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Thermochemistry

Thermochemistry. Chapters 6 and 16. Thermochemistry. Energy: “ The capacity to do work. ” In Physics, there are 2 main types of energy… Kinetic (energy of motion) = ½ mv 2 Potential (energy of position due to gravity)= mgh

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Thermochemistry

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  1. Thermochemistry Chapters 6 and 16

  2. Thermochemistry • Energy: “The capacity to do work.” • In Physics, there are 2 main types of energy… • Kinetic (energy of motion) = ½ mv2 • Potential (energy of position due to gravity)= mgh • In Chemistry, we usually concern ourselves with the heat energy gained or lost during chemical reactions.

  3. Law of Conservation of Energy: Energy can neither be created nor destroyed, but can be converted between forms The First Law of Thermodynamics: The total energy content of the universe is constant

  4. The universe • is divided into two halves. • the system and the surroundings. • The system is the part you are concerned with. • The surroundings are the rest.

  5. When a system undergoes a physical or chemical change, the change in internal energy is given by the heat added to or absorbed by the system plus the work done on or by the system: ∆E= q + w The 1st Law of Thermodynamics (E= energy, q=heat, w=work)

  6. E = q + w E= change in internal energy of a system q= heat flowing into or out of the system -qif energy is leaving tothe surroundings +qif energy is entering fromthe surroundings w= work done by, or on, the system -wif work is done bythe systemonthe surroundings +wif work is done onthe system bythe surroundings

  7. Work, Pressure, and Volume Compression Expansion +V(increase) -V(decrease) -wresults +wresults Esystemdecreases Esystemincreases Work has been done on the system by the surroundings Work has been done by the system on the surroundings

  8. State Functions depend ONLY on the present state of the system ENERGYIS A STATE FUNCTION A person standing at the top of Mt. Everest has the same potential energy whether they got there by hiking up, or by falling down from a plane  WORKIS NOT A STATE FUNCTION

  9. State Functions A state function depends only on the initial and final states of a system. Example: The altitude difference between Denver and Chicago does not depend on whether you fly or drive, only on the elevation of the two cities above sea level. Similarly, the internal energy of 50 g of H2O(l) at 25 ºC does not depend on whether we cool 50 g of H2O(l) from 100 ºC to 25 ºC or heat 50 g of H2O(l) at 0 ºC to 25 ºC.

  10. State Functions A state function does not depend on how the internal energy is used. Example: A battery in a flashlight can be discharged by producing heat and light. The same battery in a toy car produces heat and work. The change in internal energy of the battery is the same in both cases.

  11. State Functions ∆E is a state function. Enthalpy • Enthalpy, H: • Enthalpy is heat transferred between the system and surroundings carried out under constant pressure. (Open containers are under constant pressure!) • Enthalpy is also a state function. ∆H is (+) for endothermic reactions…just like q! ∆H is (−) for exothermic reactions…just like q!

  12. Endothermic and Exothermic Processes • An endothermic process is one that absorbs heat from the surroundings. (+q) An endothermic reaction feels cold. Example--an “instant” ice pack • An exothermic process is one that transfers heat to the surroundings. (-q) An exothermic reaction feels hot. Example--burning paper

  13. 2H2(g) + O2(g) 2H2O (l) + energy H2O (g) H2O (l) + energy energy + 2HgO (s) 2Hg (l) + O2(g) energy + H2O (s) H2O (l) 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. 6.2

  14. Heat Potential energy

  15. Heat Potential energy

  16. Energy Diagrams Exothermic Endothermic • Activation energy (Ea) for the forward reaction • Activation energy (Ea) for the reverse reaction (c) Delta H

  17. Direction • Every energy measurement has three parts. • A unit ( Joules of calories). • A number how many. • and a sign to tell direction. • negative - exothermic • positive- endothermic

  18. Surroundings System Energy DE <0

  19. Surroundings System Energy DE >0

  20. Chemical energy • Reaction • http://www.youtube.com/watch?v=nN8xD_bv2aQ

  21. calorimetry

  22. Calorimeters • Calorimeters measure heat flow. It measures changes in water temperature after a reaction is performed. (Constant Pressure) Bomb Calorimeter Usually studies combustion (Constant Volume)

  23. Measuring Heat, q • Heat capacity is the amount of energy required to raise the temperature of an object by 1 ºC. • Molar heat capacity, (C) is the heat capacity of 1 mol of a substance. Cp=∆H/∆T • Specific heat, (c), or specific heat capacity is the heat capacity of 1 g of a substance. • q = (grams of substance) x(specific heat) x ∆T. q=mc∆T c (water) = 4.184 J/g ºC or 1.0 cal/g ºC

  24. Measuring Heat • Practice Problems:How many joules of heat will raise the temperature of exactly 50 g of water at 25 ºC to 75 ºC q = (50 g) x (4.18J/g ºC) x (50 ºC) = *Note: q(solution) = − q(rxn) • What is the molar heat capacity for water? C = (4.18 J/g ºC) x (18.0 g/mole) = So… q=mc∆T 10450 J 75.2 J/mol ºC C = cx M

  25. Enthalpy Hesse’s law

  26. Enthalpy • abbreviated H • DH = DE + PDV • the heat at constant pressure qp can be calculated from • DE = qp + w = qp - PDV • qp = DE + P DV = DH

  27. Hess’s Law • If a reaction is carried out in a series of steps, ∆H for the reaction is the sum of ∆H for each of the steps. • The total change in enthalpy is independent of the number of steps. • Total ∆H is also independent of the nature of the path. CH4(g) + 2O2(g)-----> CO2(g) + 2H2O(g) ∆H = –802 kJ (Add) 2H2O(g)------> 2H2O(l) ∆H = –88 kJ ____________________________________________________ (Total) CH4(g) + 2O2(g) ------> CO2(g) + 2H2O(l) ∆H = –890 kJ • Hess’s Law provides a useful means of calculating energy changes that are difficult to measure…(in this case it’s getting liquid water to form instead of water vapor.)

  28. Entalphy for changes of state of matter

  29. Heat of Fusion & Heat of Vaporization • Molar heat of fusion(∆Hfus) : the amount of energy required to take 1 mole of a solid to the liquid state. Example: H2O(s) H2O(l) ∆Hfus = 6.01 kJ -Heat of fusion is usually greater for ionic solids than molecular solids since ionic solids are more strongly held together. For ice to water … ∆Hfus= 6.01 kJ/mol • Molar heat of vaporization(∆H vap): the amount of energy required to take 1 mole of a liquid to the gaseous state. Example: H2O(l) H2O(g)∆Hvap = 40.67 kJ For water to steam … ∆Hvap= 40.67 kJ/mol

  30. Entalphy of formation and reaction

  31. Enthalpies of Formation • If a compound is formed from its constituent elements, then the enthalpy change for the reaction is called the enthalpy of formation, ∆Hf . Example:C(s) + O2(g) CO2(g) ∆Hf = −393.5 kJ Standard state (standard conditions) refer to the substance at: 1 atm and 25ºC (298 K) (We will assume that reactants and products are both at 25 ºC unless otherwise stated.) • Standard enthalpy, ∆Hº, is the enthalpy measured when everything is in its standard state. • Standard enthalpy of formation of a compound, ∆Hºf , is the enthalpy change for the formation of 1 mole of compound with all substances in their standard states.

  32. DH0 (O2) = 0 DH0 (O3) = 142 kJ/mol DH0 (C, graphite) = 0 DH0 (C, diamond) = 1.90 kJ/mol f f f f Standard enthalpy of formation (DH0) 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. 6.6

  33. Enthalpies of Formation • If there is more than one state for a substance under standard conditions, the more stable one is used. -Example: When dealing with carbon we use graphite because graphite is more stable than diamond or C60. • By definition, the standard enthalpy of formation of the most stable form of an element is zero. Why? -Because there is no formation reaction needed when the element is already in its standard state. ∆Hºf C(graphite) = zero ∆Hºf O2(g) = zero ∆Hºf Br2(l) = zero

  34. C(graphite) + O2(g) CO2(g)DH0 = -393.5 kJ rxn S(rhombic) + O2(g) SO2(g)DH0 = -296.1 kJ rxn CS2(l) + 3O2(g) CO2(g) + 2SO2(g)DH0 = -1072 kJ rxn 2S(rhombic) + 2O2(g) 2SO2(g)DH0 = -296.1x2 kJ C(graphite) + 2S(rhombic) CS2 (l) C(graphite) + 2S(rhombic) CS2 (l) rxn rxn C(graphite) + O2(g) CO2(g)DH0 = -393.5 kJ + CO2(g) + 2SO2(g) CS2(l) + 3O2(g)DH0 = +1072 kJ rxn DH0 = -393.5 + (2x-296.1) + 1072 = 86.3 kJ rxn Calculate the standard enthalpy of formation of CS2 (l) given that: 1. Write the enthalpy of formation reaction for CS2 2. Add the given rxns so that the result is the desired rxn. 6.6

  35. Enthalpies of Reactions • Again, we can only measure the change in enthalpy, ∆H. ∆H = Hfinal – Hinitial • For a reaction… ∆Hrxn = H(products) – H(reactants) • The enthalpy change that accompanies a reaction is called the enthalpy of reactionor heat of reaction (∆Hrxn). • Consider the equation for the production of water: 2H2(g) + O2(g)-----> 2H2O(g) ∆Hrxn = –483.6 kJ -The equation tells us that 483.6 kJ of energy are released to the surroundings when water is formed & the reaction would feel hot. -These equations are called thermochemical equations.

  36. Enthalpies of Reactions • Enthalpy is an “extensive property” which means it depends on the amount of reactant you start with. (“Intensive properties” do not depend on the quantity of the substance, i.e. – density.) - Example:CH4(g) + 2O2(g)  CO2(g) + 2H2O(g)∆H = -802 kJ 2CH4(g) + 4O2(g)2CO2(g) + 4H2O(g) ∆H = -1604 kJ • If you reverse the reaction, change the sign on ∆H. -Example:CO2(g) + 2H2O(g) CH4(g) + 2O2(g)∆H = +802 kJ • Enthalpy change for a reaction depends on the state of the reactants and products. -Example: CH4(g) + 2O2(g) ----> CO2(g) + 2H2O(l)∆H = -890 kJ If water vapor is the product, ∆H = -802 kJ because… 2H2O(l) -------> 2H2O(g) ∆H = +88 kJ

  37. The standard enthalpy of reaction (DH0 ) is the enthalpy of a reaction carried out at 1 atm. rxn aA + bB cC + dD - [ + ] [ + ] = DH0 DH0 rxn rxn dDH0 (D) cDH0 (C) aDH0 (A) bDH0 (B) f f f f - DH0 (reactants) S DH0 (products) S = f f Hess’s 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 doesn’t matter how you get there, only where you start and end.) 6.6

  38. Using ∆Hºf to calculate ∆Hºrxn • Example: Use Hess’s Law to calculate ∆Hºrxn for… C3H8(g) + 5O2(g)-------> 3CO2(g) + 4H2O(l) • Step 1, the products…Form 3CO2 and 4H2O from their elements: 3C(s) + 3O2(g) ------>3CO2(g)∆H1 = 3∆Hºf[CO2(g)]…tripled it! 4H2(g) + 2O2(g)------> 4H2O(l)∆H2 = 4∆Hºf[H2O(l)]…quadrupled it! • Step 2, the reactants…(Note: O2 has no enthalpy of formation since it is in the elemental state, so we concern ourselves with C3H8.) 3C(s) + 4H2(g) ------>C3H8(g)∆H3 = ∆Hºf [C3H8(g)] • Step 3, Now look up the values on the charts and do the math! ∆Hºrxn = ∆Hºf(products) – ∆Hºf(reactants) ∆Hºrxn = [3(–393.5 kJ) + 4(–285.8 kJ)] – [(– 103.85 kJ)] ∆Hºrxn = –1180.5 –1143.2 + 103.85 = –2220 kJ

  39. 6.6

  40. 2C6H6(l) + 15O2(g) 12CO2(g) + 6H2O (l) - S S = DH0 DH0 DH0 - [ ] [ + ] = rxn rxn rxn [ 12 × -393.5 + 6 × -285.8 ] – [ 2 × 49.04 ] = -6535 kJ = 12DH0 (CO2) 2DH0 (C6H6) f f = - 3267 kJ/mol C6H6 6DH0 (H2O) -6535 kJ f 2 mol DH0 (reactants) DH0 (products) f f Benzene (C6H6) 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. 6.6

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

  42. The Solution Process for NaCl DHsoln = Step 1 + Step 2 = 788 – 784 = 4 kJ/mol 6.7

  43. Chapt 17 THERMODYNAMICS AH AND AG

  44. TWO Trends in Nature • Order  Disorder   • High energy  Low energy 

  45. S order disorder S H2O (s) H2O (l) Entropy (S) is a measure of the randomness or disorder of a system. If the change from initial to final results in an increase in randomness DS > 0 For any substance, the solid state is more ordered than the liquid state and the liquid state is more ordered than gas state Ssolid < Sliquid << Sgas DS > 0 18.3

  46. 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. Spontaneous process: DSuniv = DSsys + DSsurr > 0 Equilibrium process: DSuniv = DSsys + DSsurr = 0 18.4

  47. The standard entropy of reaction (DS0 ) is the entropy change for a reaction carried out at 1 atm and 250C. rxn aS0(A) bS0(B) - [ + ] cS0(C) dS0(D) [ + ] = aA + bB cC + dD - S S0(reactants) S S0(products) = DS0 DS0 DS0 DS0 rxn rxn rxn rxn What is the standard entropy change for the following reaction at 250C? 2CO (g) + O2(g) 2CO2(g) = 2 x S0(CO2) – [2 x S0(CO) + S0 (O2)] = 427.2 – [395.8 + 205.0] = -173.6 J/K•mol Entropy Changes in the System (DSsys) S0(CO) = 197.9 J/K•mol S0(CO2) = 213.6 J/K•mol S0(O2) = 205.0 J/K•mol 18.4

  48. What is the sign of the entropy change for the following reaction? 2Zn (s) + O2(g) 2ZnO (s) Entropy Changes in the System (DSsys) When gases are produced (or consumed) • If a reaction produces more gas molecules than it consumes, DS0 > 0. • If the total number of gas molecules diminishes, DS0 < 0. • If there is no net change in the total number of gas molecules, then DS0 may be positive or negative BUT DS0 will be a small number. The total number of gas molecules goes down, DS is negative. 18.4

  49. nonspontaneous spontaneous 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 0C and ice melts above 0 0C • 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 18.2

  50. Gibbs Free Energy Spontaneous process: DSuniv = DSsys + DSsurr > 0 Equilibrium process: DSuniv = DSsys + DSsurr = 0 For a constant-temperature process: Gibbs free energy (G) DG = DHsys -TDSsys DG < 0 The reaction is spontaneous in the forward direction. DG > 0 The reaction is nonspontaneous as written. The reaction is spontaneous in the reverse direction. DG = 0 The reaction is at equilibrium. 18.5

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