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Chemical Reactions / Thermochemistry / Gases

Chemical Reactions / Thermochemistry / Gases. H Advanced Chemistry Unit 2. Objective #4 Lewis vs. Bronsted -Lowery Definitions of Acids and Bases.

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Chemical Reactions / Thermochemistry / Gases

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  1. Chemical Reactions / Thermochemistry / Gases H Advanced Chemistry Unit 2

  2. Objective #4 Lewis vs. Bronsted-Lowery Definitions of Acids and Bases The traditional Arrhenius definition for acids and bases is a rather narrow definition. In this definition, acids must contain ionizablehydrogen to be an acid and ionizablehydroxide to be a base. For example: HCl and NaOH. A broader definition, the Bronsted - Lowery concept defines an acid as a species that donates a proton and defines a base as a species that accepts a proton.

  3. Example: H2S + CH3NH2 <> HS-1+ CH3NH3+1 Acid Base Conjugate Conjugate Base Acid

  4. All Arrhenius acids are also Bronsted-Lowery acids but some Bronsted-Lowery bases are not Arrhenius bases. Another broader definition for acids and bases is the Lewis concept. In this system, it is recognized that in order for a Bronsted-Lowery base to be a proton acceptor, it must also be able to donate an electron pair.

  5. Example: H+ + :NH3 > NH4+1 In this system, an acid is defined as a species that can accept an electron pair. Example: :NH3 + BF3 > NH3BF3 Lewis base Lewis acid

  6. Objective #5 Oxidation Numbers and Redox Reactions(Review of Oxidation Numbers +1 -1 Na Cl +1 +6 -2 Na2 S O4 +2 +7 -2 Mg (Cl O4)2 +4 +6 -2 Pb (Cr2 O7)2

  7. Writing Molecular and Net Ionic Equations for Displacement Reactions *a displacement reaction occurs when an ion displaced due to a change in oxidation number *example I: (oxidation state assignments)

  8. 0 +2 +5 -2 +2 +5 -2 Fe + Ni (N O3)2 > Fe (N O3)2 + 0 Ni (molecular equation) (ionic equation) Fe+ Ni+2 + 2NO3-1 > Fe+2 + 2NO3-1 + Ni

  9. (net ionic equation) Fe + Ni+2 > Fe+2 + Ni (spectator ions) NO3-1

  10. *example II: (oxidation state assignments) (molecular equation) 0 +2 +6 -2 +2 +6 -2 0 Mg + Co S O4 > Mg S O4 + Co

  11. (ionic equation): Mg + Co+2 + SO4-2 > Mg+2 + SO4-2 + Co (net ionic equation): Mg + Co+2 > Mg+2 + Co (spectator ions): SO4-2

  12. Objectives #6-9 Working with Energy and the First Law of Thermodynamics • Exothermic vs. Endothermic Reactions *exothermic reactions involve the loss of energy to the surroundings while endothermic reactions involve the gain of energy from the surroundings *the change in energy of a chemical process can be found by comparing the initial value of the energy of the reactants to the final value of the energy of the products ∆E = Efinal - Einitial

  13. Objectives #6-9 Working with Energy and the First Law of Thermodynamics *if the ∆E has a positive value, then the products of a reaction have a higher energy than the reactants, energy has been gained from the surroundings, and the sign of the energy value will be expressed as a positive value *if the ∆E has a negative value, then the products of a reaction have a lower energy than the reactants, energy has been released to the surroundings, and the sign of the energy value will be expressed as a negative value *energy diagram for endothermic vs. exothermic

  14. Endothermic vs. Exothermic Graphs

  15. Objectives #6-9 Working with Energy and the First Law of Thermodynamics *when energy is exchanged with the surroundings, it can be in the form of heat or work *the change in this energy exchange can be found using the equation: ∆E = q + w where “q” represents the heat transferred and “w” represents the work done on the reaction system or by the reaction system *when heat is added or work is done on the reaction system, the total energy of the system increases *the total energy of the system can also decrease if heat is lost to the surroundings and / or work is done on the surroundings

  16. Objectives #6-9 Working with Energy and the First Law of Thermodynamics *the following sign conventions are used to designate the q, w, and ∆t:

  17. If a reaction that produces a solid from various gaseous reactants loses 1200 J of heat to the surroundings and the external atmospheric pressure causes 500 J of work to be done on a collapsing piston, calculate the change in the internal energy of this reaction system. ΔE = q + w = -1200 J + 500 J = -700 J

  18. Calculate the change in the internal energy of the system for a process in which the system absorbs 140 J of heat from the surroundings and does 85 J of work on the surroundings. ΔE = q + w = +140 J – 85 J = +55 J

  19. Objectives #6-9 Working with Energy and the First Law of Thermodynamics *a related equation involves enthalpy which relates the heatflow of a process to its internal energy and work which is the result of pressure and volume: ∆H = ∆E + P∆V *If a reaction is performed at constant pressure, the product of P∆V will be very small as a result. Therefore, the enthalpy equals the heat gained or lost in a reaction. Enthalpy, ∆H, is an easier concept to deal with than looking at the work involved in a chemical process *If heat is gained by a reaction system, its ∆H value will be positive and the reaction is classified as endothermic *If heat is released by a reaction system, its ∆H value will be negative and the reaction is classified as exothermic

  20. *Examples: Determine if each of the following processes would have a positive or negative sign for the ΔH value: Melting ice – +ΔH Burning methane gas – -ΔH Oxidation of magnesium in oxygen – -ΔH

  21. Objective #10 Interpreting Potential Energy Diagrams for Endothermic and Exothermic Reactions *potential energy diagrams can pictorially illustrate the energy changes as reactants are converted into products and the reverse reaction as well

  22. Potential Energy Diagrams:Exothermic vs. Endothermic

  23. *diagram illustrating exothermic reaction: 2H2 + O2 > 2H2O ΔH < 0 (exothermic) Reactants at higher energy state *diagram illustrating exothermic reaction: H2O(l) > H2O(g) ΔH > 0 (endothermic) Products at higher energy state

  24. *diagram illustrating a forward and reverse reaction (example is endothermic) N2 + 3H2 <› 2NH3 ΔH > 0

  25. Objective #10 Interpreting Potential Energy Diagrams for Endothermic and Exothermic Reactions *general conventions: • Enthalpy depends on the amount of material used and produced in the reaction; for example if burning one mole of methane produces 890 kJ of heat then burning two moles will result in the production of 1780 kJ of heat • The amount of heat generated in the forward reaction will equal the amount of heat required in the reverse reaction and vice versa but the signs will be reversed. 3. Enthalpy changes depend on the states of the materials involved in the reaction. For example forming gaseous water would take up more energy than forming liquid water and therefore the final energy output to the surroundings would be less

  26. Objective #11 Problems Involving Energy Stoichiometry *if the ∆H is known for a reaction, the mole ratios of the reactants and products included in the reaction can be used to determine the amount of heat evolved or required in a reaction (examples)

  27. Calculate the amount of heat released when 5.00 g of methane burns in oxygen gas: CH4 + 2O2 › CO2 + 2H2O where the ΔH = -890 kJ 5.00 g CH4 X 1 mole CH4 / 16.0 g X -890 kJ / 1 mole CH4 = -278 kJ

  28. Calculate the amount of heat released when 5.00 g of hydrogen peroxide decomposes. 2H2O2 > 2H2O + O2 where the ΔH = 196 kJ 5.00 g H2O2 X 1 mole H2O2 / 34.0 g X 196 kJ / 2 mole H2O2 = 14.4 kJ

  29. Objectives #12-13 Problems Involving Calorimetry(bomb calorimeter)

  30. Example of a Laboratory Calorimeter

  31. Examples of Calorimeter Problems • How much heat is needed to warm 250. g of water from 22oC to 98oC if the specific heat of liquid water is 4.18 J/g.K. Q = mc∆t = (250. g) (4.18 J/g. K) (76 K) = 79400 J

  32. Calculate the quantity of heat absorbed by 50.0 kg of rocks if their temperature increases by 12.0oC and the specific heat of the rocks is .082 J/g.K. Q = mc∆t = (50000 g) (.082 J/g.K) (12K) = 49200 J

  33. What temperature change would the rocks undergo if they absorbed 450 kJ of heat? Q = mc∆t Q / mc = ∆t 450 kJ / (.082 J/g.K) (50000 g) = ∆t 110 K = ∆t

  34. When 4.00 g of CH6N2 is combusted in a bomb calorimeter (in other words it goes BOOM!), the temperature of the calorimeter increases from 25.00 oC to 39.50 oC. If the heat capacity of the calorimeter is 7.794 kJ/oC determine the heat of reaction per mole of CH6N2. (so we need to find kJ of heat per mole of material)

  35. Find heat of rx. from change in temp and specific heat of calorimeter: ∆t = 39.50oC – 25.00oC = 14.50oC qrxn = -Ccal X ∆t = -(7.794 kJ/oC) (14.50oC) = -113.0 kJ *the heat produced came from the 4.00 g of material

  36. -113.0 kJ / 4.00 g CH6N2 X 46.1 g CH6N2/ 1 mole = -1.30 X 103 kJ/mole

  37. When 50.0 ml of .100 M AgNO3 and 50.0 ml of .100 M HCl are mixed in a constant pressure calorimeter, the temperature of the mixture increases from 22.20oC to 23.11oC. Calculate the ∆H for this reaction in kJ/mole AgNO3, assuming that the combined solution has a mass of 100.0 g and a specific heat of 4.18 J/g.C.

  38. *once again we need heat per mole of material Q = mc∆t = (100.0 g) (4.18 J/goC) (.91oC) = 380 J = .380 kJ Moles AgNO3 = (.100 M) (.0500 L) = .00500 moles .380 kJ / .00500 moles = -76.0 kJ / mole AgNO3

  39. Germain Henri Hess1802-1850

  40. Objectives #14-15 Hess’s Law *if a reaction is carried out in a series of steps, the ∆H for the reaction will equal the sum of the enthalpy changes for the individual steps

  41. Objectives #14-15 Hess’s Law

  42. Objectives #14-15 Hess’s Law *Hess’s Law makes it possible to calculate the energy changes for a reaction indirectly when it is not possible to measure the energy changes directly for a particular reaction *Hess’s Law problems often involve manipulating the step by step reactions that can lead to the overall reaction desired; these manipulations can include: reversing a reaction multiplying or dividing a reaction by some factor canceling like terms When these changes are made, the sign and the of the energy term must also be adjusted (examples)

  43. *For example, let’s say we want to calculate the energy change for the following reaction: C(s) + 1/2O2(g) › CO(g) given the following thermochemical data: • C(s) + O2(g) › CO2(g) ∆H1 = -393.5 kJ 2. CO(g) + 1/2O2(g) › CO2(g) ∆H2 = -283.0 kJ

  44. Step I Rx.#1 has the carbon on the correct reactant side and so we can just recopy that reaction as is: C(s) + O2(g) › CO2(g) ∆H1 = -393.5 kJ Step II Rx.#2 needs to be reversed so that the carbon monoxide appears on the product side as shown in the desired equation: CO2(g) › CO(g) + 1/2O2(g)

  45. since the reaction is reversed, the sign on the energy term needs to switch as well ∆H = +283.0 kJ Step III The carbon dioxide molecules are on opposite sides of the equation and so they can now be canceled out and the remaining substances will now match the desired equation. C(s) + 1/2O2(g) › CO(g)

  46. Step IV Add the energy terms to obtain the final energy value desired. ∆H = +283.0 kJ -110.5 kJ = -110.5 kJ

  47. Example II: Calculate the ∆H for the reaction: 1. 2C(s) + 2O2(g) › 2CO2(g) ∆H = -787 kJ • H2(g) + 1/2O2(g) › H2O(l) ∆H = -285.8 kJ • 2CO2(g) + H2O(l) › C2H2(g) + 5/2O2(g) ∆H = 1299.6 kJ --------------------------------------------- 2C(s) + H2(g) › C2H2(g) ∆H = 226.8 kJ

  48. Example III: Calculate the ∆H for the reaction: • NO(g)+ O3(g)›NO2(g) + O2(g) ∆H=-198.9 kJ • O(g) › 1/2O2(g) ∆H = -247.5 kJ • 3/2O2(g) › O3(g) ∆H = +142.3 kJ --------------------------------------------- NO(g) + O(g) › NO2(g) ∆H = -304.1 kJ

  49. Example IV Calculate the ∆H for: • SO3(g) › SO2(g) + 1/2O2(g) ∆H = +98 kJ • S(s) + 3/2O2(g) › SO3(g) ∆H = -395 kJ ---------------------------------------------- S(s) + O2(g) › SO2(g) ∆H = -297 kJ

  50. Objectives #14-15 Hess’s Law Part II Using Enthalpies of Formation to Calculate Enthalpies of Reaction *standard enthalpies of formation can also be used to determine the ∆H for a reaction *these enthalpy values represent the energy change that results when 1 mole of a compound is formed from its elements in their standard states at 25oC; when ∆H values are determined in this manner, they are referred to as standard enthalpy change and are denoted by the symbol ∆H0

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