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Unit 3-1 thermodynamics

Chapter 5. Unit 3-1 thermodynamics. Classification of energy. Energy : The capacity to do work or produce heat There are two types of energy: Potential E: Energy due to position or composition. For exmple, electron energy level (n).

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Unit 3-1 thermodynamics

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  1. Chapter 5 Unit 3-1 thermodynamics

  2. Classification of energy • Energy: The capacity to do work or produce heat • There are two types of energy: • Potential E: Energy due to position or composition. For exmple, electron energy level (n). • Kinetic E: Energy due to the motion of the object. For example, heat.

  3. System and Surroundings • The system in chemistry includes the molecules going through reactions, we want to study (here, the H2and O2molecules). • The surroundings are everything else (including the cylinder and piston). • There is no matter exchange between the system and surroundings (law of conservation of mass), only energy exchange.

  4. State Functions depend ONLY on the present state of the system ENERGYIS A STATE FUNCTION • A person standing at the top of mountain has the same potential energy whether they got there by hiking up, or by falling down from a plane.  • States of the system : T, P, V WORKIS NOT A STATE FUNCTION

  5. The first law of thermodynamics • The First Law of Thermodynamics: The total energy content of the universe is constant. • Law of Conservation of Energy: Energy can neither be created nor destroyed, but can be converted between forms. • Potential E can be changed into kinetic E or vice versa.

  6. Internal Energy The internal energy of a system is the sum of all kinetic and potential energies of all mthe system; we call it E. Use Fig. 5.5

  7. Internal Energy By definition, the change in internal energy, E, is the final energy of the system minus the initial energy of the system: E = Efinal−Einitial Use Fig. 5.5

  8. Changes in Internal Energy • If E > 0, Efinal > Einitial • Therefore, the system absorbed energy from the surroundings. • This energy change is called endergonic.

  9. Changes in Internal Energy • If E < 0, Efinal < Einitial • Therefore, the system released energy to the surroundings. • This energy change is called exergonic.

  10. Changes in Internal Energy • When energy is exchanged between the system and the surroundings, it is exchanged as either heat (q) or work (w). • That is, E = q + w.

  11. E = q + w E(change in internal energy of a system) = Efinal – E initial q= heat flowing into or out of the system -qif energy is leaving tothe surroundings +qif energy is entering fromthe surroundings w= work done by, or on, the system -wif work is done bythe systemonthe surroundings +wif work is done onthe system bythe surroundings

  12. E = q + w • E(change in internal energy of a system) = Efinal – E initial • Work: Energy used to cause an object that has mass to move. • w = F d = -P ΔV where w is work, F is the force, and d is the distance over which the force is exerted. • Heat: Energy used to cause the temperature of an object to rise. • q = Heat lost or gained • s = Specific Heat Capacity • T = Temperature change

  13. Work, Pressure, and Volume Compression Expansion +V(increase) -V(decrease) -wresults +wresults Esystemdecreases Esystemincreases Work has been done on the system by the surroundings Work has been done by the system on the surroundings

  14. First law of thermodynamics: Page 177 5.3 (a) and (c). Page 178 5.15 (a) and (b), 5.19 (c) Homework

  15. Enthalpy (H) • If a process takes place at constant pressure (as the majority of processes we study do) and the only work done is this pressure-volume work • Enthalpy is the internal energy plus the product of pressure and volume: H = E + PV

  16. Enthalpy • When the system changes at constant pressure (as the majority of processes we study do) , the change in enthalpy, H, is H = E + PV • Since E = q + w and w = −PV, we can substitute these into the enthalpy expression: H = E + PV H = (q+w) −w H = q • So, at constant pressure, the change in enthalpy of system is the heat gained or lost.

  17. Endothermicity and Exothermicity • A process is endothermic, Q >0 then, when H is positive • A process is exothermic, Q<0, when H is negative.

  18. Enthalpies of Reaction (H) The change of H, H, is called the enthalpy of reaction, or the heat of reaction. H = Hproducts−Hreactants

  19. The Truth about Enthalpy • Enthalpy is an extensive property. • H for a reaction in the forward direction is equal in size, but opposite in sign, to H for the reverse reaction. • H for a reaction depends on the state of the products and the state of the reactants.

  20. Enthalpy: Page 179 5.21(b) and (c), 5.25, 5.27 Homework

  21. Calorimetry Since we cannot know the exact enthalpy of the reactants and products, we measure H through calorimetry, the measurement of heat flow, usually by the change in temperature of a known quantity of water in a calorimeter.

  22. Calculations Involving Specific Heat s = Specific Heat Capacity s· m = heat capacity T= Temperature change

  23. The amount of heat required to raise the temperature of one gram of substance by one degree Celsius. Specific Heat

  24. Constant Pressure Calorimetry • For reaction in aqueous solution in a simple calorimeter under constant P, ΔHrxn = q • Q is absorbed by surround water. • Mass of surrounding = mass of calorimeter + mass of rxn • S of water = 4.184 J/g·K, we can measure H for the reaction with this equation: q = m  s  T

  25. Bomb Calorimetry • Reactions can be carried out in a sealed “bomb” • The volume in the bomb is constant. We still assume ΔHrxn = q • But mass of surrounding = mass of calorimeter. The mass of rxn bomb is counted as system mass.

  26. Calorimetry: Page 179 5.39, Page 180 5.41, 5.43, 5.45 Homework

  27. Hess’s law

  28. Enthalpies of Formation (Hf) • An enthalpy of formation, Hf, is defined as the enthalpy change for the reaction in which a compound is made from its constituent elements in their elemental forms. • Na (s) + ½ Cl2 (g)  NaCl(s) • Ag+(aq) + Cl-(aq) AgCl (s)

  29. Standard Enthalpies of Formation  Standard enthalpies of formation, Hf, are measured under standard conditions (298K and 1.00 atm pressure).

  30. Hess’s Law “In going from a particular set of reactants to a particular set of products, the change in enthalpy is the same whether the reaction takes place in one step or a series of steps.”

  31. Calculation of Heat of Reaction based on Hf Hrxn = Hf(products) - Hf(reactants) Hrxn = [-393.50kJ + 2(-285.83kJ)] – [-74.80kJ] Hrxn = -890.36 kJ

  32. Calculating the ΔHorxn without using Hfo If you know the ΔHorxn of the broken down rxns! A different Hess’s Law Problem

  33. Hess’s Law Example Problem CH4 C + 2H2 +74.80 kJ Step #1: CH4 must appear on the reactant side, so we reverse reaction #1 and change the sign on H.

  34. Hess’s Law Example Problem CH4 C + 2H2 +74.80 kJ C + O2 CO2 -393.50 kJ Step #2: Keep reaction #2 unchanged, because CO2 belongs on the product side

  35. Hess’s Law Example Problem CH4 C + 2H2 +74.80 kJ C + O2 CO2 -393.50 kJ 2H2 + O2 2 H2O -571.66 kJ Step #3: Multiply reaction #2 by 2

  36. Hess’s Law Example Problem CH4 C + 2H2 +74.80 kJ C + O2 CO2 -393.50 kJ 2H2 + O2 2 H2O -571.66 kJ CH4 + 2O2 CO2 + 2H2O -890.36 kJ Step #4: Sum up reaction and H

  37. Page 180 5.51, 5.53 Page 181 5.61(c), 5.65, 5.67 (a)&(b). Homework

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