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Ch 7 Energy and Chemical Change

Ch 7 Energy and Chemical Change. Brady & Senese, 5th Ed. Index. 6.1 An object has energy if it is capable of doing work 6.2 Internal energy is the total energy of an object’s molecules 6.3 Heat can be determined by measuring temperature changes

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Ch 7 Energy and Chemical Change

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  1. Ch 7 Energy and Chemical Change Brady & Senese, 5th Ed.

  2. Index 6.1 An object has energy if it is capable of doing work 6.2 Internal energy is the total energy of an object’s molecules 6.3 Heat can be determined by measuring temperature changes 6.4 Energy is absorbed or released during most chemical reactions 6.5 Heats of reaction are measured at constant volume or constant pressure 6.6 Thermochemical equations are chemical equations that quantitatively include heat 6.7 Thermochemical equations can be combined because enthalpy is a state function 6.8 Tabulated standard heats of reaction can be used to predict any heat of reaction using Hess’s law

  3. Energy Is The Ability To Do Work • Energy is the ability to do work (move mass over a distance) or transfer heat • Types: kinetic and potential • kinetic: the energy of motion • potential: the stored energy in matter • Internal energy (E): the sum of the kinetic and potential energy for each particle in the system 6.1 An object has energy if it is capable of doing work

  4. Kinetic Energy: The Energy Of Motion • KE=½mv2 • Exists in two major forms: • Heat is random particle motion • Work is directed motion • Heat can be transferred by moving particles • Collision of fast particles with slower particles causes the slow particle to speed up while the fast molecule slows 6.1 An object has energy if it is capable of doing work

  5. Potential Energy Depends on Position • Potential energy inc when: objects that attract move apart, or objects that repel move toward each other • Stronger bonds tend to be shorter (opposite charges closer together) • Chemical potential energy is released when reactants have weaker bonds than products formed by a reaction • Stored energy can be converted to kinetic energy 6.1 An object has energy if it is capable of doing work

  6. Energy cannot be created or destroyed but can be transformed from one form of energy to another Also known as the first law of thermodynamics How does water falling over a waterfall demonstrate this law? Energy crisis and water powered cars??? Law Of Conservation Of Energy 6.1 An object has energy if it is capable of doing work

  7. Heat And Temperature Are Not The Same • The temperature of an object is proportional to the average kinetic energy of its particles—the higher the average kinetic energy, the higher the temperature • Heatis energy (also called thermal energy) transferred between objects caused by differences in their temperatures until they reach thermal equilibrium 6.1 An object has energy if it is capable of doing work

  8. What Is Temperature? • Temperature (T) is proportional to the average kinetic energy of all particle units: °C, °F, K • Avg KE= ½ mvavg2 • At a high temperature, most molecules are moving at higher average 6.2 Internal energy is the total energy of an object’s molecules

  9. SI unit is the Joule, J J = kg·m2/s2 If the calculated value is greater than 1000 J, use the kJ British unit is the calorie, cal Energy to change the temp of 1g of water by 1°C 1 cal = 4.184 J (exact) Nutritional unit is the Calorie (note capital c), which is one kilocalorie 1 Cal = 1 kcal = 4.184 kJ = 1000 cal Units Of Energy 6.1 An object has energy if it is capable of doing work

  10. Heat Transfer, q • Heat (q or ∆H) = Heat absorbed or released during temperature, physical or chemical change • units: J, cal • A calorie is the amount of energy needed to raise the temperature of 1.00 g water by 1°C • A metal spoon at 25°C is placed in boiling water. What happens? 6.3 Heat can be determined by measuring temperature changes

  11. Surroundings / System / Universe • System: the reaction or area under study • Surroundings: the rest of the universe • Open systemscan gain or lose mass and energy across their boundaries • i.e. the human body • Closed systemscan absorb or release energy, but not mass, across the boundary • i.e. a light bulb • Isolated systemscannot exchange matter or energy with their surroundings Adiabatic • i.e. a stoppered Thermos bottle • No truly isolated systems exist 6.3 Heat can be determined by measuring temperature changes

  12. Internal Energy is Conserved • 1st Law of Thermodynamics: For an isolated system the internal energy (E) is constant because no energy may enter or leave the system • E = work + heat • Internal energy is difficult to measure, so we measure changes in energy Δ E = Ef - Ei = 0 Δ E = Eproduct - Ereactant = 0 6.2 Internal energy is the total energy of an object’s molecules

  13. The Sign Convention Endothermic systems require energy to be added to the system, thus the q is (+) Exothermic reactions release energy to the surroundings. Their q is (-) Energy changes are measured from the point of view of the system 6.3 Heat can be determined by measuring temperature changes

  14. Heat Capacity and Transfer • Heat capacity (C or Cp)- the (extensive) ability of an object with constant mass to absorb heat. • a/k/a calorimeter constant • Varies with the sample mass and the identity of the substance • Common Units: J/°C or cal/°C • q=C×Δt • q= heat transferred • C= heat capacity of object • Δt= Change in Temperature (tfinal-tinitial) 6.3 Heat can be determined by measuring temperature changes

  15. Learning Check: A cup of water is used in an experiment. Its heat capacity is known to be 720 J/ °C. How much heat will it absorb if the experimental temperature changed from 19.2 °C to 23.5 °C? 6.3 Heat can be determined by measuring temperature changes

  16. Heat Transfer and Specific Heat • Specific heat (s or Csp)- The (intensive) ability of a substance to store heat. • C = amount×s • Units : J/g·°C; J/g · K; J/mol · K; cal/g·°C; cal/g · K; cal/mol · K • q=amount×Δt×s • q= heat transferred • amount= amount of object (in g or mol) • Δt= Change in Temperature (tfinal-tinitial) 6.3 Heat can be determined by measuring temperature changes

  17. Specific Heats Substances with high specific heats resist temperature changes Note that water has a very high specific heat (this is why coastal temperatures are different from inland temperatures) 6.3 Heat can be determined by measuring temperature changes

  18. Learning Check: Calculate the specific heat of a metal if it takes 235 J to raise the temperature of a 32.91 g sample by 2.53°C. 6.3 Heat can be determined by measuring temperature changes

  19. The First Law Of Thermodynamics Explains Heat Transfer If we monitor the heat transfers (q) of all materials involved and all work processes, we can predict that their sum will be zero By monitoring the surroundings, we can predict what is happening to our system (for example, energy lost by our system = energy gained by the surroundings) Heat transfers until thermal equilibrium, thus the final temperature is the same for all materials 6.3 Heat can be determined by measuring temperature changes

  20. Learning Check: A 43.29 g sample of solid is transferred from boiling water (T=99.8°C) to 152 g water at 22.5°C in a coffee cup. The Twater rose to 24.3°C. Calculate the specific heat of the solid. qsample+ qwater + qcup=0 qcup is neglected in problem qsample = - qwater 6.3 Heat can be determined by measuring temperature changes

  21. Chemical Potential Energy • Chemical bond: net attractive forces that bind atomic nuclei and electrons together • Exothermic reactions form stronger bonds in the product than in the reactant and release energy (-E) • Endothermic reactions break stronger bonds than they make and require energy (+E) 6.4 Energy is absorbed or released during most chemical reactions

  22. Work and Pistons Pressure = force/area If the container volume changes, the pressure changes Work = -P×ΔV units: L • atm 1 L • atm = 101 J In expansion, ΔV>0, and is exothermic Work is done by the system in expansion 6.5 Heats of reaction are measured at constant volume or constant pressure

  23. How does work relate to reactions? Work = Force · Distance is often due to the expansion or contraction of a system due to changing moles of gas. w = -Patm×ΔV The deployment of an airbag is one example of this process.2 NaN3(s) →2 Na(s) + 3 N2(g) 0 moles of gas→ 3 moles of gas 6.5 Heats of reaction are measured at constant volume or constant pressure

  24. Learning Check: P-V work Ethyl chloride is prepared by reaction of ethylene with HCl. How much PV work (in J) is done if 89.5g ethylene and 125g of HCl are allowed to react at atmospheric pressure and the volume change is -71.5 L? Calculate the work (in kilojoules) done during a synthesis of ammonia in which the volume contracts from 8.6 L to 4.3L at a constant external pressure of 44 atm w = - 1.00 atm × -71.5L =71.5 L·atm w = 7240 J w = - 44atm × (4.3-8.6)L =189.2L·atm w = 19 kJ 1 L·atm = 101.325 J 6.5 Heats of reaction are measured at constant volume or constant pressure

  25. Energy Can Be Transferred as Heat and Work ΔE= q + w internal energy changes are state functions 6.5 Heats of reaction are measured at constant volume or constant pressure

  26. Calorimetry Is Used To Measure Heats Of Reaction Heat of reaction: the amount of heat absorbed or released in a chemical reaction Calorimeter: an apparatus used to measure temperature changes in materials surrounding a reaction that result from a chemical reaction From the temperature changes we can calculate the heat of the reaction, q qv heat measured under constant volume conditions qp heat measured under constant pressure conditions 6.5 Heats of reaction are measured at constant volume or constant pressure

  27. Internal Energy Is Measured With A Bomb Calorimeter Used for reactions in which there is change in the number of moles of gas present Measures qv Immovable walls mean that work is zero ΔE = qv 6.5 Heats of reaction are measured at constant volume or constant pressure

  28. Learning Check: Bomb Calorimeter 500. mg naphthalene (C10H8) is combusted in a bomb calorimeter containing 1000. g of water. The temperature of the water increases from 20.00°C to 24.37°C. The calorimeter constant is 420 J/°C. What is the change in internal energy for the reaction? swater = 4.184 J/g°C qv reaction + qwater + qcal = 0 by the first law qwater= 1000. g ×(24.37-20.00)°C× 4.184 J/g°C qcal= 420 J/°C ×(24.37-20.00)°C qv = ΔE = -2.01×104 J 6.5 Heats of reaction are measured at constant volume or constant pressure

  29. Enthalpy Of Combustion • When one mole of a fuel substance is reacted with elemental oxygen, a combustion reaction can be written whose enthalpy is ΔΗc˚ • Is always negative • Learning Check: What is the equation associated with the enthalpy of combustion of C6H12O6(s)? C6H12O6(s) + 9O2(g)→6CO2(g) + 6H2O(l) 6.5 Heats of reaction are measured at constant volume or constant pressure

  30. Your Turn! 252 mg of benzoic acid, C6H5CO2H,is combusted in a bomb calorimeter containing 814 g water at 20.00ºC. The reaction increases the temperature of the water to 21.70 deg C. What is the internal energy released by the process? • -711 J • -2.85 J • +711 J • +2.85 J • None of these swater = 4.184 J/g°C 6.5 Heats of reaction are measured at constant volume or constant pressure

  31. Enthalpy Measured in a Coffee Cup Calorimeter when no change in moles of gas is expected, we may use a coffee cup calorimeter the open system allows the pressure to remain constant thus we measure qp ΔE=q + w since there is no change in the moles of gas present, there is no work thus we also are measuring ΔE 6.5 Heats of reaction are measured at constant volume or constant pressure

  32. Learning Check: Coffee Cup Calorimetery When 50.0 mL of .987 M H2SO4 is added to 50.0 mL of 1.00 M NaOH at 25.0 °C in a calorimeter, the temperature of the aqueous solution increases to 31.7 °C. Calculate heat for the reaction per mole of limiting reactant. Assume that the specific heat of the solution is 4.18 J/g°C, the density is 1.00 g/mL, and that the calorimeter itself absorbs a negligible amount of heat moles H2SO4 = 0.04935 moles NaOH = 0.0500, is limiting qvrxn = - qsoln H2SO4(aq) + 2 NaOH(aq)→ 2H2O(l) + Na2SO4(aq) qsoln = 100.gsoln×(31.7-25.0)°C×4.18J/g°C qvrxn = -5.6×104 J/mol qvrxn = -2.8×103 J 6.5 Heats of reaction are measured at constant volume or constant pressure

  33. Your Turn! A sample of 50.00mL of 0.125M HCl at 22.36 ºC is added to a 50.00mL of 0.125M Ca(OH)2 at 22.36 ºC. The calorimeter constant was 72 J/ ºC. The temperature of the solution (s=4.184 J/g ºC, d=1.00 g/mL) climbed to 23.30 ºC. Which of the following is not true? • qcal=68 J • qsolution= 393 J • qrxn = 461 J • qrxn = -461 J • None of these 6.5 Heats of reaction are measured at constant volume or constant pressure

  34. Calorimetry Overview The equipment used depends on the reaction type. If there will be no change in the moles of gas (work = 0) we may use a coffee-cup calorimeter or a closed system. Under these circumstances, we measure qp. If there is a large change in the moles of gas, we use a bomb calorimeter to measure qv. The bomb has rigid walls preventing a change in volume (work = 0) 6.5 Heats of reaction are measured at constant volume or constant pressure

  35. Thermochemical Equations (Heat Stoichiometry) Relate the energy of a reaction to the quantities involved Must be balanced, but may use fractional coefficients quantities are presumed to be in moles 6.6 Thermochemical equations are chemical equations that quantitatively include heat

  36. Learning Check: • 2C2H2(g) + 5 O2(g) → 4 CO2(g) + 2 H2O(g) ΔE = -2511 kJ • The reactants (acetylene and oxygen) have 2511 kJ more energy than the products • How many KJ are released for 1.000 mol C2H2? -1,256 kJ 6.6 Thermochemical equations are chemical equations that quantitatively include heat

  37. 6CO2(g) + 6H2O(l) → C6H12O6(s) + 6O2(g) Δ H = +2816 kJ How many kJ are required for 44 g CO2 (MM=44.00986 g/mol)? If 100. kJ are provided, what mass of CO2 can be converted to glucose? Learning Check: +470 kJ 9.38 g 6.6 Thermochemical equations are chemical equations that quantitatively include heat

  38. Learning Check: Calorimetry of Chemical Reactions The meals-ready-to-eat (MRE) in the military can be heated on a flameless heater. Assume the reaction in the heater is Mg(s) + 2H2O(l) → Mg(OH)2(s) + H2(g) ΔH = -353kJ What quantity of magnesium is needed to supply the heat required to warm 25 mL of water from 25 to 85 °C? Specific heat of water = 4.184 J/g °C. Assume the density of the solution is the same as for water at 25°C, 1.00 g/mL masssoln = 25 mL×1.00g/mL= 25 g qsoln = 25 g ×(85-25) °C×4.184 J/g °C= 6.276 ×103 J massMg =0.43 g 6.6 Thermochemical equations are chemical equations that quantitatively include heat

  39. Your Turn! Consider the thermite reaction. The reaction is initiated by the heat released from a fuse or reaction The enthalpy change is –852 kJ/mol Fe2O3 at 298K. 2 Al(s) + Fe2O3(s)→2Fe(s) + Al2O3(s) What mass of Fe (MM=55.847 g/mol) is made when 500. kJ are released? 65.5 g 0.587 g 32.8 g None of these 6.6 Thermochemical equations are chemical equations that quantitatively include heat

  40. Learning Check: Ethyl Chloride Reaction Revisited Ethyl chloride is prepared by reaction of ethylene with HCl: C2H4(g) + HCl(g)→ C2H5Cl(g) ΔH° = -72.3kJ What is the value of ΔE if 89.5 g ethylene and 125 g of HCl are allowed to react at atmospheric pressure and the volume change is -71.5 L? mol HCl: 3.4283 mol C2H4: 3.1903, is Limiting ΔHrxn =-230.66 kJ w = -1atm × -71.5L =71.5 L·atm = 7.24 kJ ΔE= -223 kJ MM ethylene=28.054 g/mol; MM HCl=36.4611 g/mol 6.6 Thermochemical equations are chemical equations that quantitatively include heat

  41. Enthalpy Diagram 6.7 Thermochemical equations can be combined because enthalpy is a state function

  42. Hess’s Law The overall enthalpy change for a reaction is equal to the sum of the enthalpy changes for individual steps in the reaction For example: Fe2O3(s ) →2 Fe + 3/2 O2ΔH= +822.2 kJ 3/2 O2 + 2Al(s) →Al2O3(s) ΔH= -1647 kJ Fe2O3(s) + 2Al(s) →Al2O3(s) + 2Fe ΔH= -852 kJ 6.7 Thermochemical equations can be combined because enthalpy is a state function

  43. Rules for Adding Thermochemical Reactions • When an equation is reversed—written in the opposite direction—the sign of H must also be reversed. • Formulas canceled from both sides of an equation must be for the substance in identical physical states. • If all the coefficients of an equation are multiplied or divided by the same factor, the value of H must likewise be multiplied or divided by that factor. 6.7 Thermochemical equations can be combined because enthalpy is a state function

  44. Strategy for Adding Reactions Together: • Choose the most complex compound in the equation (1) • Choose the equation (2 or 3 or…) that contains the compound • Write this equation down so that the compound is on the appropriate side of the equation and has an appropriate coefficient for our reaction • Look for the next most complex compound 6.7 Thermochemical equations can be combined because enthalpy is a state function

  45. Hess’s Law (Cont.) • Choose an equation that allows you to cancel intermediates and multiply by an appropriate coefficient • Add the reactions together and cancel like terms • Add the energies together, modifying the enthalpy values in the same way that you modified the equation • if you reversed an equation, change the sign on the enthalpy • if you doubled an equation, double the energy 6.7 Thermochemical equations can be combined because enthalpy is a state function

  46. Learning Check: How can we calculate the enthalpy change for the reaction 2 H2(g) + N2(g)→ N2H4(g) using these equations ? N2H4(g) + H2(g)→ 2 NH3(g) ΔH° = -187.6 kJ 3H2(g) + N2(g)→ 2NH3(g) ΔH° = -92.2 kJ Reverse the first reaction (and change sign) 2 NH3(g)→ N2H4(g) + H2(g) ΔH° = +187.6 kJ Add the second reaction (and add the enthalpy) 3H2(g) + N2(g)→ 2NH3(g) ΔH° = -92.2 kJ 2 NH3(g)+ 3H2(g) + N2(g)→ N2H4(g) + H2(g)+2NH3(g) 2 H2(g) + N2(g)→ N2H4(g) (187.6-92.2)= +95.4 kJ 2 6.7 Thermochemical equations can be combined because enthalpy is a state function

  47. Learning Check: Calculate ΔH for 2C(s) + H2(g)→ C2H2(g) using: 2 C2H2(g) + 5O2(g)→ 4 CO2(g) + 2H2O(l) ΔH° = -2599.2 kJ C(s) + O2(g)→ CO2(g) ΔH° = -393.5 kJ H2O(l)→ H2(g) + ½ O2(g) ΔH° = +285.9 kJ • -½(2 C2H2(g) + 5O2(g)→ 4 CO2(g) + 2H2O(l) ΔH° = -2599.2) • 2 CO2(g) + H2O(l)→ C2H2(g) + 2-½ O2(g) ΔH° = +1299.6 • 2(C(s) + O2(g)→ CO2(g) ΔH° = -393.5 kJ) • 2C(s)+ 2O2(g)→ 2CO2(g) ΔH° = -787.0 kJ) • -1(H2O(l)→ H2(g) + ½ O2(g) ΔH° = +285.9 kJ) • H2(g) + ½ O2(g)→H2O(l) ΔH° = -285.9 kJ • 2C(s) + H2(g)→ C2H2(g) ΔH° = +226.7 kJ 6.7 Thermochemical equations can be combined because enthalpy is a state function

  48. Your Turn! What is the energy of the following process: 6A + 9B + 3D + 1 F→2 G Given that: • C → A + 2B ∆H=20.2 kJ/mol • 2C + D →E + B ∆H=30.1 kJ/mol • 3E + F →2G ∆H=-80.1 kJ/mol • 70.6 kJ • -29.8 kJ • 111.0 kJ • -111.0 kJ (20.2×-6)+(30.1×3)+(-80.1) kJ =-111.0 6.7 Thermochemical equations can be combined because enthalpy is a state function

  49. State matters! • C3H8(g) + 5 O2(g)→ 3 CO2(g) + 4 H2O(g) • ΔH= -2043 kJ • C3H8(g) + 5 O2(g)→ 3 CO2(g) + 4 H2O(l) • ΔH= -2219 kJ • note that there is a difference in energy because the states do not match • If H2O(l) → H2O(g) ΔH = 44 kJ/mol • 4H2O(l) → 4H2O(g) ΔH = 176 kJ/mol • -2219 +176 kJ = -2043 kJ 6.8 Tabulated standard heats of reaction can be used to predict any heat of reaction using Hess’s law

  50. Standard State Most stable form of the pure substance at • 1 atm pressure • Stated temperature. If temperature is not specified, assume 25 °C • Solutions are 1M in concentration. • Measurements made under standard state conditions have the ° mark: ΔH° • Most ΔH values are given for the most stable form of the compound or element. 6.8 Tabulated standard heats of reaction can be used to predict any heat of reaction using Hess’s law

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