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Thermodynamics

Thermodynamics. Terms/ Definitions. Thermodynamics Deals with the interconversion of heat an other forms of energy Thermochemistry Deals with heat change in chemical reactions State Function Function that depends only on the conditions (state) not on how the state was obtained. Energy(E)

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Thermodynamics

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  1. Thermodynamics

  2. Terms/ Definitions • Thermodynamics • Deals with the interconversion of heat an other forms of energy • Thermochemistry • Deals with heat change in chemical reactions • State Function • Function that depends only on the conditions (state) not on how the state was obtained

  3. Energy(E) • Internal energy = kinetic + potential energy • Kinetic energy comes from molecular motion, electron motion etc. • Potential energy comes from attractive and repulsive forces in nuclei, and interactions between molecules • Enthalpy(H) • H = E + PV • Extensive property • State function • Can only measure the difference in enthalpy between two states

  4. Heat (Q,q) • Transfer of thermal energy between two bodies at different temperatures • Work (W,w) • Form of energy, can be mechanical or non mechanical • Mechanical work is normally pressure - volume work • W = -PexDV • Work done by the system on the surrounding = negative • Work done by the surroundings on the system = positive

  5. System • Specific thing we are looking at • Surroundings • Everything outside the system • Universe • System + surroundings • Open system • Can exchange both matter and energy with the surroundings • Closed system • Can exchange energy but not matter with the surroundings • Isolated system • No exchange of matter or energy with the surroundings • Q = ? • Q = 0

  6. Thermochemical Equation • Chemical equation that includes the enthalpy change • e.g. H2O(s) --> H2O(l) DH = 6.01 kJ • Heat flow • Endothermic, Q = +, heat absorbed by the system • Exothermic, Q = -, heat given off by the system • Units of energy • Joules(J), kilojoules(kJ) • Calories(cal), kilocalories(kcal) • 1 cal = 4.184 J • (Liter)(atmosphere) 1 L atm = 101.3 J

  7. Enthalpy • State function • Heat energy of state • Look at change between states H • Extensive Property - depends on the amount 1 mole of the reaction as written

  8. Changes sign as reverse reaction: • Follows Hess’ Law

  9. Hess’ Law If a reaction can be considered to go by a series of steps, DH of the reaction is the sum of DH of the steps. DH kJ/mol Reaction +90.4 1/2 N2(g) + 1/2 O2(g) NO(g) +56.5 NO2(g) NO(g) + 1/2 O2(g) What is the value of DH for the reaction: 1/2 N2(g) + O2(g) NO2(g) -56.5 NO(g) + 1/2 O2(g) NO2(g) +90.4 1/2 N2(g) + 1/2 O2(g) NO(g) +33.9 1/2 N2(g) + O2(g) NO2(g)

  10. Standard States Most stable form at 1 atmosphere and usually at 25oC Indicate standard state by DHo H2(g) N2(g) Br2(l) e.g. C(s)graphite Fe(s) S(s) Hg(l) I2(s) Allotropes More than one stable form at 1 atmosphere C(s) graphite and C(s) diamond O2 and O3 Standard heat of formation (Standard enthalpy of formation) Heat change that occurs when 1 mole of a single product is formed from elements in their standard states. CO2 = -393.5 kJ/mol C(s) graph + O2(g) CO2(g) -393.5 kJ

  11. Write the formation reaction for the following: NH4NO3(s) 2 NH4NO3(s) 3/2 N2(g) + H2(g) + O2(g) C6H6(l) benzene 3 C6H6(l) C(s)graphite + H2(g) 6 aniline C6H5NH2(l) C(s)graphite + H2(g) + N2(g) 6 1/2 C6H5NH2(l) 7/2

  12. Value of standard state of elements: By definition, DHo formation of an element in its standard state = 0. Which of the following will NOT have DH formation = 0 H2(g) Ne(g) Cl2(g) I2(l) Hg(s) Br2(l) Ca(s) Fe(l)

  13. Calculate Heat of a reaction: 1. Heat of reaction can be calculated from Heat of Formation data 2. Heat of reaction can be calculated using Hess’ Law Using Heat of formation data If data for all the reactants and products is given as , then is equal to the sum of of the products minus the sum of of the reactants.

  14. Calculate Horxn for the reaction: C2H6(g) + O2(g) CO2(g) + H2O(l) 2 3 7/2 = {2(-90.1) + 3(-68.3)} - {1(-20.2) + 7/2 (?)} = {2(-90.1) + 3(-68.3)} - {1(-20.2) + 7/2 (0)} = -364.9 kcal/ mol

  15. Using Hess’ Law If all information is not given as Hf, then need to go a different way Use Hess’ Law Given: ReactionHo(kJ/mol) Find H for the formation of C2H2 1. Write a target reaction: --> C2H2(g) 2C(s) graph + H2(g) 2. Rearrange reactions so that when added together you get the target --> 2C2H2(g) 4C(s) graph + 2H2(g) +453.2 --> C2H2(g) 2C(s) graph + H2(g) +453.2/2 = 226.6 kJ

  16. Which of the following would have = 0? C(s) diamond Hg(g) Xe(g) Br2(g) Cl - (aq)

  17. Energy(E) - Internal Energy (KE + PE) Kinetic energy: comes from the molecular motion and the electronic motion Potential Energy: comes from attractive and repulsive forces in nuclei and from interactions between molecules. Energy can be converted from one form to another, it can not be created or destroyed. i.e. The energy of the universe is constant. First Law: Euniverse = Esystem + Esurroundingss = 0 All energies do not have to be the same form Esystem = - Esurroundingss Since we are primarily interested in what happens to a chemical system, we use the form: Q = heat E = q + w W = work

  18. Sign convention - + Heat(q) from system to from surroundings surroundings to system (exothermic) (endothermic) Work(w) by the system on by surroundings the surroundings on the system Work includes all kinds of work, both mechanical and non-mechanical. We limit to mechanical work at this time. Mechanical work: W = -PexV = - PV at constant P A gas expands against a constant external pressure of 5.0 atm from 2.0 L to 4.0 L. How much work is done?

  19. A gas expands against a constant external pressure of 5.0 atm from 2.0 L to 4.0 L. How much work(in J) is done? 10.0 J - 10.0 J 1.01 x 103 J 20.0 J - 1.01 x 103 J

  20. A gas expands against a constant external pressure of 5.0 atm from 2.0 L to 4.0 L. How much work is done? W = - PV = - (5.0 atm)(4.0 L - 2.0 L) = - 10.0 L atm 1 L atm = 101.3 J = (- 10.0 L atm) (101.3 J/ Latm) = - 1013 J = - 1.01 x 103 J

  21. Relationship between E and q E = q + w w = - PexV At constant volume DV = ? = 0 E = q - PexV E = qv Relationship between E and H H = E + PV By definition: At constant P: (PV) = P V So: H = E + (PV) H = E + P V When you are looking at a reaction dealing with gases: So: P V = nRT PV = nRT And: H = E + nRT

  22. Relationship between H and q H = E + P V E = q + w At constant P: Pex = P H = q + w + P V W = - PexV W = - PV H = q - PV + P V H = qP

  23. A system at 1 atm and 11.0 L absorbs 300 J of heat and has 700 J of work performed on it. What is E for the process? + 1000 J + 400 J - 400 J - 1000 J

  24. A system at 1 atm and 11.0 L absorbs 300 J of heat and has 700 J of work performed on it. What is E for the process? E = q + w q = + 300 J w = + 700 J E = 300J + 700 J = 1000 J

  25. Calorimetry Deals with the transfer of heat energy Heat Capacity - Capacity of a system to store heat - Heat needed to raise the temperature of a system by 1oC ( 1K) = J/K q = (capacity of system)(T) Specific Heat (c) - Heat needed to raise the temperature of 1 gram of material by 1oC (1K) = J/g K q = msT Molar Heat Capacity (C) - Heat needed to raise the temperature of 1 mole of material by 1oC (1K) = J/ mol K q = nCT

  26. Constant volume Calorimetry Bomb Calorimeter Q = E Isolated system qsystem = ? = 0 qsystem = qsample + qcalorimeter qsample = - qcalorimeter qsample not directly measurable qcalorimeter can be measured qcalorimeter =heat capacity of bomb(T) = - qsample Since reaction carried out at constant volume qsample = E

  27. Constant Pressure Calorimetry Q = H Coffee Cup Calorimeter Isolated system qsystem = ? = 0 Generally look at two different kinds of systems 1. Reaction in aqueous solution 2. Metal solid in water qsystem = qmetal + qwater 2. 1. qsystem = qreaction + qsolution qreaction = - qsolution - qmetal = qwater qreaction not directly measurable • (m)(c)T = (m)(c)T • metal water qsolution can be measured qsolution = mcT = -qreaction Since at constant pressure qreaction = Hreaction

  28. 0.500 L of 1.0 M Ba(NO3)2 at 25oC was mixed with 0.500 L of 1.0 M Na2SO4 at 25oC. A white precipitate forms (BaSO4) and the Temperature rises to 28.1oC. What is H/ mol of BaSO4 formed? Specific heat mix = 4.184 J/g K d of mix = 1.0 g/ mL Assume a coffee cup calorimeter, P = constant q transfer from system to surroundings = ? = 0 Why? Coffee cup calorimeter assumes an isolated system qsys = qrxn + qmix = 0 Can’t measure heat of the ppt’n directly but can measure heat of mixing qrxn = - qmix qmix = msT = (1000mL)(1.0g/mL)(4.184J/g K)(28.1-25.0) = 12970J = 12.970kJ But this is only for the amount of BaSO4 made in this reaction and we want per mole qppt = - qmix = - 12.970kJ

  29. Need to determine the amount of BaSO4 that was formed in the reaction 0.500 L of 1.0 M Ba(NO3)2 at 25oC was mixed with 0.500 L of 1.0 M Na2SO4 at 25oC. A white precipitate forms (BaSO4) and the Temperature rises to 28.1oC. What is H/ mol of BaSO4 formed? Specific heat mix = 4.184 J/g K d of mix = 1.0 g/ mL Rxn: Ba(NO3)2 + Na2SO4 --> BaSO4 + 2NaNO3 Mole Ba(NO3)2 = VM = (0.500L)(1.0M) = 0.50 mol Mole Na2SO4 = VM = (0.500L)(1.0M) = 0.50 mol Ba(NO3)2 + Na2SO4 --> BaSO4 + 2NaNO3 0.50 mol 0.50 mol --> 0.50 mol So, the reaction produced 0.50 mol BaSO4 and that released - 12.97kJ Therefore the production of 1 mol of BaSO4 will have H = ? H = 2(- 12.97) = - 25.94 kJ = - 25.9 kJ

  30. 28.2 g Ni metal at 99.8 oC was placed in 150. g water at 23.50 oC. The final temperature at equilibrium was 25.00 oC. Find the specific heat of Ni in cal/ g K. 0.107 cal/ g K 0.1066 cal/ g K - 0.1066 cal/ g K - 0.107 cal/g K 0.1267 cal/g K

  31. 28.2 g Ni metal at 99.8 oC was placed in 150. g water at 23.50 oC. The final temperature at equilibrium was 25.00 oC. Find the specific heat of Ni in cal/ g K. qmetal = mcT for the Ni qsystem = qmetal + qsolution qmetal = (28.2 g)coC qsolution = mcT for the water qsolution = (150 g)(1 cal/g oC)(25.0-23.5oC) - qmetal = qsolution - (28.2 g)coC = (150 g)(1 cal/g oC)(25.0-23.5oC) cNi = 0.1066 cal/goC = 0.107 cal/ g K

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