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Thermochemistry

Thermochemistry. Chapter 6 Dr. Ali Bumajdad. Chapter 6 Topics. Thermochemistry. Study of heat change in a chemical reaction. Nature and Type of Energy Energy Changes in Chemical Reaction Intro. to Thermodynamics: 1 st Law, Wark, Heat, Enthalpy Calorimetry, Specific Heat, Heat Capacity

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Thermochemistry

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  1. Thermochemistry Chapter 6 Dr. Ali Bumajdad

  2. Chapter 6 Topics Thermochemistry Study of heat change in a chemical reaction • Nature and Type of Energy • Energy Changes in Chemical Reaction • Intro. to Thermodynamics: 1st Law, Wark, Heat, Enthalpy • Calorimetry, Specific Heat, Heat Capacity • Sstandard enthalpy of formation and reaction • Hess Law • Enthalpy of solution Dr. Ali Bumajdad

  3. Energyis the capacity to do work Unit = J = Kgm2s-2 • Work = Fd • Radiant energycomes from the sun (also called solar energy) • Thermal energy is the energy associated with the random motion of atoms and molecules • Chemical energy is the energy stored within the bonds of chemical substances • Nuclear energy is the energy stored within the collection of neutrons and protons in the atom • Potential energyis the energy object position relative to other objects (e.g. gravity result in pot. E to objects) • Kinetic Energy is the energy of motion Energy convert from one form to another (Law of Conservation of Energy)

  4. Temperature = Thermal Energy 900C 400C Energy Changes in Chemical Reactions Heat is the transfer of thermal energy between two bodies that are at different temperatures. Q) Water cup in the fridge, what is the direction of heat transfer? Temperature is a measure of the thermal energy. greater thermal energy

  5. Open system Closed system Isolated system Type: Exchange: mass & energy energy nothing Thermochemistry is the study of heat change in chemical reactions. • System is the specific part of the universe that is of interest in the study • Surrounding is every things except the thing we are studying

  6. 2H2(g) + O2(g) 2H2O (l) + energy H2O (g) H2O (l) + energy energy + 2HgO (s) 2Hg (l) + O2(g) energy + H2O (s) H2O (l) Exothermic process is any process that gives off heat or transfers thermal energy from the system to the surroundings. Endothermic process is any process in which heat has to be supplied to the system from the surroundings.

  7. DP = Pfinal - Pinitial DV = Vfinal - Vinitial DT = Tfinal - Tinitial Thermodynamics ‘The study of the interconversion of heat and other kinds of energy’ • State functions properties determined by the current state of the system, regardless of how that condition was achieved e.g. energy , pressure, volume, temperature DE = Efinal - Einitial Work and Heat Not State Function Potential energy of hiker 1 and hiker 2 is the same even though they took different paths.

  8. C3H8 + 5O2 3CO2 + 4H2O (1) Work done on (+) or by (-) the system W=-PV DEsys = q + w 2) (2) Heat transfer to (+) or from (-) the system Change in internal energy of a system • First law of thermodynamics: • 1) energy can be converted from one form to another, but cannot be created or destroyed DEsystem + DEsurroundings = 0 DEsystem = -DEsurroundings OR Exothermic chemical reaction! Chemical energy lost by combustion = Energy gained by the surroundings

  9. q=-ve Combustion reaction release heat Vaporization absorbed heat q=+ve Condensation release heat q=-ve Freezing release heat q=-ve Melting absorbed heat q=+ve

  10. DV > 0 -PDV < 0 wsys < 0 P x V = x d3 = F x d = w pressure vol Work is not a state function! F d2 Dw = wfinal - winitial Work Done On the System w = F x d w = -P DV initial final

  11. (2) w = -P DV (a) (b) DV = 5.4 L – 1.6 L = 3.8 L DV = 5.4 L – 1.6 L = 3.8 L 101.3 J = -1430 J w = -14.1 L•atm x 1L•atm Q) A sample of nitrogen gas expands in volume from 1.6 L to 5.4 L at constant temperature. What is the work done in joules if the gas expands (a) against a vacuum and (b) against a constant pressure of 3.7 atm? 1L atm =101.3 J P = 0 atm W = -0 atm x 3.8 L = 0 L•atm = 0 joules P = 3.7 atm w = -3.7 atm x 3.8 L = -14.1 L•atm  Work is not state Function

  12. (3) DH = DE + PDV Enthalpy and the First Law of Thermodynamics • Enthalpy (H) used to quantify heat flow into or out of a system in a process that occurs at constant pressure. DE = q + w At constant volume: w= 0 and DE = qv At constant pressure: DH = qp and w = -PDV DE = DH - PDV State Function

  13. (4) DH = H (products) – H (reactants) Hproducts < Hreactants Hproducts > Hreactants DH < 0 DH > 0 Enthalpy of Reactions DH = heat given off or absorbed during a reaction at constant pressure

  14. H2O (s) H2O (l) H2O (s) H2O (l) DH = 6.01 kJ/mol Is DH negative or positive? System absorbs heat Endothermic DH > 0, H =+ve 6.01 kJ are absorbed for every 1 mole of ice that melts at 00C and 1 atm.

  15. DH = -890.4 kJ/mol CH4(g) + 2O2(g) CO2(g) + 2H2O (l) CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l) Is DH negative or positive? System gives off heat Exothermic DH < 0, H =+ve 890.4 kJ are released for every 1 mole of methane that is combusted at 250C and 1 atm.

  16. Thermochemical Equations 2H2O (s) 2H2O (l) H2O (l) H2O (g) H2O (s) H2O (l) H2O (l) H2O (s) H2O (s) H2O (l) DH = -6.01 kJ/mol DH = 6.01 kJ/mol DH = 2 x 6.01= 12.0 kJ/mol DH = 6.01 kJ/mol DH = 44.0 kJ/mol 1) If you reverse a reaction, the sign of DH changes  Hforward =-Hbackward 2) If you multiply both sides of the equation by a factor n, then DH must change by the same factor n. H extensive property 3) The physical states of all reactants and products must be specified in thermochemical equations H H2O(g) > H H2O(l)>H H2O(l)

  17. P4(s) + 5O2(g) P4O10(s)DH = -3013 kJ/mol Q) How much heat is evolved when 266 g of white phosphorus (P4) burn in air? = 6470 kJ

  18. 2Na (s) + 2H2O (l) 2NaOH (aq) + H2 (g) DH = -367.5 kJ/mol A Comparison of DH and DE DE = DH - PDV At 25 0C, 1 mole H2 = 24.5 L at 1 atm and VVgas PDV = 1 atm x 24.5 L = 2.5 kJ DE = -367.5 kJ/mol – 2.5 kJ/mol = -370.0 kJ/mol E > H Some of Internal energy released used to do gas expansion work, this is why For reactions involving gases E slightly different than H For reactions do not involve gasesE  H (V is very small)

  19. (6) (5) q q C= s= mT T q = m x s x DT J/g.°C q = C x DT J/°C From (5) and (6): (7) C = m x s • Calorimetry, Specific Heat, Heat Capacity • Calorimetry The measurement of heat flow • Specific heat (s) is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius. • Heat capacity (C) is the amount of heat (q) required to raise the temperature of a given quantity (m) by one degree Celsius.

  20. (5) q s= mT q = m x s x Dt J/g.°C Q) How much heat is given off when an 869 g iron bar cools from 940C to 50C? s of Fe = 0.444 J/g •0C = -34,000 J

  21. q = m x s x Dt

  22. (8) qrxn = -Ccal xDT DH = qrxn (1) Constant-Volume Calorimetry (Good for combustion reaction) • Heat produced by combustion and absorbed by surrounding (calorimeter) qsys = qcal + qrxn qsys = 0 qrxn = - qcal qcal = Ccal xDT For Reaction at Constant V DE=qrxn No heat enters or leaves (Adiabatic)

  23. (8b) qrxn = -Ccal xDt molar n • Molar heat of combustion: Heat of combustion of one mole

  24. (2) Constant-Pressure Calorimetry • Heat produced (or absorbed) by reaction and absorbed (or produced) by the surrounding (solution) Endo. qsys = qsolution + qrxn Exoth. qsys = 0 qrxn = - qsolution qcal = Ccal xDT (8) qrxn = - Ccal xDT Reaction at Constant P DEqrxn DH = qrxn No heat enters or leaves! (Adiabatic)

  25. (8b) qrxn = -Ccal xDt molar n

  26. (5) q s= mT q = m x s x DT J/g.°C q n Cmolar= nT (7) C = m x s Sa. Ex.5.6a: How much heat is needed to warm 250g of water from 22°C to 98°C. (water specific heat =4.184 J/g.°C) =79kJ Sa. Ex.5.6b: What is the molar heat capacity of water? or

  27. qwater = m x s x DT =-qPb qPb = m x s x DT

  28. Reaction is exothermic • qrxn = - qsolution • when density 1 volume = mass • mol of NaOH = MNaOH VNaOH • mol of HCl = MHCl VHCl =-56.2kJ/mol

  29. DH0 (O2) = 0 DH0 (O3) = 142 kJ/mol DH0 (C, graphite) = 0 DH0 (C, diamond) = 1.90 kJ/mol f f f f Standard Enthalpy of Formation (H0f) (It is the heat change when 1 mole of a compound formed from its elements at 1 atm) • It is a reference point for all enthalpy expressions (similar to the sea level for measuring heights) • The standard enthalpy of formation of any element in its most stable form is zero

  30. Standard Enthalpy of Reaction (H0 ) rxn (9) aA + bB cC + dD - [ + ] [ + ] = - S S = DH0 DH0 rxn rxn dDH0 (D) nDH0 (products) mDH0 (reactants) bDH0 (B) cDH0 (C) aDH0 (A) f f f f f f (10) Hrxn= Hstep1 + Hstep2+ Hstep3+…… (It is the enthalpy of a reaction carried out at 1 atm) Hess’s Law • When reactants are converted to products, the change in enthalpy is the same whether the reaction takes place in one step or in a series of steps. Enthalpy is a state function It doesn’t matter how you get there, only where you start and end. • If a reaction is carried out in steps, H will equal the sum of the enthalpy changes for the individual steps.

  31. C (graphite) + 1/2O2(g) CO (g) CO (g) + 1/2O2(g) CO2(g) C (graphite) + O2(g) CO2(g)

  32. C(graphite) + O2(g) CO2(g)DH0 = -393.5 kJ rxn S(rhombic) + O2(g) SO2(g)DH0 = -296.1 kJ rxn CS2(l) + 3O2(g) CO2(g) + 2SO2(g)DH0 = -1072 kJ rxn 2S(rhombic) + 2O2(g) 2SO2(g)DH0 = -296.1x2 kJ C(graphite) + 2S(rhombic) CS2 (l) C(graphite) + 2S(rhombic) CS2 (l) rxn rxn C(graphite) + O2(g) CO2(g)DH0 = -393.5 kJ + CO2(g) + 2SO2(g) CS2(l) + 3O2(g)DH0 = +1072 kJ rxn DH0 = -393.5 + (2x-296.1) + 1072 = 86.3 kJ rxn Q) Calculate the standard enthalpy of formation of CS2 (l) given that: 1. Write the enthalpy of formation reaction for CS2 2. Add the given rxns so that the result is the desired rxn.

  33. 2C6H6(l) + 15O2(g) 12CO2(g) + 6H2O (l) - S S = DH0 DH0 DH0 - [ ] [ + ] = (9) rxn rxn rxn [ 12x–393.5 + 6x–285.8 ] – [ 2x49.04 ] = -6535 kJ = 12DH0 (CO2) 2DH0 (C6H6) f f = - 3268 kJ/mol C6H6 6DH0 (H2O) -6535 kJ f 2 mol nDH0 (products) mDH0 (reactants) f f Q) Benzene (C6H6) burns in air to produce carbon dioxide and liquid water. How much heat is released per mole of benzene combusted? The standard enthalpy of formation of benzene is 49.04, for CO2 is -393.5 and for H2O is -285.8 kJ/mol

  34. (10) Hrxn= Hstep1 + Hstep2+ Hstep3+……

  35. - S S = DH0 (9) rxn nDH0 (products) mDH0 (reactants) f f Hf°Fe2O3(s)=-822.2kJ/mol, Hf°Al2O3(s)=-1669.8 kJ/mol Divide by 2 to have the molar value then divide by the molar mass of Al to get it kJ/g

  36. Enthalpy of solution (Hsoln) The heat generated or absorbed when a certain amount of solute dissolves in a certain amount of solvent. DHsoln = Hsoln - Hcomponents

  37. The Solution Process for NaCl DHsoln = Step 1 + Step 2 = 788 – 784 = 4 kJ/mol

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