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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)

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Enthalpy (H)

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  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 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 System Energy Before reaction After reaction Myers, Oldham, Tocci, Chemistry, 2004, page 41

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

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

  6. 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

  7. 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

  8. 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)

  9. 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

  10. 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

  11. 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 2 ( ) 2 ( ) 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 3 2 CH4(g) + 4 O2(g)  2 CO2(g) + 4 H2O(g)ΔHrxn = -1605 kJ 2 CH3OH + 3 O2 2 CO2 + 4 H2O ΔHrxn = -1277 kJ

  12. 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 (i.e. CH3OH, CO2) • 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

  13. 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) - ½ ( ) - ½ ( ) * S(s) + 3/2O2(g) → SO3(g) ΔH = -395.2 kJ 2SO3(g) → O2(g) + 2SO2(g) ΔH = +198.2 kJ * SO3(g) → ½ O2(g) + SO2(g) ΔH = +99.1 kJ ΔH = -296.1 kJ S(s) + O2(g) → SO2(g)

  14. 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) -( ) 2( ) 2( ) * 2C(s) + 2O2(g) → 2CO2(g) ΔH = -788 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 = +226 kJ

  15. 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) -½( ) -½( ) -½( ) -½( ) * NO(g) + O3(g) → NO2(g) + O2(g) ΔH = - 199 kJ * O(g) → ½ O2(g) ΔH = - 247.5 kJ * 3/2 O2(g) → O3(g) ΔH = + 213.5 kJ NO(g) + O(g) → NO2(g) ΔH = - 233 kJ

  16. 4. Given the following data: Fe2O3(s) + 3CO(g) → 2Fe(s) + 3CO2(g) ΔH = -23 kJ 3Fe2O3(s) + CO(g) → 2Fe3O4(s) + CO2(g) ΔH = -39 kJ Fe3O4(s) + CO(g) → 3FeO(s) + CO2(g) ΔH = 18 kJ Calculate ΔH for the following reaction: FeO(s) + CO(g) → Fe(s) + CO2(g)

  17. Hess’s Law HW Questions 1. A B ∆H = + 30 kJ C B ∆H = - 60 kJ Calculate ΔH for the following reaction: A  C 2. Suppose you are given the following reactions: 4X 2Y DH = - 40 kJ X ½ Z DH = - 95 kJ Calculate ΔH for the following reaction: Y Z

  18. 3. From the following heats of reaction: 2 H2 (g) + O2 (g) 2 H2O (g) DH = -483.6 kJ 3 O2 (g) 2 O3 (g) DH = +284.6 kJ Calculate the heat of the reaction (∆H): 3 H2 (g) + O3 (g) 3 H2O (g)

  19. 4. From the following enthalpies of reaction: H2 (g) + F2 (g) 2 HF (g) DH = - 537kJ C (s) + 2 F2 (g) CF4 (g) DH = - 680 kJ 2 C (s) + 2 H2 (g) C2H4 (g) DH = + 52.3 kJ Calculate the DH for the reaction of ethylene with F2. C2H4 (g) + 6F2 (g) 2 CF4 (g) + 4 HF(g)

  20. 5. Given the following data: N2 (g) + O2 (g) 2 NO (g) DH = + 180.7 kJ 2 NO (g) + O2 (g) 2 NO2 (g) DH = - 113.1 kJ 2 N2O (g) 2 N2 (g) + O2 (g) DH = - 162.3 kJ Calculate DH for the reaction below: N2O(g) + NO2 (g) 3 NO (g) DH = ?

  21. 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 (remember M = mol/L) • 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

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

  23. 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)

  24. 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 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

  25. 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 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

  26. Enthalpy Review #2. Calculate H for the following reaction: N2H4 (g) + O2 (g) → N2 (g) + 2H2O (g) Given: H (kJ/mol) 2 NH3(g) + 3 N2O (g) → 4 N2(g) + 3 H2O (g) -1010 N2O (g) + 3 H2 (g) → N2H4 (g) + H2O (g) -317 2 NH3(g) + ½ O2 (g) → N2H4 (g) + H2O (g) -143 H2(g) + ½ O2 (g) → H2O (g) -286

  27. Bond Energies • Chemical reaction⇔ Bond breakage & bond formation • Bond energy = energy required to break a bond • Bond breaking is endothermic (raises potential energy) • Bond formation is exothermic (lowers PE) • Average energy for one type of bond in different molecules • Common Bond Energies C-H : 413 kJ/mol C=O : 799 kJ/mol O=O : 495 kJ/mol O-H : 467 kJ/mol • ΔHrxn = (bonds broken) – (bonds formed) Energy required Energy released

  28. Bond Energies • ΔHrxn = (bonds broken) – (bonds formed) ex. CH4 + 2 O2 CO2 + 2 H2O • C-H : 413 kJ/mol C=O : 799 kJ/mol • O=O : 495 kJ/mol O-H : 467 kJ/mol ΔHrxn = [4(C-H) + 2(O=O)] – [2(C=O) + 4(O-H)] ΔHrxn = [4(413kJ) + 2(495kJ)] – [2(799kJ) + 4(467kJ)] ΔHrxn = -824 kJ Compare to ΔHrxn = -802.5 kJ

  29. Enthalpy Summary • Enthalpy (ΔH) rxn = heat = q • All rxns have some ΔH • ΔH = +…endo • ΔH = - …exo • If given ΔH, can use stoich to quantify • Three ways to estimate ΔH (if not given) • Hess’s Law • known eq’s manipulated into desired eq.  ΔHrxn • Heats of Formation (ΔHf): values from appendix • ΔHrxn = nΔH°f(products) - nΔH°f(reactants) • Bond energies • ΔHrxn= bonds broken – bonds formed

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