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Thermochemical Principles

Thermochemical Principles. Data book Enthalpies. We can’t measure enthalpies of specific reactants or products directly, but we can measure enthalpy changes.

emma-wood
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Thermochemical Principles

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  1. Thermochemical Principles

  2. Data book Enthalpies • We can’t measure enthalpies of specific reactants or products directly, but we can measure enthalpy changes. • Rather than listing the enthalpy change for every possible reaction, data books list the standard enthalpy of combustion and the standard enthalpy of formation of compounds, and we can use these figures to work out the enthalpy change for other reactions.

  3. Enthalpy change of combustion Hc Enthalpy change of combustion Hc the enthalpy change for the complete combustion of 1 mole of a compound in excess pure oxygen under standard conditions of 298k and 1 atmosphere pressure. C2H5OH(l) + 3O2(g)2CO2(g) +3H2O(l) Hc (ethanol) = - 1367 kJ mol-1 Hc are always negative - an exothermic reaction. A fuel contains lots of energy which is released to the environment when more stable products are formed.

  4. C2H5OH(l) + 3O2(g)2CO2(g) +3H2O(l) Hc (ethanol) = - 1367 kJ mol-1 A fuel has high chemical potential Energy C2H5OH(l) + 3O2(g) Energy Hc (ethanol) = - 1367 kJ mol-1 stable products 2CO2(g) +3H2O(l)

  5. Enthalpy of Formation of a Compound Hf Enthalpy change of formation Hf : the enthalpy change for the formation of 1 mole of a compound from its elements under standard conditions of 298k and 100 kPa atmosphere pressure. 2C(s) +3H2(g) + 0.5O2  C2H5OH(l) Hf (ethanol) = -1367 kJ mol-1 Mg(s) + Cl2(g)  MgCl2(s) Hf (MgCl2(s) ) = - 641 kJ mol-1 Hf are almost always negative, since stable compounds have less energy, are more stable, than their elements Hf of elements are all of course 0, since to form themselves from themselves takes no energy !!!

  6. Mg(s) + Cl2(g) MgCl2(s) Hf (MgCl2(s) ) = - 641 kJ mol-1 Elements have high chemical potential Energy Mg(s) + Cl2(g) Energy Hf (MgCl2(s) ) = - 641 kJ mol-1 stable products MgCl2(s)

  7. If Hf of the reactants and products of a reaction are known, thenHrof the reaction can be calculated For the reaction below a = moles of reactant A etc aA + bB cC + dDHr Then H= c.Hf (C) + d.Hf (D) - a.Hf (A) - b.Hf (B) Hr = sumHfof products - sumHfreactants (Use correct number of moles as in the equation, becauseallHf are per MOLE )

  8. An example of the equation Hr = sumHfof products - sumHfreactants Hr= sumHfof products - sumHfreactants Hf for reactants and products in the equation below are known. Calculate Hcombustion ethanol. Show that: We use Hess’s Law, and construct a cycle. If we know the energy by 1 route we can can calculate the energy by another route that ends at the same chemical. Step 1 Write down the equation for the reaction which we are asked to calculate the unknown H Hc (ethanol) C2H5OH(l) + 3O2(g)2CO2(g) +3H2O(l)

  9. STEP 2 We are givenHf for the reactants. We can connect the combustion equation to the elements contained in the reactants. Make sure that the equation forHf is balanced for moles of each element Hc (ethanol) C2H5OH(l) + 3O2(g)2CO2(g) +3H2O(l) Hf(ethanol) 2C(s) +3H2(g) + 0.5O2

  10. STEP 3 We are givenHf for the products too; we can connect the products to the elements contained in the compounds Hc (ethanol) C2H5OH(l) + 3O2(g)2CO2(g) +3H2O(l) 2. Hf (CO2) Hf(ethanol) 3.Hf (H2O) 2C(s) +3H2(g) + 0.5O2

  11. STEP 4 We can now calculate Hc (ethanol); by applying Hess’s law. Route 1 Hc (ethanol) C2H5OH(l) + 3O2(g)2CO2(g) +3H2O(l) 2. Hf (CO2) 3.Hf (H2O) Route 2 Hf(ethanol) 2C(s) +3H2(g) + 0.5O2 Enthalpy Change by Route 1 = Enthalpy Change byRoute 2 Hc (ethanol) = - Hf(Ethanol) + 2. Hf (CO2) + 3.Hf (H2O)

  12. Enthalpy change of neutralisation Ho(neutralisation) Enthalpy change of neutralisation: the enthalpy change when 1 mole of water is produced by the neutralisation of an acid by an alkali under standard conditions of 100 kPa pressure and at 298 k NaOH(aq) +HCl(aq) = NaCl(aq) + H2O(l) Ho(neut.)= - 57.6 kJmol-1 KOH(aq)+ 0.5H2SO4(aq) =KHSO4(aq)+H2O(l) Ho(n)= -57.6 kJ mol-1 These enthalpies of neutralization are the same because the reaction is the same. The acids and alkalis are strong, so they are totally ionised; Na+, K+, Cl-, SO42- and HSO4- are spectator ions. H+(aq) + OH-(aq) H2O(l) Ho(n) = - 57.6 kJ mol-1

  13. Calculate Hro for the reaction below, given the enthalpies of combustion / kJ mol-1; C2H4 -1411, H2 - 286, C2H6 - 1560 Route 1 Hro C2H4 + H2 C2H6 Route 2 - 2 x 1411 - 286 -1560 2 CO2 + 2 H2O H2O Enthalpy Change by Route 1 = Enthalpy Change by Route 2 Hro = - 2 x 1411 - 286 - ( - 1560) = -1548 kJ

  14. Enthalpy change of solution Hsolution Enthalpy change of solution Hsolution: the enthalpy change when 1 mole of ionic solid dissolves to form a dilute 1 molar solution in water at 298 k and 100 kPa pressure. Dissolving an ionic solid is a chemical process. Ionic bonds between the positive and negative ions must be broken, to free the ions from the giant lattice structure; this is a highly endothermic process. This breaking of the ionic lattice is only possible in water as strong ion - dipole bonds are formed between water and the ions. This is a highly exothermic process, and is called hydration.

  15. Hydration of ions The breaking of the lattice as an ionic solid dissolves in water is possible because strong ion - dipole bonds are formed between water molecules and free ions, releasing energy.This is called hydration, and is highly exothermic. Hydration enthalpy offsets the endothermic breaking of the ionic bonds in the lattice. Water has a dipole, the O atom carries a slight negative charge and the H atom a slight positive charge • + • - H one ion -dipole chemical bond • Na+ O in Na+(H2O)6 i.e. Na+(aq) • H • + • + • - H Cl- one ion -dipole chemical bond • O in Cl-(H2O)6 i.e. Cl-(aq) • H • +

  16. Enthalpy of solution of sodium chloride is slightly endothermic, because the lattice enthalpy is slightly bigger than the hydration enthalpy Na+(g) + Cl-(g) + aq Hydration of ions, forms ion-dipole strong bonds Lattice enthalpy (breaks up ions) Energy Na+(aq) + Cl-(aq) Hsolution(NaCl) NaCl(s) + aq hydration Na+ + 6H2O Na+(H2O)6 = Na+(aq) hydration Cl- + 6H2O Cl-(H2O)6 = Cl-(aq)

  17. Solid Sodium Chloride in water Na+ ion Water molecule Cl- ion

  18. One Cl- is being removed by water molecules. Water molecules do not yet completely surround the Cl- ion

  19. Sodium chloride crystal dissolving in water Chloride ion hydrated by water. Note that the positive H atom is orientated to the Cl- ion - it isnow a Cl-(aq) ion A water molecule

  20. Experimental Determination of enthalpy changes If in a reaction in solution, m g of water is warmed through T oC and c is the specific heat capacity of water Then the heat absorbed by the water is q, and q = m. c. T • If n moles of product are produced then the standard molar enthalpy change, i.e. the heat change per mole is Hr, Hr = - q n = - m.c. T J per mole n and Hr = - m.c. T kJ per mole n .1000 If the temperature falls, the reaction is endothermic and H has a positive sign.

  21. Enthalpy of Neutralisation of NaOH and HCl Initial Temperature = 22.35 oC

  22. Enthalpy of Neutralisation of NaOH and HCl Final temperature = 29.85 oC

  23. Calculation of the Enthalpy of Neutralisation H+(aq) + OH-(aq) = H2O(l) Mass of liquid (water) being warmed = 50 + 20 = 70 g Temperature rise of liquid & calorimeter = 29.85 - 22.35 = 7.5 oC Heat capacity of calorimeter = C(cal) = 78.2 J oC-1 Heat to warm calorimeter = C(cal) .T = 78.2 x 7.5 = 586.5 J Specific Heat capacity of water = 4.184 J g-1oC-1 Heat to warm up water = q = m.c. T q = 70 x 4.184 x 7.5 = 2196.6 J Total heat out in reaction = qtot=586.5 + 2196.6 = 2783.1 J

  24. Calculation of the Enthalpy of Neutralisation H+(aq) + OH-(aq) = H2O(l) 20 mL of 3M HCl and 50 mL of 1 M NaOH were mixed Moles of H+ = 0.02 x 3 = 0.06 ; Moles of OH- = 0.05 x 1 = 0.05 Thus H+ is in excess, and 0.05 moles of water will be produced Hneutralisation= total heat out per mole of water produced H is negative, since the temperature has risen Hneutralisation = - qtot / n= - 2783.1 / 0.05 J mol-1 Hneutralisation = - 55662 J mol-1 Hneutralisation = - 55.6 kJ mol-1

  25. An Energy level diagram for the Enthalpy of Neutralisation of hydrochloric acid by sodium hydroxide Ionic Equation NaOH(aq) +HCl(aq) H+(aq) + OH-(aq) Hneutralisation = - 55.6 kJ mol-1 Energy H2O(l) NaCl(aq) +H2O(l) The neutralisation equation can be written: Na+ (aq) + OH- (aq) + H+ (aq) + Cl- (aq) = Na+ (aq) + Cl- (aq) +H2O(l) But Na+ (aq) and Cl- (aq) are spectator ions and contribute nothing to the enthalpy change; they are ignored in the ionic equation.

  26. Enthalpy of Solution of Ammonium Nitrate The initial temperature =23.15 oC ammonium nitrate in glass vial Crushing device

  27. Enthalpy of Solution of Ammonium Nitrate The final temperature is 19.05 oC - the reaction is endothermic

  28. Calculation of the Enthalpy of Solution NH4NO3 (s) +aq = NH4+(aq) + NO3-(aq) Mass of liquid (water) being cooled = 60 g Temperature fall of liquid & calorimeter = 23.15 - 19.05 = 4.1 oC Heat capacity of calorimeter = C(cal) = 153 J oC-1 Heat to cool calorimeter = C(cal) x T = 153 x 4.1 = 627.3 J Specific Heat capacity of water = 4.184 J g-1oC-1 Heat to cool water = q = m.c. T q = 60 x 4.184 x 4.1 = 1029.3 J Total heat removed in reaction = qtot=627.3 + 1029.3= 1656.6 J

  29. Calculation of the Enthalpy of Solution NH4NO3 (s) +aq = NH4+(aq) + NO3-(aq) 60 g of water was mixed with 5.00 g of pure ammonium nitrate - Mr= 80.04 Moles of NH4NO3 = 5 / 80.04 = 0.0625 H is positive, since the temperature has fallen. Heat has been taken from the water and calorimeter by the chemicals. Hsolution = total heat absorbed per mole of NH4NO3 Hsolution = + qtot / n= + 1656.6 / 0.0625 J mol-1 Hsolution = + 26505.6 J mol-1 Hsolution = + 26.5kJ mol-1

  30. An Energy level diagram for the Enthalpy of Solution of ammonium nitrate NH4+(aq) + NO3-(aq) Hsolution = + 26.5 kJ mol-1 Energy NH4NO3 (s) + aq The ionic lattice in the giant structure has to be broken. With ammonium nitrate this endothermic process is larger than the exothermic hydration enthalpies of the ammonium and nitrate ions. Sodissolving the solid is overall endothermic.

  31. The Bomb CalorimeterHow accurate enthalpy changes are measured Ignition wires Thermometer Stirrer High pressure oxygen Insulated container Steel Bomb Ignition heater Reactants in sample cup Water

  32. The operation of a Bomb Calorimeter • Benzoic acid is used to determine the heat capacity of the apparatus. It has an accepted value for its enthalpy of combustion. • A known mass of benzoic acid is burnt and the temperature rise measured. The heat capacity of the ‘Bomb’ can be calculated. • To determine the enthalpy of combustion of a new substance a known mass is burnt. The conditions are kept as similar as possible to the standardization experiment; i.e. same temperature rise. • The Enthalpy of Combustion is then calculated

  33. Bond Energies (enthalpies) The bond energy is the enthalpy change to break 1 mole of specified covalent bonds in a gaseous molecule under standard conditions of 298k and 100 kPa pressure. H2(g) 2H(g) H = + 435.9 kJ mole-1 H-H 2H.H=+ 435.9 kJ mole-1 The H-H Bond Energy is + 435.9 kJ mole-1 Bond energies are always positive, since they are the energies to break covalent bonds. I.e to break the electrostatic attraction between a shared pair of electrons and 2 adjacent positive nuclei.

  34. Average bond energies These are averages of bond energies for a particular bond, taken from a wide range of compounds containing that bond. The Average Bond Energy is the average energy to break 1 mole of specified bonds from a wide number of compounds at 298 k and 1 atmosphere pressure. • They are averages, and the actual bond energy in a molecule may not equal the average bond energy. • Factors that change actual bond energies from the average are the electro-negativities of adjacent atoms. • When one breaks up methane into C(g) + 4 H(g) atoms 4 C-H bonds are broken; the average bond energy is the total enthalpy for breaking up methane /4 • The average bond energy for a C-H bond is quoted as 413 kJ mole-1, which is the average for methane, ethane and a host of other hydrocarbons etc.

  35. If all the Bond energies (B.E.) are known for the reactants and products of a reaction then Hr can be calculated. If B.E.(A-B) is the bond energy in the molecule A-B etc. Hr A-B + C-D A-C + B-D A B C D B.E.(C-D) B.E.(A-C) B.E.(A-B) B.E.(B-D) Gaseous elements By Hess’s law; energy change route 1 = energy change route 2 Hr =B.E.(A-B) + B.E.(C-D) - B.E.(A-C) - B.E.(B-D)

  36. Calculation of Hr from the bond energies of reactants and products Estimate the H for the reaction, given the Average Bond Enthalpies / kJ mol-1 C=C 699, C-C 368, C-H 435, C-F 484, F-F 155 H H C=C H H H H F C C F H H Hr F F Break ALL bonds 4 C-H = + 4 x 435 1 C-C = + 369 2 C-F = + 2 x 484 4 C-H = + 4 x 435 1 C=C = + 699 1 F-F = +155 2 C + 4 H + 2F

  37. Enthalpy change by route 1 = change by route 2 Hr = + 4 x 435 + 699 + 155 - (4 x 435) - 369 - (2 x 484) = - 478 kJ An estimate, as average bond energies used Route 1 H H C=C H H H H F C C F H H Hr F F Route 2 Break ALL bonds 4 C-H = + 4 x 435 1 C-C = + 369 2 C-F = + 2 x 484 4 C-H = + 4 x 435 1 C=C = + 699 1 F-F = +155 2 C + 4 H + 2F The - sign arise as bonds are formed in 1,2-difluoroethane

  38. Using Bond Enthalpies (B.E.) to calculate the enthalpy of combustion of ethanol 2 C + 6H + 7O BREAK 1 C-C, 5 C-H, 1 C-O and 1 O-H BONDS Endothermic(+) H H H-C-C-O-H H H 3 O=O MAKE 4 C=O, and 6O-H BONDS Exothermic (-) Hc(ethanol) 2 O=C=O + 3 H-O-H B.E./ kJ: C-H +413, C-C 348, O=O 498, C-O 360, C=O 743,O-H 463 Hc(ethanol) = + 348 + 5 x 413 +360 + 463 - 4 x 743 - 6 x 463 = -1368 kJ mol-1

  39. Enthalpy and Changes of State H FUSION (+ve) SOLID LIQUID - H FUSION (+ve) - H VAP H SUBLIMATION H VAPORISATION (+ve) GAS Liquids have higher kinetic energies than solids. Heat energy must be supplied to change a solid into a liquid, H FUSION is thus+ve. Similarly gases have more KE than liquids, and so H VAPORISATION isalso+ve.

  40. Enthalpy and Changes of State H = -- 242 kJ mol-1 H2(g) + 1/2 O2(g) H2O(g) H Vaporisation (Water) H = -- 286 kJ mol-1 H2O(l) By Hess’s law - 242 = - 286 + H Vaporisation (Water) H Vaporisation (Water) = - 242 +286 = + 44 kJ mol-1

  41. 4th April 2012Enthalpy of combustion • AIM – to calculate enthalpy of combustion from experimental data

  42. Complete investigation 4.2 from p.168

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