 O (s) H 2 O (l)

5 Enthalpy (H) is used to quantify the heat flow into or out of a system in a process that occurs at constant pressure. H = H (products) – H (reactants) H = heat given off or absorbed during a reaction at constant pressure H products < H reactants H < 0 H products > H reactants H > 0 6.4

6 Thermochemical Equations H 2 O (s) H 2 O (l) H = 6.01 kJ Is H negative or positive? System absorbs heat Endothermic H > 0 6.01 kJ are absorbed for every 1 mole of ice that melts at 0 0 C and 1 atm. 6.4

7 Thermochemical Equations CH 4 (g) + 2O 2 (g) CO 2 (g) + 2H 2 O (l) H = -890.4 kJ Is H negative or positive? System gives off heat Exothermic H < 0 890.4 kJ are released for every 1 mole of methane that is combusted at 25 0 C and 1 atm. 6.4

8 H 2 O (s) H 2 O (l) H = 6.01 kJ/mol ΔH = 6.01 kJ The stoichiometric coefficients always refer to the number of moles of a substance Thermochemical Equations If you reverse a reaction, the sign of H changes H 2 O (l) H 2 O (s) H = - 6.01 kJ If you multiply both sides of the equation by a factor n, then H must change by the same factor n. 2H 2 O (s) 2H 2 O (l) H = 2 mol x 6.01 kJ/mol = 12.0 kJ 6.4

9 H 2 O (s) H 2 O (l) H = 6.01 kJ The physical states of all reactants and products must be specified in thermochemical equations. Thermochemical Equations 6.4 H 2 O (l) H 2 O (g) H = 44.0 kJ How much heat is evolved when 266 g of white phosphorus (P 4 ) burn in air? P 4 (s) + 5O 2 (g) P 4 O 10 (s) H reaction = -3013 kJ 266 g P 4 1 mol P 4 123.9 g P 4 x 3013 kJ 1 mol P 4 x = 6470 kJ

10 Standard enthalpy of formation ( H 0 ) is the heat change that results when one mole of a compound is formed from its elements at a pressure of 1 atm. f The standard enthalpy of formation of any element in its most stable form is zero. H 0 (O 2 ) = 0 f H 0 (O 3 ) = 142 kJ/mol f H 0 (C, graphite) = 0 f H 0 (C, diamond) = 1.90 kJ/mol f 6.6

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12 The standard enthalpy of reaction ( H 0 ) is the enthalpy of a reaction carried out at 1 atm. rxn aA + bB cC + dD H 0 rxn d H 0 (D) f c H 0 (C) f = [+] - b H 0 (B) f a H 0 (A) f [+] H 0 rxn H 0 (products) f = H 0 (reactants) f - 6.6 Hesss Law: When reactants are converted to products, the change in enthalpy is the same whether the reaction takes place in one step or in a series of steps. (Enthalpy is a state function. It doesnt matter how you get there, only where you start and end.)

13 Benzene (C 6 H 6 ) burns in air to produce carbon dioxide and liquid water. How much heat is released per mole of benzene combusted? The standard enthalpy of formation of benzene is 49.04 kJ/mol. 2C 6 H 6 (l) + 15O 2 (g) 12CO 2 (g) + 6H 2 O (l) H 0 rxn H 0 (products) f = H 0 (reactants) f - H 0 rxn 6 H 0 (H 2 O) f 12 H 0 (CO 2 ) f = [+] - 2 H 0 (C 6 H 6 ) f [] H 0 rxn = [ 12 × -393.5 + 6 × -285.8 ] – [ 2 × 49.04 ] = -6535 kJ -6535 kJ 2 mol = - 3267 kJ/mol C 6 H 6 6.6

14 Calculate the standard enthalpy of formation of CS 2 (l) given that: C (graphite) + O 2 (g) CO 2 (g) H 0 = -393.5 kJ rxn S (rhombic) + O 2 (g) SO 2 (g) H 0 = -296.1 kJ rxn CS 2 (l) + 3O 2 (g) CO 2 (g) + 2SO 2 (g) H 0 = -1072 kJ rxn 1. Write the enthalpy of formation reaction for CS 2 C (graphite) + 2S (rhombic) CS 2 (l) 2. Add the given rxns so that the result is the desired rxn. rxn C (graphite) + O 2 (g) CO 2 (g) H 0 = -393.5 kJ 2S (rhombic) + 2O 2 (g) 2SO 2 (g) H 0 = -296.1x2 kJ rxn CO 2 (g) + 2SO 2 (g) CS 2 (l) + 3O 2 (g) H 0 = +1072 kJ rxn + C (graphite) + 2S (rhombic) CS 2 (l) H 0 = -393.5 + (2x-296.1) + 1072 = 86.3 kJ rxn 6.6

15 Chemistry in Action: Fuel Values of Foods and Other Substances C 6 H 12 O 6 (s) + 6O 2 (g) 6CO 2 (g) + 6H 2 O (l) H = -2801 kJ/mol 1 cal = 4.184 J 1 Cal = 1000 cal = 4184 J

16 The enthalpy of solution ( H soln ) is the heat generated or absorbed when a certain amount of solute dissolves in a certain amount of solvent. H soln = H soln - H components 6.7 Which substance(s) could be used for melting ice? Which substance(s) could be used for a cold pack?

17 The Solution Process for NaCl H soln = Step 1 + Step 2 = 788 – 784 = 4 kJ/mol 6.7

18 Energy Diagrams ExothermicEndothermic (a)Activation energy (Ea) for the forward reaction (b)Activation energy (Ea) for the reverse reaction (c) Delta H 50 kJ/mol300 kJ/mol 150 kJ/mol100 kJ/mol -100 kJ/mol+200 kJ/mol

19 Entropy (S) is a measure of the randomness or disorder of a system. orderS disorder S If the change from initial to final results in an increase in randomness S > 0 For any substance, the solid state is more ordered than the liquid state and the liquid state is more ordered than gas state S solid < S liquid << S gas H 2 O (s) H 2 O (l) S > 0 18.3

20 First Law of Thermodynamics Energy can be converted from one form to another but energy cannot be created or destroyed. Second Law of Thermodynamics The entropy of the universe increases in a spontaneous process and remains unchanged in an equilibrium process. S univ = S sys + S surr > 0 Spontaneous process: S univ = S sys + S surr = 0 Equilibrium process: 18.4

21 Entropy Changes in the System ( S sys ) aA + bB cC + dD S 0 rxn dS 0 (D) cS 0 (C) = [+] - bS 0 (B) aS 0 (A) [+] S 0 rxn S 0 (products) = S 0 (reactants) - The standard entropy of reaction ( S 0 ) is the entropy change for a reaction carried out at 1 atm and 25 0 C. rxn 18.4 What is the standard entropy change for the following reaction at 25 0 C? 2CO (g) + O 2 (g) 2CO 2 (g) S 0 (CO) = 197.9 J/K mol S 0 (O 2 ) = 205.0 J/K mol S 0 (CO 2 ) = 213.6 J/K mol S 0 rxn = 2 x S 0 (CO 2 ) – [2 x S 0 (CO) + S 0 (O 2 )] S 0 rxn = 427.2 – [395.8 + 205.0] = -173.6 J/K mol

22 Entropy Changes in the System ( S sys ) 18.4 When gases are produced (or consumed) If a reaction produces more gas molecules than it consumes, S 0 > 0. If the total number of gas molecules diminishes, S 0 < 0. If there is no net change in the total number of gas molecules, then S 0 may be positive or negative BUT S 0 will be a small number. What is the sign of the entropy change for the following reaction? 2Zn (s) + O 2 (g) 2ZnO (s) The total number of gas molecules goes down, S is negative.

23 Spontaneous Physical and Chemical Processes A waterfall runs downhill A lump of sugar dissolves in a cup of coffee At 1 atm, water freezes below 0 0 C and ice melts above 0 0 C Heat flows from a hotter object to a colder object A gas expands in an evacuated bulb Iron exposed to oxygen and water forms rust spontaneous nonspontaneous 18.2

24 S univ = S sys + S surr > 0 Spontaneous process: S univ = S sys + S surr = 0 Equilibrium process: Gibbs Free Energy For a constant-temperature process: G = H sys -T S sys Gibbs free energy (G) G < 0 The reaction is spontaneous in the forward direction. G > 0 The reaction is nonspontaneous as written. The reaction is spontaneous in the reverse direction. G = 0 The reaction is at equilibrium. 18.5

25 G = H - T S 18.5

26 aA + bB cC + dD G 0 rxn d G 0 (D) f c G 0 (C) f = [+] - b G 0 (B) f a G 0 (A) f [+] G 0 rxn G 0 (products) f = G 0 (reactants) f - The standard free-energy of reaction ( G 0 ) is the free- energy change for a reaction when it occurs under standard- state conditions. rxn Standard free energy of formation ( G 0 ) is the free-energy change that occurs when 1 mole of the compound is formed from its elements in their standard states. f G 0 of any element in its stable form is zero. f

27 2C 6 H 6 (l) + 15O 2 (g) 12CO 2 (g) + 6H 2 O (l) G 0 rxn G 0 (products) f = G 0 (reactants) f - What is the standard free-energy change for the following reaction at 25 0 C? G 0 rxn 6 G 0 (H 2 O) f 12 G 0 (CO 2 ) f = [+] - 2 G 0 (C 6 H 6 ) f [] G 0 rxn = [ 12x–394.4 + 6x–237.2 ] – [ 2x124.5 ] = -6405 kJ Is the reaction spontaneous at 25 0 C? G 0 = -6405 kJ < 0 spontaneous 18.5

28 Recap: Signs of Thermodynamic Values NegativePositive Enthalpy (ΔH)ExothermicEndothermic Entropy (ΔS)Less disorderMore disorder Gibbs Free Energy (ΔG) SpontaneousNot spontaneous

29 Gibbs Free Energy and Chemical Equilibrium G = G 0 + RT lnQ R is the gas constant (8.314 J/K mol) T is the absolute temperature (K) Q is the reaction quotient At Equilibrium G = 0 Q = K 0 = G 0 + RT lnK G 0 = RT lnK 18.6

30 G 0 = RT lnK 18.6

31 The specific heat (s) [most books use lower case c] of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius. The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius. C = ms Heat (q) absorbed or released: q = ms t q = C t t = t final - t initial 6.5

32 How much heat is given off when an 869 g iron bar cools from 94 0 C to 5 0 C? s of Fe = 0.444 J/g 0 C t = t final – t initial = 5 0 C – 94 0 C = -89 0 C q = ms t = 869 g x 0.444 J/g 0 C x –89 0 C= -34,000 J 6.5

33 Constant-Pressure Calorimetry No heat enters or leaves! q sys = q water + q cal + q rxn q sys = 0 q rxn = - (q water + q cal ) q water = ms t q cal = C cal t 6.5 Reaction at Constant P H = q rxn

34 6.5

35 The boiling point is the temperature at which the (equilibrium) vapor pressure of a liquid is equal to the external pressure. The normal boiling point is the temperature at which a liquid boils when the external pressure is 1 atm. 11.8 Phase Changes

36 The critical temperature (T c ) is the temperature above which the gas cannot be made to liquefy, no matter how great the applied pressure. The critical pressure (P c ) is the minimum pressure that must be applied to bring about liquefaction at the critical temperature. 11.8

37 Wheres Waldo? Can you find… The Triple Point? Critical pressure? Critical temperature? Where fusion occurs? Where vaporization occurs? Melting point (at 1 atm)? Boiling point (at 6 atm)? Carbon Dioxide

38 Melting 11.8 Freezing H 2 O (s) H 2 O (l) The melting point of a solid or the freezing point of a liquid is the temperature at which the solid and liquid phases coexist in equilibrium

39 Sublimation 11.8 Deposition H 2 O (s) H 2 O (g) Molar heat of sublimation ( H sub ) is the energy required to sublime 1 mole of a solid. H sub = H fus + H vap ( Hesss Law)

40 Molar heat of fusion ( H fus ) is the energy required to melt 1 mole of a solid substance. 11.8

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42 Sample Problem How much heat is required to change 36 g of H 2 O from -8 deg C to 120 deg C? Step 1: Heat the iceQ=mcΔT Q = 36 g x 2.06 J/g deg C x 8 deg C = 593.28 J = 0.59 kJ Step 2: Convert the solid to liquidΔH fusion Q = 2.0 mol x 6.01 kJ/mol = 12 kJ Step 3: Heat the liquidQ=mcΔT Q = 36g x 4.184 J/g deg C x 100 deg C = 15063 J = 15 kJ

43 Sample Problem How much heat is required to change 36 g of H 2 O from -8 deg C to 120 deg C? Step 4: Convert the liquid to gasΔH vaporization Q = 2.0 mol x 44.01 kJ/mol = 88 kJ Step 5: Heat the gasQ=mcΔT Q = 36 g x 2.02 J/g deg C x 20 deg C = 1454.4 J = 1.5 kJ Now, add all the steps together 0.59 kJ + 12 kJ + 15 kJ + 88 kJ + 1.5 kJ = 118 kJ


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