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Thermodynamics Standard 7

Thermodynamics Standard 7. Chemistry. Ms. Siddall. Standard 7a: ‘heat flow’. Chemical Thermodynamics = the movement of heat in a chemical reaction. Temperature = a measure of the average kinetic energy of particle motion

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Thermodynamics Standard 7

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  1. ThermodynamicsStandard 7 Chemistry. Ms. Siddall.

  2. Standard 7a: ‘heat flow’ Chemical Thermodynamics = the movement of heat in a chemical reaction. • Temperature= a measure of the average kinetic energy of particle motion • Heat = The transfer of energy from a hotter object to a colder object (sometimes called ‘heat flow’) • temperature measures energy • Heat measures energy transfer

  3. Summary 1 • Describe the difference between heat and temperature

  4. Energy transfer • Particle vibrations increase when a particle gains energy • Vibrations are transferred to surrounding particles Describe how energy is transferred between atoms. Summary 2

  5. Identifying heat transfer: • System: experiences a change • Surroundings: causes a change • e.x. hot coffee (system) cools because it transfers heat to the air, the cup, the table & the whole universe! (surroundings)

  6. Summary 3 Consider an ice cube dropped into a glass of warm water. • Ice cube = system • Water = surroundings • Does heat flow into the system or out of the system? • What is gaining energy (system or surroundings)?

  7. Standard 7b: exothermic & endothermic process Summary 4 Endothermic Process: A process in which energy is absorbed. • Example: Water boiling • H2O(l) + heat  H2O(g) • In an endothermic process heat is a reactant. product reactants In an endothermic process which has more energy; reactants or products?

  8. Summary 5 Exothermic Process: A process in which energy is released. • Example: • A fire • 3C + 2O2 heat + 2CO + CO2 • In an exothermic process heat is a product products reactants In an exothermic process which has more energy, reactants or products?

  9. Energy diagram Summary 6 • Draw an energy diagram for the campfire reaction. • Show reactants and products. • Draw only one arrow from reactants to products and label the arrow (endothermic or exothermic) H2O(g) exothermic endothermic Increasing energy H2O(l)

  10. Transition State energy diagram activation energy = energy needed to form transition state (activated complex) Transition state Energy released when products form reactants Total energy released during reaction energy products

  11. Summary 7 Transition State: An intermediate state that can occur during a reaction • Also called an ‘activated complex’ • An exothermic reaction is not always spontaneous because energy is needed to form a transition state. • e.x. a spark is needed to start a fire • Draw a transition state energy diagram for an endothermic reaction

  12. Summary 8 Measuring heat flow. • Energy is measured in joules (J) or calories (cal) • Example: 334J of energy are needed to melt 1g of ice. • 1 calorie (c) = 4.18J • 1 food calorie (C) = 1000 calories = 4180J If your body burns about 2,000 food calories a day, approximately how many joules of energy is that?

  13. Energy released = exothermic KJ = kilojoules = 1000J Showing a change in energy: • S(s) + O2(g) SO2(g) + energy • S(s) + O2(g)  SO2(g) + 297KJ • S(s) + O2(g)  SO2(g) ∆H = -297KJ -∆H = exothermic +∆H = endothermic ∆H = change in enthalpy Enthalpy = energy/heat

  14. N2(g) + 2O2(g) 2NO2(g) ∆H = + 68KJ • N2(g) + 2O2(g) + 68KJ  2NO2(g) Endothermic reaction Energy is a reactant

  15. Summary 9 • Write an equation to show water melting. Use ∆H to show energy. (it takes 5.9kJ of energy to melt ice) • Is ∆H negative or positive? Why?

  16. Standard 7c: energy of phase change Phase Change: The physical state of a compound changes • The same compound is observed before and after the change • Example: ice melting H2O(s) H20(l) • There is no temperature change. • Energy is used to overcome intermolecular attractions.

  17. Summary 10 • Is the example of ice melting an endothermic process or an exothermic process?

  18. Physical state gas evaporating break hydrogen bonds Condensing endothermic liquid exothermic Energy released intermolecular attractions take over melting freezing break lattice structure solid

  19. Summary 11 • In which phase do the molecules have the most energy? (solid, liquid, or gas) • Is the process of condensing endothermic or exothermic? • Is the process of vaporization endothermic or exothermic?

  20. Standard 7d: solving problems Freezing/boiling point graph for water. Energy absorbed = no temp change = physical change boiling ΔHvap 110 100 steam meltingΔHfus Water (CH2O(l)) Temperature (°C) Energy absorbed = Change in temperature = Change in K.E. 0 ice -10 energy

  21. Summary 12 • Which two sections of the graph show no temperature change. • Why is there no temperature change in these sections?

  22. Standard 7d: solving problems Latent Heat of fusion.(latent heat = hidden heat) ΔHfus = The energy released when 1g of a substance is frozen OR the energy needed when 1g of a substance is melted. ΔHfus= enthalpy of fusion (J/g) • Fusion = freezing (liquid  solid) • also used for melting (solid  liquid) Summary 13: What does ‘fusion’ mean?

  23. Example: freezing water • How much energy is released when 10g water freezes? (ΔHfusH2O = 334J/g) 10g H2O(s) 334J = 3340J = 3.34kJ 1g H2O(s) Summary 14 How much energy is needed to melt 100g of water? (show calculation)

  24. Latent Heat of vaporization ΔHvap = The energy needed when 1g of a substance is evaporated OR the energy released when 1g of a substance is condensed. ΔHvap= enthalpy of vaporization (J/g) • vaporization = evaporating (liquid  gas) • also used for condensing (gas liquid)

  25. Summary 15 • What does vaporization mean? • What does condensation mean?

  26. Example: Boiling water Summary 16 • How much energy is needed to boil 10g water? (ΔHvapH2O = 2260J/g) 10g H2O(l) 2260J = 22600J = 22.6kJ = J 1g H2O(l) How much energy is released when 100g of water vapor is condensed? (show work)

  27. Heat Capacity. • C = specific heat capacity • The amount of heat energy needed to raise the temperature of 1g of a substance by 1°C • Example: CH2O(l) = 4.18J/g°C • It takes 4.18J of energy to raise the temperature of 1g of water by 1°C • 1 calorie = 4.18J

  28. Summary 17 • How much energy is needed to raise the temperature of 1g of water by 1°C? (give your answer in joules and calories)

  29. Example. • How much energy is needed to raise the temperature of 5g water from 22°C to 24°C? (CH2O(l) = 4.18J/g°C) 5g H2O(l) 4.18J H2O(l) 2°C = 41.8J = J g °C

  30. Summary 18 • How much energy is released when 10g water cools from 40°C to 30°C? 10g H2O(l) 4.18J H2O(l) 10°C = 418.J = J g °C

  31. Measuring specific heat capacity for different compounds Thermometer: Measures temperature change for water ‘q’=energy released by metal =energy absorbed by water Unknown compound: heated to 100°C and placed in the cold water H2O

  32. Summary 19 • How much energy (q) is released by a metal if the temperature of 100g of water in the calorimeter rises from 20°C to 30°C?

  33. Measuring the heat of a reaction: ‘q’ (q = energy released or absorbed by water) • Thermometer: measures temperature change for water • T  = exothermic • T  = endothermic Reaction chamber: 3H2 + N2 NH3 heat of reaction is absorbed by water 100g H2O

  34. Example: 10g NH3 are produced in the above reaction. The temperature rises from 20.0°C to 30.0°C. • Calculate ‘q’ (energy) for the reaction. • Is the reaction endothermic or exothermic? • Calculate ΔH (J/g) for this reaction • Calculate ΔH (mol/g) for this reaction

  35. Kinetic energy distribution diagram

  36. Kinetic energy distribution diagram • T1 = low temperature = low energy • T2 = higher temperature = higher energy • Emin = minimum energy needed to escape. • More T2 particles have Emin • Less T1 particles have Emin

  37. Summary 20 • Explain why more particles evaporate from a cup of hot water compared to a cup of cold water.

  38. Standard 7e: Apply Hess’s Law to calculate enthalpy change in a reaction • Hess’s Law: If a series of reactions are added together the enthalpy change for the net reaction will be the sum of the enthalpy changes for the individual steps. • E.x. N2(g) + 2O2(g) 2NO2(g) • N2(g) + O2(g) 2NO(g) ΔH = +181kJ • 2NO(g) + O2(g) 2NO2(g) ΔH = -113kJ • Find the sum of the 2 equations…

  39. N2(g) + O2(g) 2NO(g) ΔH = +181kJ • 2NO(g) + O2(g) 2NO2(g) ΔH = -113kJ N2(g) + O2(g) + 2NO(g) + O2(g) 2NO(g) +2NO2(g) N2(g) + 2O2(g)  2NO2(g) ΔH = notes: • You can reverse reactions (change sign of ΔH) • You can multiply or divide equations (do same to ΔH) + (-113kJ) = +68kJ +181kJ

  40. Hess summary • Complete questions 66, 74, 81 & 84 on page 536 & 537

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