1 / 48

Enthalpy (H)

Enthalpy (H). The heat transferred sys ↔ surr during a chemical rxn @ constant P Can’t measure H, only Δ H At constant P, Δ H = q = mC Δ T, etc. Literally, Δ H = H products - H reactants Δ H = + (endothermic) Heat goes from surr into sys Δ H = - (exothermic)

royal
Download Presentation

Enthalpy (H)

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Enthalpy (H) • The heat transferred sys ↔ surr during a chemical rxn @ constant P • Can’t measure H, only ΔH • At constant P, ΔH = q = mCΔT, etc. • Literally,ΔH = Hproducts - Hreactants • ΔH = +(endothermic) • Heat goes from surr into sys • ΔH = - (exothermic) • Heat leaves sys and goes into surr

  2. Reactant + Energy Product Surroundings System In this example, the energy of the system (reactants and products) ↑, while the energy of the surroundings ↓ Notice that the total energy does not change Endothermic Reaction Surroundings Notice that E must be added, and thus is like a reactant Energy System Before reaction After reaction Myers, Oldham, Tocci, Chemistry, 2004, page 41

  3. Reactant Product + Energy Surroundings System In this example, the energy of the system (reactants and products) ↓, while the energy of the surroundings ↑ Notice again that the total energy does not change Exothermic Reaction Surroundings Notice that E is released and thus is like a product System Energy Before reaction After reaction Myers, Oldham, Tocci, Chemistry, 2004, page 41

  4. Burning of a Match D(PE) (Products) System Surroundings (Reactants) Potential energy Energy released to the surrounding as heat Exothermic Reaction Zumdahl, Zumdahl, DeCoste, World of Chemistry2002, page 293

  5. Reaction Coordinate Diagrams: Endothermic Reaction Activation Energy Products Ea D(PE) ΔHrxn= + Reactants PE Progress of the Reaction

  6. Reaction Coordinate Diagrams: Exothermic Reaction Activation Energy ΔHrxn= - Ea D(PE) Reactants Products PE Progress of the Reaction

  7. Reaction Coordinate Diagrams Activation Energy ΔHrxn= -458.1 kJ Ea D(PE) C + O2 CO2 Draw the reaction coordinate diagram for the following rxn: C(s) + O2(g)  CO2 + 458.1kJ EXOTHERMIC PE Progress of the Reaction

  8. Enthalpies of Reaction - 483.6 kJ - 483.6 kJ - 483.6 kJ 1 mol H2O ½ mol O2 1 mol H2 • All reactions have some ΔH associated with it H2(g) + ½ O2(g) → H2O(l)ΔH = - 483.6 kJ • How can we interpret this ΔH? • Amount of energy released or absorbed per specific reaction species • Use balanced equation to find several definitions or or Able to use like conversion factors in stoichiometry

  9. Enthalpies of Reaction • Formation of water H2(g) + ½ O2(g) → H2O(l)ΔH = - 483.6 kJ • ΔH is proportional to amount used and will change as amount changes 2H2(g) + O2(g) → 2H2O(l) • For reverse reactions, sign of ΔH changes 2H2O(l)→ 2H2(g) + O2(g) • Treat ΔH like reactant or product H2(g) + ½ O2(g) → H2O(l) ΔH = - 483.6 kJ ΔH = - 967.2 kJ ΔH = + 967.2 kJ H2(g) + ½ O2(g) → H2O(l) + 483.6 kJ (exo)

  10. Enthalpies of Reaction Practice Consider the following rxn: C(s) + 1/2O2(g)  CO+ 458.1kJ Is the ΔH for this reaction positive or negative? NEGATIVE (E released as a product) What is the ΔH for 2.00 moles of carbon, if all the carbon is used? 2.00 mol C - 458.1 kJ = - 916 kJ 1 mol C What is the ΔH if 50.0g of oxygen is used? 50.0 g O2 1 mol O2 - 458.1 kJ = -1430 kJ 32.0 g O2 0.5 mol O2 What is the ΔH if 50.0 g of carbon monoxide decompose, in the reverse reaction? 50.0 g CO 1 mol CO 458.1 kJ = 818 kJ 1 mol CO 28.0 g CO

  11. Hess’s law • Hess’s Law states that the enthalpy of a whole reaction is equivalent to the sum of it’s steps. • For example: C + O2 CO2 This can occur as 2 steps C + ½O2  CO H = – 110.5 kJ CO + ½O2  CO2 H = – 283.0 kJ C + CO + O2  CO + CO2 H = – 393.5 kJ • I.e. C + O2  CO2 H = – 393.5 kJ • Hess’s law allows us to add equations. • We add all reactants, products, & H values.

  12. Hess’s Law Reactants  Products The change in enthalpy is the samewhether the reaction takes place in one step or a series of steps Why? Because enthalpy is a state function Victor Hess To review: 1. If a reaction is reversed, ΔH is also reversed 2 CH4 + O2  2 CH3OH ΔHrxn = -328 kJ 2 CH3OH  2 CH4 + O2ΔHrxn = +328 kJ 2. If the coefficients of a reaction are multiplied by an integer, ΔH is multiplied by that same integer CH4 + 2 O2  CO2 + 2 H2O ΔHrxn = -802.5 kJ 2(CH4 + 2 O2  CO2 + 2 H2O) ΔHrxn = -1605 kJ

  13. Example: Methanol-Powered Cars 2 CH3OH(l) + 3 O2(g) 2 CO2(g) + 4 H2O(g)ΔHrxn = ? 2 CH4(g) + O2(g)  2 CH3OH(l)ΔHrxn = -328 kJ CH4(g) + 2 O2(g)  CO2(g) + 2 H2O(g)ΔHrxn = -802.5 kJ 2 CH3OH(l) 2 CH4(g) + O2(g)ΔHrxn = +328 kJ 2(CH4(g) + 2 O2(g)  CO2(g) + 2 H2O(g)) ΔHrxn = -1605 kJ 2 CH3OH(l) + 2 CH4(g) + 4 O2(g)  2 CH4(g) + O2(g) + 2 CO2(g) + 4 H2O(g) 3 2 CH3OH + 3 O2 2 CO2 + 4 H2O ΔHrxn = -1277 kJ

  14. Tips for applying Hess’s Law… Look at the final equation that you are trying to create first… • Find a molecule from that eq. that is only in one of the given equations • Make whatever alterations are necessary to those • Once you alter a given equation, you will not alter it again • Continue to do this until there are no other options • Next, alter remaining equations to get things to cancel that do not appear in the final equation

  15. 1. Given the following data: S(s) + 3/2O2(g) → SO3(g) ΔH = -395.2 kJ 2SO2(g) + O2(g) → 2SO3(g) ΔH = -198.2 kJ . Calculate ΔH for the following reaction: S(s) + O2(g) → SO2(g) S(s) + 3/2O2(g) → SO3(g) ΔH = -395.2 kJ 2SO2(g) + O2(g) → 2SO3(g) ΔH = -198.2 kJ 2SO3(g) → O2(g) + 2SO2(g) ΔH = +198.2 kJ SO3(g) → ½ O2(g) + SO2(g) ΔH = +99.1 kJ S(s) + O2(g) → SO2(g) ΔH = -296.1kJ

  16. 2. Given the following data: C2H2(g) + 5/2O2(g) → 2CO2(g) + H2O(l) ΔH = -1300 kJ C(s) + O2(g) → CO2(g) ΔH = -394 kJ H2(g) + 1/2O2(g) → H2O(l) ΔH = -286 kJ Calculate ΔH for the following reaction: 2C(s) + H2(g) → C2H2(g) C(s) + O2(g) → CO2(g) ΔH = -394 kJ 2C(s) + 2O2(g) → 2CO2(g) ΔH = -788 kJ C2H2(g) + 5/2O2(g) → 2CO2(g) + H2O(l) ΔH = -1300 kJ 2CO2(g) + H2O(l) → C2H2(g) + 5/2O2(g) ΔH = +1300 kJ H2(g) + 1/2O2(g) → H2O(l) ΔH = -286 kJ 2C(s) + H2(g) → C2H2(g) ΔH = +226kJ

  17. 3. Given the following data:   2O3(g) → 3O2(g) ΔH = - 427 kJ O2(g) → 2O(g) ΔH = + 495 kJ NO(g) + O3(g) → NO2(g) + O2(g) ΔH = - 199 kJ Calculate ΔH for the following reaction: NO(g) + O(g) → NO2(g) NO(g) + O3(g) → NO2(g) + O2(g) ΔH = - 199 kJ O2(g) → 2O(g) ΔH = + 495 kJ O(g) → ½ O2(g) ΔH = - 247.5 kJ 2O3(g) → 3O2(g) ΔH = - 427 kJ 3/2 O2(g) → O3(g) ΔH = + 213.5 kJ NO(g) + O(g) → NO2(g) ΔH = +233kJ

  18. Heats of Formation, ΔH°f The enthalpy change when one mole of a compound is formed from the elements in their standard states ° = standard conditions • Gases at 1 atm pressure • All solutes at 1 M concentration • Pure solids and pure liquids f = a formation reaction • 1 mole of product formed • From the elements in their standard states (1 atm, 25°C) For all elements in their standard states, ΔH°f= 0 What’s the formation reaction for adrenaline, C9H12NO3(s)?  C9H12NO3(s) Cgr + H2(g) + N2(g) + O2(g) 9 6 1/2 3/2

  19. Thermite Reaction Welding railroad tracks Fe2O3(s) + 2 Al(s) Al2O3(s) + 2 Fe(l) ΔHrxn = ?

  20. Thermite Reaction Reactants Elements (standard states) Products Fe2O3(s) + 2 Al(s) Al2O3(s) + 2 Fe(l) Fe2O3(s) 2 Fe(s) 2 Fe(l) 2 Al(s) 3/2 O2(g) Al2O3(s) 2 Al(s) ΔHrxn =2ΔH°f(Fe(l))+ΔH°f(Al2O3(s)) - ΔH°f(Fe2O3(s)) - 2ΔH°f(Al(s)) ΔHrxn =[2(15 kJ)+(-1676 kJ)] – [(-822 kJ)– 2(0)] ΔHrxn = -824 kJ ΔHrxn = nΔH°f(products) - nΔH°f(reactants)

  21. ΔH°f Example Problems ∆Hrxn = Σn∆Hof Products - Σn∆Hof Reactants 1. CH4(g) + 2 Cl2(g) CCl4(g) + 2 H2(g) ΔHrxn = ? (0) 2 (-106.7) (-74.8) (0) 2 ∆H = [(-106.7) + 0] – [(-74.8)+0] = -106.7 + 74.8 = -31.9 kJ/mol 2. 2 KCl(s) + 3 O2(g) 2KClO3(s) ΔHrxn = ? (0) 3 (-435.9) (-391.2) 2 2 ∆H = [(2)(-391.2)] – [(2)(-435.9) + (3)(0)] = -782.4 + 871.8 = 89.4 kJ/mol

  22. ΔH°f Example Problems ∆Hrxn = Σ n∆Hof Products - Σn∆Hof Reactants 3. AgNO3(s) + NaCl(aq) AgCl(s) + NaNO3(aq) ΔHrxn = ? (-446.2) (-127.0) (-124.4) (-407.1) ∆H = [(-127.0) + (-446.2)] – [(-124.4) + (-407.1)] = -573.2 + 531.5 = - 41.7 kJ/mol 4. C2H5OH(l) + 7/2 O2(g) 2CO2(g) + 3H2O(g) ΔHrxn = ? (7/2) (0) (-277.7) (3) (2) (-393.5) (-241.8) ∆H = [(2)(-393.5) + (3)(-241.8)] – [(-277.7) + (7/2)(0)] = -1512.4 + 277.7 = -1234.7 kJ/mol

  23. THIS MEANS THAT THINGS FALL. THEY FALL FROM HEIGHTS OF ENERGY AND STRUCTURED INFORMATION INTO MEANINGLESS, POWERLESS DISORDER. MATTER IS ENERGY. ENERGY IS INFORMATION. EVERYTHING IS INFORMATION. PHYSICS SAYS THAT STRUCTURES... BUILDINGS, SOCIETIES, IDEOLOGIES... WILL SEEK THEIR POINT OF LEAST ENERGY. THIS IS CALLED ENTROPY. Entropy (S) = a measure of randomness or disorder

  24. Entropy: Tendency toward disorder

  25. Entropy: Tendency toward disorder

  26. Second Law of Thermodynamics • In any spontaneous process, the entropy of the universe increases or is + ΔSuniverse > 0 occurs without outside intervention ΔSuniverse = ΔSsystem + ΔSsurroundings

  27. Entropy of the Universe ΔSuniverse = ΔSsystem + ΔSsurroundings Positional disorder Energetic disorder ΔSuniverse > 0 spontaneous process Both ΔSsys and ΔSsurr positive Both ΔSsys and ΔSsurr negative ΔSsys negative, ΔSsurr positive ΔSsys positive, ΔSsurr negative spontaneous process nonspontaneous process depends depends

  28. ΔSsys: Positional Disorder and Probability Probability of 1 particle in left bulb = ½ " 2 particles both in left bulb = (½)(½) = ¼ " 3 particles all in left bulb = (½)(½)(½) = 1/8 " 4 " all " = (½) 4 = 1/16 " 10 " all " = (½)10 = 1/1024 " 20 " all " = (½)20 = 1/1048576 " a mole of " all " = (½)6.02 x1023 The arrangement with the greatest entropy is the one with the highest probability (most “spread out”).

  29. Entropy of the System: Positional Disorder Ssolid < Sliquid <<<< Sgas Ludwig Boltzmann Ordered state Low probability (few ways) Ludwig Boltzmann Low S Disordered state High probability (many ways) High S • Ssystem is proportional to positional disorder • S increases with increasing # of possible positions

  30. Entropy of the Surroundings(Energetic Disorder) System Entropy Heat ΔSsurr > 0 ΔHsys < 0 Surroundings System Surroundings Heat Entropy ΔSsurr < 0 ΔHsys > 0 Low T → large entropy change (surroundings) High T→ small entropy change (surroundings)

  31. The Third Law of Thermodynamics The Third Law: The entropy of a perfect crystal at 0 K is zero • Everything locked into place • No molecular motion whatsoever Crystallization of Water into Ice

  32. Entropy Curve Solid Liquid Gas ΔHvap (l ↔ g) S (J/K) Δfus (s ↔ l) 0 Temperature (K) 0 S° (absolute entropy) can be calculated for any substance

  33. Entropy Increases with... • Melting (fusion) Sliquid > Ssolid ΔHfus/Tfus = ΔSfus • Vaporization Sgas > Sliquid ΔHvap/Tvap = ΔSvap • Increasing ngas in a reaction • Heating ST2 > ST1 if T2 > T1 • Dissolving (usually) Ssolution > (Ssolvent + Ssolute) • Molecular complexity more bonds, more entropy • Atomic complexity more e-, protons, neutrons

  34. Entropy Practice C6H12O6(s) + 6 O2(g) → 6 CO2(g) + 6 H2O(g) For the above reaction, predict the sign of ΔSsys We can predict S values based on phases… Remember that Ssolid< Sliquid <<<<<< Sgas and that ΔSsys = Sproducts – Sreactants ΔSsys. = S (12 mol gas) – S (6 mol gas + 1 mol solid) Solid is negligible, more gas products, ↑ disorder Therefore…ΔSsys = +

  35. Entropy Practice 4 Al (s) + 3 O2 (g) → 2 Al2O3 (s) For the above reaction, predict the sign of ΔSsys We can predict S values based on phases… Remember that Ssolid< Sliquid <<<<<<< Sgas ΔSsys = Sproducts – Sreactants ΔSsys = S (2 mol solid) – S (4 mol solid + 3 mol gas) Solid is negligible, gas R → solid P , ↓ disorder Therefore…ΔSsys = -

  36. Predicting ΔSsys sign summary • Use relative S values for phases Ssolid< Saqueous< Sliquid < Sgas • Gases always have greater entropy • Consider number of moles, especially with gases • You cannot predict for some reactions H2(g) + Cl2(g) → 2HCl(g) 2 mol gas →2 mol gas Unable to predict sign ΔSsys for this reaction

  37. C6H12O6(s) + 6 O2(g) 6 CO2(g) + 6 H2O(g) Calculating Entropy Quantitatively Calculate the value of ΔS°sys: Compound C6H12O6(s) O2(g) CO2(g) H2O(g) S° (J/mol K) 212 205 214 189 ΔS°sys = ΣnS°(products) - ΣnS°(reactants) = [6 S°(CO2(g)) + 6 S°(H2O(g))] – [S°(C6H12O6(s)) + 6 S°(O2(g))] = [6(214) + 6(189)] – [(212) + 6(205)] J/K ΔS°sys= 976 J/K

  38. C6H12O6(s) + 6 O2(g) 6 CO2(g) + 6 H2O(g) YES! Is this reaction spontaneous at 298K? ΔSuniverse = ΔSsys + ΔSsurr ΔSsurr = - ΔH/T and Compound C6H12O6(s) O2(g) CO2(g) H2O(g) ΔH°f (kJ/mol) -1275 0 -393.5 -242 S° (J/mol K) 212 205 214 189 ΔH°rxn = ΣnΔH°f (products) - ΣnΔH°f(reactants) = [6 ΔH°f(CO2(g)) + 6 ΔH°f(H2O(g))] – [ΔH°f(C6H12O6(s)) + 6 ΔH°f(O2(g))] = [6(-393.5) + 6(-242)] – [(-1275) + 6(0)] kJ ΔH°rxn = -2538 kJ ΔSsys = 976 J/K from last problem 9.49 kJ/K ΔSuniverse = 0.976 kJ/K + -(-2538 kJ)/298K =

  39. Entropy (S) Review • ΔSuniverse > 0 for spontaneous processes • ΔSuniverse = ΔSsystem + ΔSsurroundings positional energetic • We can find the absolute entropy value for a substance • S° values for elements & compounds in their standard states are tabulated (Thermodynamic Appendix) • For any chemical reaction, we can calculate ΔS°rxn: • ΔS°rxn = ΣnS°(products) - ΣnS°(reactants)

  40. Recap: Characteristics of Entropy • S is a state function (we can use final – initial) • S is extensive (more stuff, more entropy) • At 0 K, S = 0 (we can determine absolute entropy) • S > 0 for elements and compounds in their standard states • Raise T  increase S • Increase ngas  increase S • More complex systems  larger S

  41. Gibbs Free Energy (G) ΔSuniverse = ΔS – ΔH/T ΔG = ΔH – TΔS Divide both sides by –T -ΔG/T = -ΔH/T + ΔS G = H – TS At constant temperature, ΔG = ΔH – TΔS (system’s point of view) –ΔGmeans+ΔSuniv A process (at constant T, P) is spontaneous if free energy decreases Josiah Gibbs

  42. ΔG and Chemical Reactions • ΔG = ΔH – TΔS • If ΔG < 0, the reaction is spontaneous • If ΔG > 0, the reaction is not spontaneous • (The reverse reaction is spontaneous) • If ΔG = 0, the reaction is at equilibrium • Neither the forward nor the reverse reaction is favored • Both reactions are occurring simultaneously and at equal rates • ΔG is an extensive, state function Depends on final and initial states only Depends on how much stuff

  43. Ba(OH)2(s) + 2NH4Cl(s) BaCl2(s) + 2NH3(g) + 2 H2O(l) Is the following reaction spontaneous at 298 K? ΔH°rxn = 50.0 kJ (per mole Ba(OH)2) ΔS°rxn = 328 J/K (per mole Ba(OH)2) ΔG = ΔH - TΔS ΔG° = 50.0 kJ – 298 K(0.328 kJ/K) ΔG° = – 47.7 kJ Spontaneous At what T does the reaction stop being spontaneous? The T where ΔG = 0 ΔG = 0 = 50.0 kJ – T(0.328 kJ/K) 50.0 kJ = T(0.328 kJ/K) T = 152 K not spontaneous below 152 K

  44. Effect of ΔH and ΔS on Spontaneity ΔG = ΔH – TΔS ΔG (-) → spontaneous reaction ΔH – + – + ΔS + + – – • Spontaneous? • Spontaneous at all temps • Spontaneous at high temps • Reverse reaction spontaneous at low temps • Spontaneous at low temps • Reverse reaction spontaneous at high temps • Not spontaneous at any temp

  45. Ways to Calculate ΔG°rxn 1.ΔG° = nΔG°f(products) - nΔG°f(reactants) • ΔG°f = free energy change when forming 1 mole of compound from elements in their standard states (see Thermodynamics Appendix for values) 2.ΔG° = ΔH° - TΔS° 3.ΔG° can be calculated by combining ΔG° values for several reactions Using Hess’s Law! Your favorite!

  46. 2 H2(g) + O2(g) 2 H2O(g) Calculate ΔG° for the following reaction: 1.ΔG° = ΔG°f(products) - ΔG°f(reactants) ΔG°f(O2(g)) = 0 ΔG°f(H2(g)) = 0 ΔG°f(H2O(g)) = -229 kJ/mol ΔG° = (2(-229 kJ) – 2(0) – 0) kJ = -458 kJ 2.ΔG° = ΔH° - TΔS° ΔH° = -484 kJ ΔS° = -89 J/K ΔG° = -484 kJ – 298 K(-0.089 kJ/K) = -457 kJ

  47. 2H2(g) + O2(g) 2 H2O(g) 3.ΔG° = combination of ΔG° from other reactions (using Hess’s Law) 2H2O(l) 2H2(g) + O2(g)ΔG°1 = 475 kJ H2O(l) H2O(g)ΔG°2 = 8 kJ ΔG° = - ΔG°1 + 2(ΔG°2) ΔG° = -475 kJ + 16 kJ = -459 kJ Method 1: Method 2: Method 3: Method 1: -458 kJ Method 2: -457 kJ Method 3: -459 kJ

  48. What is Free Energy, Really? • NOT just “another form of energy” • Free Energy is the energy available to do useful work • If ΔG is negative, the system can do work (wmax = ΔG) • If ΔG is positive, then ΔG is the work required to make the process happen • Example: Photosynthesis 6 CO2 + 6 H2O  C6H12O6 + 6 O2 ΔG = 2870 kJ/mol of glucose at 25°C Thus, 2870 kJ of work is required to photosynthesize 1 mole of glucose

More Related