Lect w10 abbrev_ thermochemistry_alg

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  • Lect w10 abbrev_ thermochemistry_alg

    1. 1. The orbitals of 2p electrons are often represented as being? <ul><li>elliptical </li></ul><ul><li>tetrahedral </li></ul><ul><li>dumbbell shaped </li></ul><ul><li>spherical </li></ul>
    2. 2. General Chemistry II CHEM 152 Unit 3 Week 10
    3. 3. Week 10 Reading Assignment Chapter 6 – Sections 6.2 – 6.8 (thermochemistry)
    4. 4. HYDROCHLORIC ACID SODIUM HYDROXIDE Make a Prediction . Make a Prediction Will a reaction take place? . HCl + NaOH  NaCl + H 2 O ??????????? ?????????? YES!!!!!
    5. 5. What are the chances of the car just by itself Turning into the car on the right? Will this reaction occur by itself? 2 Fe 2 O 3  4 Fe + 3O 2 CHEMICAL REACTIONS have a natural DIRECTION How can we predict that direction???
    6. 6. Which way will it go? <ul><li>We’ve studied: </li></ul><ul><li>KINETICS or the SPEED of reactions </li></ul><ul><li>EQUILIBRIUM or the position of a reaction when the forward and reverse rates are equal </li></ul><ul><li>Now we’d like to PREDICT DIRECTIONALITY </li></ul><ul><ul><li>Based on the stability of products versus reactants (Stored energy within the bonds) </li></ul></ul>Energy is a key element in directionality
    7. 7. During a chemical reaction chemical bonds are broken and new bonds are formed. The chemical nature of the substances present in the system changes. ENERGY IS ABSORBED OR RELEASED IN THE PROCESS Thermochemistry How can we determine whether the process requires or releases energy? Macro Micro
    8. 8. Surroundings System q > 0 heat ENDOTHERMIC Heat is transferred from surroundings to the system The surroundings get cooler The surroundings get warmer EXOTHERMIC Surroundings q < 0 heat System Heat is transferred from the system to the surroundings
    9. 9. <ul><li>1 c alorie = energy required to raise the temperature of 1.00 g of H 2 O by 1.0 o C. </li></ul>Units of Energy The energy transferred between two systems with different temperatures ( heat ) is measured using the same units 1000 cal = 1 kilocalorie = 1 kcal 1 kcal = 1 C alorie (a food “calorie”) But we use the unit called the JOULE 1 cal = 4.184 Joules (exactly)
    10. 10. <ul><li>Most chemical reactions occur at constant pressure P, so heat is transferred at constant P </li></ul><ul><li>q p =  H where H = Enthalpy </li></ul><ul><li> H = enthalpy change= heat transferred at constant P </li></ul>HEAT VS ENTHALPY If  H is positive (  H>0 ) Process is ENDOTHERMIC If  H is negative (  H<0) Process is EXOTHERMIC
    11. 11. Exothermic H 2 (g) + ½ O 2 (g)  H 2 O(l)  H = - 285.8 kJ/mol In this case, products have a lower potential energy than reactants. H 2 + ½ O 2 PE H 2 O(l) Reactants Products ENERGY RELEASED  H
    12. 12. Endothermic NH 4 NO 3 (s) + H 2 O(l)  NH 4 + (aq) + NO 3 - (aq) In this case, products have a higher potential energy than reactants. NH 4 NO 3 + H 2 O PE NH 4 + + NO 3 - Reactants Products ENERGY ABSORBED  H
    13. 13. Q (Calorimeter) = − Q (reaction) How could we MEASURE the energy transferred? Heat coming out of the reaction all goes into the water. CALORIMETER What happens to the temperature of the water?
    14. 14. 0.0150 mol of a substance was combusted in a bomb calorimeter. 500.0 g of water in a calorimeter had an increase of 4.90 ºC. What was the energy change? C p of water is 4.184 J/(g ºC) What was the  H of the combustion reaction in units of kJ/mol?
    15. 15. Energy is never lost so…. Estimating Δ H using lab experiments H Sn + Cl 2 SnCl 2 + Cl 2 SnCl 4 -325 kJ -186 kJ -511 kJ Sn + Cl 2  SnCl 2 Δ H = -325 kJ SnCl 2 + Cl 2  SnCl 4 Δ H= -186 kJ Sn + 2Cl 2  SnCl 4 Δ H = -511 kJ Hess’ Law – If a reaction can be written as a series of steps, the The Δ H of the overall reaction equals the sum of the the Δ H’s Of the steps.
    16. 16. Calculate  H  for:   Ca(OH) 2(aq) + 2HCl (aq)  CaCl 2(aq) + 2H 2 O (l) using:   CaO (s) + 2HCl (aq)  CaCl 2(aq) + H 2 O (l)  H  = ‒186 kJ CaO (s) + H 2 O (l)  Ca(OH) 2(s)  H  = ‒62.3 kJ Ca(OH) 2(s)  Ca(OH) 2(aq)  H  = ‒12.6 kJ
    17. 17. The limiting reactant in a reaction can be recognized because it is the reagent that? <ul><li>Has the smallest coefficent in the balanced equation </li></ul><ul><li>Has the smallest mass in the reaction mixture </li></ul><ul><li>Is present in the smallest molar quantity </li></ul><ul><li>Would be used up first </li></ul>
    18. 18. What if we can’t do experiments?
    19. 19. Standard Enthalpy Values Most  H values for a chemical reaction are labeled  H o measured under standard conditions P = 1 bar (close to 1 atmosphere) T = usually 25 o C (298.15 K) All species in standard states e.g., C = graphite and O 2 = gas  H o f = standard molar enthalpy of formation This is the enthalpy change when 1 mol of a compound is formed from elements under standard conditions.
    20. 20.  H o f , Standard Enthalpy of Formation H 2 (g) + 1/2 O 2 (g)  H 2 O(l)  H o f = -285.8 kJ/mol H 2 (g) + O 2 (g)  H 2 O 2 (l)  H o f = -187.8 kJ/mol H 2 O 2 (l)  H 2 O(l) + ½ O 2 (g) ? By convention,  H o f = 0 for elements in their standard states H 2 , O 2 H 2 O(l) H 2 O 2 (l) PE -285.8 kJ -187.8 kJ
    21. 21. Molar Enthalpies of Formation Substance Name  H o f (kJ/mol) CH 4 (g) Methane -74.8 CO(g) Carbon monoxide -110.5 CO 2 (g) Carbon dioxide -393.5 H 2 O(g) Water vapor -241.8 H 2 O(l) Liquid water -285.8 CH 3 OH(l) Methanol -238.7 C 2 H 5 OH(l) Ethanol -277.7 BaCO 3 (s) Barium carbonate -1216.3 CaO(s) Calcium oxide -635.1 NH 3 (g) Ammonia -46.1
    22. 22. Using Standard Enthalpy Values <ul><li>H 2 O(g) + C(graphite)  H 2 (g) + CO(g) </li></ul><ul><li>From tables we find </li></ul><ul><li> H ° f of H 2 O vapor = - 241.8 kJ/mol </li></ul><ul><li>H 2 (g) + 1/2 O 2 (g)  H 2 O(g) </li></ul><ul><li> H ° f of CO = - 110.5 kJ/mol C(s) + 1/2 O 2 (g)  CO(g) </li></ul><ul><li>and we know  H ° f =0 for C(graph) and H 2 (g) </li></ul>We can use  H f ° ’s to calculate the heat of reaction of different processes:
    23. 23. Using Standard Enthalpy Values <ul><li>In general, when ALL enthalpies of formation are known, </li></ul><ul><li> H o rxn =   H o f (products) -   H o f (reactants) </li></ul>Calculate  H of reaction?
    24. 24. Using Standard Enthalpy Values H 2 O(g) + C(graphite)  H 2 (g) + CO(g)  H o rxn =   H o f (products) -   H o f (reactants)  H o rxn =  H o f (H 2 ) +  H o f (CO) -  H o f (H 2 O) –  H o f (C)  H o rxn = 0 +(–110.5 kJ) – (-241.8 kJ) – 0  H o rxn = + 131.3 kJ (Endothermic)
    25. 25. H 2 O(g) + C(graphite)  H 2 (g) + CO(g)  H o rxn =   H o f (products) -   H o f (reactants)  H o rxn = + 131.3 kJ (Endothermic) If we find ∆H for the heats of formation If we find ∆H from a calorimeter measuring the reaction ∆ H = + 131.3 kJ ∆ H is the same no matter what route we take.
    26. 26. Using Standard Enthalpy Values Nitroglycerin is a powerful explosive because it decomposes exothermically and four different gases are formed: 2 C 3 H 5 (NO 3 ) 3 (l)  3 N 2 (g) +½ O 2 (g) +6 CO 2 (g) +5 H 2 O(g) For liquid nitroglycerin  H o f =-364.0 kJ/mol. Calculate the energy transfer when 1 mole of nitroglycerin explodes Substance Name  H o f (kJ/mol) CO 2 (g) Carbon dioxide -393.5 H 2 O(g) Water vapor -241.8 H 2 O(l) Liquid water -285.8
    27. 27. Using Bond Energy Data We can estimate  H by comparing the amount of energy needed to separate atoms in the reactants ( BREAKING BONDS ) and the energy released in forming new compounds ( MAKING BONDS ) Remember ∆H is a function of state BOND ENERGY (kJ/mol) H—H 436 C—C 346 C=C 602 C  C 835 N  N 945 The energy required to break a bond is a measure of “bond strength” The same energy is released in making the bond.
    28. 28. Using Bond Energies <ul><li>Bond energies are useful to estimate the energy required or generated by a chemical reaction: </li></ul><ul><li> 2 H—O—O—H  O=O + 2 H—O—H </li></ul>How much energy is released/absorbed in this process? How many bonds of each type are broken/formed?
    29. 29. Using Bond Energies <ul><li>2 H—O—O—H  O=O + 2 H—O—H </li></ul><ul><li>Energy required to break bonds: break 4 mol of O—H bonds = ?? break 2 mol O—O bonds = ?? </li></ul>ENERGY to break bonds = 2124 kJ Energy evolved on making bonds: make 1 mol of O=O bonds = ?? make 4 mol O—H bonds = ?? ENERGY evolved on making bonds = -2339 kJ O=O 499 kJ/mol O—H 460 kJ/mol O—O 142 kJ/mol
    30. 30. Using Bond Energies <ul><li>2 H—O—O—H  O=O + 2 H—O—H </li></ul><ul><li>Net energy =+2124 kJ- 2339 kJ = - 215 kJ </li></ul>or -215/2= -107.5 kJ/mol (compared to -98.0 kJ/mol ) More energy is evolved on making bonds than is used in breaking bonds The reaction is exothermic!
    31. 31. Using Bond Energies <ul><li>Estimate the energy needed or generated by this chemical reaction </li></ul><ul><li> H—H + Cl—Cl  2 H—Cl </li></ul>H—H = 436 kJ/mol Cl—Cl = 242 kJ/mol H—Cl = 432 kJ/mol Net energy =  H o = = energy required to break bonds + energy evolved when bonds are made
    32. 32. How many atoms are in 1.50 g of Al? <ul><li>0.0556 </li></ul><ul><li>18.0 </li></ul><ul><li>3.35 x 10 22 </li></ul><ul><li>2.44 x 10 25 </li></ul>
    33. 33. Using Bond Energies <ul><li>Estimate the energy needed or generated by this chemical reaction </li></ul><ul><li> H—H + Cl—Cl  2 H—Cl </li></ul> H o = -186 kJ Notice that two homogeneous bonds break to form heterogeneous bonds  These tend to be exothermic
    34. 34. Using Bond Energies Which of these two molecules will generate more energy during combustion? Oxygenated bonds require more energy to break than other types of bonds (C-H) Sucrose Stearic Acid
    35. 35. Does burning several logs release more Or less energy than burning one log? There is a relationship between the enthalpy of a reaction And the moles of material. Given by the balanced equation
    36. 36. Energy and Moles What if I use up 6.00 moles of Oxygen in the following reaction? How much energy is released? H 2 (g) + 1/2O 2 (g)  H 2 O(l) Δ H= - 286kJ Conversion factor 6.00mol O 2 ( -286kJ ) = -3430 kJ ( ½ mol O 2 ) Proportions 6.00 mol O 2 = X solving x = -3430kJ ½ mol O 2 -286kJ Δ H= - 3430kJ OR +3430kJ are RELEASED
    37. 37. Which Way does it go? <ul><li>If a reaction gives off heat (  H = -), there is a tendency for the reaction to go forward (K > 1) – but that is not the complete story… </li></ul>WHICH WAY? KCl(s)  KCl(aq)  H = + And yet K > 1 – it goes forward anyways
    38. 38. Summary Activity 12C (s,graphite) + 11H 2(g) + O 2(g)  C 12 H 22 O 11(s) No one has been able to get this reaction to run… so how could we get a value for the heat of formation of sucrose? We can combust sucrose:   C 12 H 22 O 11(s) + 12O 2(g)  12CO 2(g) + 11H 2 O (l) The standard heat of combustion  H c ° of sucrose is – 5640.9 kJ/mol. Can you now calculate the heat of formation of sucrose using Hess’s Law?  H f º(CO 2 (g)) = -393.5 kJ/mol  H f º(H 2 O(l)) = -285.8 kJ/mol 11 2
    39. 39. Summary Activity Nitrogen monoxide has recently been found to be involved in a wide range of biological processes. The gas reacts with oxygen to give brown NO 2 gas.   2NO (g) + O 2(g)  2NO 2 (g) Is the reaction endothermic or exothermic? If 1.25 g of NO is converted completely to NO 2 , what quantity of heat is absorbed or released?  Hº = -114.1 kJ M (NO) = 30.01 g/mol

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