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Chapter 19

Chapter 19. Thermodynamics . Chemical Systems. A system refers to any part of the universe under observation. Everything else is called the surroundings. An open system can transfer energy and matter to the surroundings.

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Chapter 19

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  1. Chapter 19 Thermodynamics

  2. Chemical Systems • A system refers to any part of the universe under observation. • Everything else is called the surroundings. • An open systemcan transfer energy and matter to the surroundings. • A closed system is one where energy can be transferred to the surroundings but matter cannot.

  3. State Functions • State functions describe an aspect of a chemical system. • The value of a state functions does not depend on how the given state was obtained. • A system heated from 215 K to 228 is the same as a system that was cooled from 485 K to 228 K. • The thermodynamic state functions we will see are internal energy (E), enthalpy (H), entropy (S), and Gibbs free energy (G).

  4. Standard State • The thermodynamic quantities, ΔH, ΔS, ΔG, and ΔE are all extensive properties. • Meaning they change as the amount of the sample changes. • A system is in the standard state when the pressure is 1 atmosphere and the temperature is 25 oC. • When thermodynamic quantities are determined at standard state they are written with a O as a superscript. ΔHo, ΔSo, ΔGo, and ΔEo

  5. Exo- and Endo- • The prefix Exo, as in exothermic indicates that energy is being lost from the system to the surroundings. • Mathematically this corresponds to a negative sign. • The prefix Endo-, as in endothermic, indicates that energy is being gained by the system from the surroundings. • Mathematically this corresponds to a positive sign.

  6. Types of Energy • Energy takes many forms: • Heat, Light, Chemical, Nuclear, Electrical… • The law of conservation of energy states that energy cannot be created or destroyed. • Energy can also be categorized as either kinetic energy (KE) or potential energy (PE). • Kinetic energy is the energy that matter possesses because of its motion. • KE = ½ mv2

  7. Potential Energy • Potential energy is stored energy. • Two forms of potential energy are gravitational potential energy: • PEgrav = Kgrav(m1m2/r) • and electrostatic potential energy • PEelect = Kelect(q1q2/r) • Total energy = PE + KE

  8. Measuring Energy • Heat is often called “The lowest form of energy” because energy is easily measured in this form. • The SI unit for energy is the Joule. • We use the idea of Specific Heat to measure energy as heat. • Specific heat is the amount of energy required to raise the temperature of 1 gram of a substance by 1 oC. • For water that number is 4.184 J/goC • Dulong and Petit Law: • Specific heat x molar mass = 25 J/mol oC

  9. Using Specific Heat • The following equation allows chemists to calculate the heat energy of any process by measuring the change in a known mass of water. • q = (Specific heat)(Mass)(Temperature change) • An insulated cup contains 75.0 g of water at 24.00 oC. A 26.00 g sample of a metal at 85.25 oC is added. The final temperature of the water and metal is 28.34 oC. • (a) What is the specific heat of the metal? • (b) According the the Dulong and Petit law what is the approximate molar mass of the metal? • (c) what is the apparent identity of the metal?

  10. First Law of Thermodynamics • The first law of thermodynamics states that energy must be conserved. • Two common forms of energy we use in chemistry are heat (q) and work (w). • The heat and work transferred to or by a system must equal the total energy change of the system. • ΔE = q + w • Signs of q and w:

  11. Work • Work is defined as the force applied to an object as it moves a certain distance: • Work = Force x Distance • Force can be defined as the pressure exerted over a given area so… • Work = Pressure x Area x Distance • Multiplying area and distances gives us volume • work = Pressure x ΔV • W = -pΔV

  12. Work is not a state function • Calculate the work involved in expanding a gas from an initial state of 1.00 L and 10.0 atm of pressure to (a) 10.0 L and 1.0 atm, (b) 5.00 L and 2.00 atm and then to 10.0 L and 1.o atm.

  13. Using R to convert units of Energy • So far we have 3 units of energy calories, Joules, and now L-atm. • x J = -13.0 L-atm

  14. Enthalpy, H • Enthalpy is defined as the heat content of a substance. • The first law of thermodynamics can be written as: • ΔE = q – pΔV • In most chemical reactions we do not have a volume change, which makes –pΔV = 0 • ΔE = qp • The heat energy generated by a chemical reaction at constant pressure (qp) is called the enthalpy change ΔH. • We can rewrite the first law of thermodynamics as: • ΔE = ΔH – pΔV

  15. Enthalpy change and heat of reaction • The heat energy, or enthalpy change (ΔH), produced by a chemical reaction is an extensive property. • If we react larger amounts of chemical reactants we will produce a greater amount of heat. • CH3CH2CH3(g) + 5O2(g)  3CO2(g) + 4H2O(g) • The more propane we burn the more heat we will get.

  16. CH3CH2CH3(g) + 5O2(g)  3CO2(g) + 4H2O(g) • In order to make the heat produced by a reaction an intensive property the amount of chemicals that react must be specified. • The standard heat of a reaction, ΔHo, is defined as the heat produced when the number of moles specified in the balanced chemical equation react. • When one mole of propane reacts with five moles of oxygen: • ΔHo = -2044 kJ • The negative sign indicates that heat is released by the reaction. • We can determine the heat per mole of any reactant or product by dividing the total heat of the reaction by the number of moles of the substance

  17. Hess’s Law • Hess’s Law states that, whatever mathematical operations are performed on a chemical equation, the same operations are applied to the heat of the reaction. • If the coefficients of a chemical equation are all multiplied by a constant, ΔHo will also be multiplied by the same constant. • If two or more equations are added together to obtain an overall reaction, the heats of these equations are also added together.

  18. The first rule of Hess’s Law states that if we multiply a reaction by a certain factor we must also multiply ΔHoreact by that factor. • CH3CH2CH3(g) + 5O2(g)  3CO2(g) + 4H2O(g) • ΔHoreact = -2044 kJ • If we reverse the reaction: • 3CO2(g) + 4H2O(g)  5O2(g) + CH3CH2CH3(g) • ΔHoreact = +2044 kJ • The fact that ΔHoforwardreact = -ΔHoreverse react is a very important part of Hess’s Law.

  19. If we multiply the equation for the combustion of propane by 2 we get: • 2CH3CH2CH3(g) + 10 O2(g)  6CO2(g) + 8H2O(g) • We then must multiply ΔHoreactby 2: • ΔH = - 2044 kJ x 2 = - 4088 kJ • ½ CH3CH2CH3(g) + 2.5 O2(g)  1.5 CO2(g) + 2 H2O(g) • ΔH = - 2044 kJ x ½ = - 1022 kJ • 6 CH3CH2CH3(g) + 30 O2(g)  18 CO2(g) + 24 H2O(g) • ΔH = - 2044 kJ x 6 = -12,264 kJ

  20. The second rule of Hess’s Law states that if we add two chemical equations to obtain an overall equation we must add the heats of both reactions. • N2(g) + O2(g)  2NO(g) ΔHo1 = +180.5 kJ • 2 NO(g) + O2(g)  2 NO2(g) ΔHo2 = -114.1 kJ • N2(g) + 2 O2(g)  2 NO2(g) ΔHototal = + 66.4 kJ • ΔHo1 + ΔHo2 = ΔHototal

  21. Calculate the ΔHoreact for the synthesis of propane from carbon and hydrogen: • 3 C(s) + 4 H2(g)  CH3CH2CH3(g) • Use the following combustion reactions for propane, hydrogen, and carbon. • CH3CH2CH3(g) + 5 O2(g)  3 CO2(g) + 4 H2O(g) • ΔHo= - 2044 kJ • 2 H2(g) + O2(g)  2 H2O(g) • ΔHo = -483.6 kJ • C(s) + O2(g)  CO2(g) • ΔHo = - 393.5 kJ

  22. Formation Reactions and Heats of Formation • A heat of formation is defined as one in which the reactants are elements in their standard state (1 atm and 25 oC) and there is only 1 mole of product. • Fe(s) + ½ O2(g) FeO(s) • Fractional coefficients need to be used to ensure that when only 1 mole of product is formed the reaction is still balanced. • ΔHof is tabulated as heat produced per mole of product. • The heat of formation for any element is always 0.00 kJ.

  23. Using Heat of Formation • The total heat of a reaction ΔHo can be determined using the heats of formation of all of the components. • Calculate the ΔHofor the combustion of propane using the following data: • 3 C(s) + 4 H2(g)  CH3CH2CH3(g) ΔHof= -103.8 kJ • C(s) + O2(g)  CO2(g) ΔHof = -393.5 kJ • H2(g) + ½ O2(g)  H2O(g) ΔHof = -241.8 kJ

  24. Entropy and The Second Law of Thermodynamics • Entropy is related to the number of different ways in which a system can arrange the particles within it. • If we cut the volume of a system into 8 different pieces and label them 1-8 we can arrange them in 56 different ways (8 x 7 = 56) • If we cut the volume into 16 different pieces we can arrange them in 240 different ways (16 x 15 = 240) • This represents an increase in entropy. • The second law of thermodynamics states that any physical or chemical change must result in an increase in the entropy of the universe.

  25. This all leads to the idea that an increase in the number of particles in a chemical system leads to an increase in entropy, a volume increase will increase entropy and change in physical state (say from liquid to gas) will increase entropy. • Entropy increase as you go from solid to liquid to gas. • Entropy is assigned the symbol, S, and has the units J/oC. • Standard entropy is based on the mole and is written as So with units of J/oC-mol • Unlike E and H where we can only calculate the change in energy and enthalpy we are able to calculate the actual value of entropy, S.

  26. Calculating S • A perfect crystal at absolute 0 (0 K or -273 oC) would have an entropy of 0 because there is only one possible arrangement of the atoms. • The Boltzmann entropy equation: • S = kln(w), where k is the Boltzmann constant, and w is the number of possible microstates.

  27. Entropy Changes, ΔS • Changes in entropy due to a chemical process are calculated in the same fashion as the heats of reaction. • Just like ΔHof tables of standard entropy changes for 1 mole of many substances are tabulated. • Use the given data to calculate the standard entropy change for the combustion of propane • ΔSCO2 = 213.6 J/mol-k • ΔSH2O = 188.7 J/mol-k • ΔSO2 = 205.0 J/mol-k • ΔSCH3CH2CH3 = 270.2 J/mol-k

  28. Entropy in General • Formation of a gas increase entropy greatly • A great value of Δn will result in a greater entropy increase. • If Δn = 0 phase changes from solid to liquid become the leading contributors to an increase in entropy. • A solute dissolving leads to an increase in entropy. • An increase in temperature leads to an increase in entropy.

  29. Gibbs Free-Energy, ΔG • The free-energy change is the maximum amount of energy available from any chemical reaction. • Driving forces behind ΔG: • ΔH, which represents the change in internal potential energy • The drive towards and increase in entropy of the system. • If ΔH is negative it means that the overall internal energy of the system is decrease, this favors a spontaneous reaction. • If ΔS is positive and the disorder (Entropy) is increased this also favors a spontaneous reaction.

  30. Gibbs Free-Energy Equation • ΔGo = ΔHo - TΔSo

  31. Calculating ΔG • Just like ΔHo and ΔSo we can calculate ΔGo from tabulated data of free energies of formation. • Temperature is an important factor that affects the value of free energy and must be specified. • Most tables will list ΔGovalues at 298 K. • The equation for finding ΔGototal is very similar to the ones for finding ΔHoand ΔSo.

  32. Free Energy at different Temperatures • In order to find ΔGofor a reaction at temperatures different from 298K we must know ΔHoreact and ΔSoreact. • For a certain reaction ΔHo= + 2.98 kJ and ΔSo = +12.3 J/K. What is the ΔGoof this reaction at 298 K, 200 K, and 400 K?

  33. For a certain reaction, ΔHo= -13.65 kJ and ΔSo = -75.8 J/K. • What is the ΔGoat 298 K? • Will increasing or decreasing the temperature make the reaction spontaneous? If so at what temperature will the reaction become spontaneous?

  34. Free-Energy and Equilibrium • When a system is not at standard state, the free energy change is represented as ΔGnot ΔGo. • The following equation shows the relationship between ΔGand ΔGo • ΔG = ΔGo+ RT ln(Q) • If ΔGis negative the reaction will proceed in the forward direction. • If ΔG is positive the reaction will proceed in the reverse direction. • If ΔG is 0… • The reaction is at equilibrium

  35. At equilibrium… • ΔGo = -RTlnK • The value of the equilibrium constant for a certain reaction is 45 at 298 K. At the same temperature Q = 35. Determine the value of ΔGo at 298 K, and predict in which direction the reaction will proceed.

  36. Extra Stuff • Bond energies and ΔHo • ΔGofor a phase change • Entropy S is the only thermochemical quantity that we can measure directly. • Recognition of combustion reactions.

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