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Thermochemical principles includes: transfer of heat between the system and the surroundings

Thermochemical principles includes: transfer of heat between the system and the surroundings the enthalpy change for any process is the sum of the enthalpy changes for the steps into which the process can be divided

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Thermochemical principles includes: transfer of heat between the system and the surroundings

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  1. Thermochemical principles includes: • transfer of heat between the system and the surroundings • the enthalpy change for any process is the sum of the enthalpy changes for the steps into which the process can be divided • definition of the following terms: cH, fH, rH, vapH and fusH.     

  2. Heat Heat is the transfer of energy from regions of high temperature to regions of low temperature. As temperature is a measure of the movement of molecules (average kinetic energy), heat moves from regions of high thermal motion to a region where there is less thermal motion.

  3. Enthalpy When considering the heat change in chemical reactions we use a quantity called enthalpy (H)which is a measure of the chemical potential energy stored in the bonds of the substances involved.

  4. Enthalpy A change in enthalpy of a system is equal to the heat released or absorbed at constant pressure. It is measured in kilojoules, kJ (or joules, J). The change in enthalpy is given by rH, where rH = Hfinal – H initial = H products - H reactants

  5. Enthalpy Exothermic and Endothermic processes For an exothermic reaction, heat is released and rH < 0 In this case the products are more stable because they contain less enthalpy than they started with.

  6. This means the temperature of all speciesand the surroundings increases. All combustion reactions are exothermic. This is shown in an energy profile diagram  rH < 0 Products Enthalpy, H

  7. Enthalpy The chemical potential energy is not lost but is converted into increased kinetic energy of all the particles (both reactants and products) and also transferred to the surroundings.

  8. An endothermic reaction absorbs heat from the surroundings because the enthalpy of the products is more than reactants and rH > 0. This means the temperature of the system gets colder. Dissolving may be endothermic or exothermic depending on the nature of the solute and solvent e.g. dissolving NH4NO3(s) in water is endothermic while dissolving conc H2SO4 is exothermic.

  9. Exercise: In the space below draw a labelled diagram for an endothermic reaction, showing reactants, products and rH. PRODUCTS Enthalpy, H • rH > 0 Products REACTANTS

  10. At both its melting point and boiling point, even though the sample is being heated, temperature remains constant. All heat being added is used to break the forces holding the molecules in their solid or liquid states, rather than being converted into kinetic energy.

  11. Could you explain looking at the graph why a drink stays cold while it contains ice cubes?

  12. Molar heat of Fusion (∆fusH) This is the energy required to change one mole of a substance from a solid to a liquid at the melting point. H2O (solid at 0 oC) H2O (liquid at 0 oC) ∆fusH=6.0 kJmol-1 The ∆fusH gives an indication of the strength of forces holding the particles together in the solid phase

  13. The melting and boiling points of a substance indicate the strength of the forces between the particles in the solid or liquid phases

  14. Molar heat of vaporisation (∆vapH) This is the energy required to change one mole of a substance from a liquid to a gas at the boiling point. H2O (liquid at 100 oC) H2O (gas at 100 oC) ∆vapH = 41.0kJmol-1 The ∆vapH gives an indication of the strength of forces holding the particles together in the liquid phase

  15. Enthalpies We can’t measure enthalpies of reactants and products directly all we can measure are enthalpy changes. Data books list the standard enthalpies of combustion and /or the standard enthalpy of formation for compounds. These figures can be used to calculate the enthalpy change for other reactions

  16. Enthalpy Definitions (∆cH 0) – Standard enthalpy (heat) of combustion The standard enthalpy of combustion is the enthalpy change when one mole of the substance is completely burnt with all reactants and products in their standard states Standard states are the states at room temperature (25oC) and pressure - that is O2 is a gas and water is H2O (l) C2H6(g) + 3 ½ O2(g) 2CO2(g) + 3H2O(l) ∆cH0 = -1557kJmol-1 Heats of combustion are always exothermic

  17. (∆fH 0) – Standard enthalpy (heat) of formation The standard enthalpy of formation of a compound is the enthalpy change when one mole of the substance is formed from its elements, with all the reactants and products in their standard states. 2C(s) + 3 H2(g) + ½ O2 (g) C2H5 OH(l) ∆fHo= -278kJmol-1 Do you think these enthalpies of formation look strange if so why? Data tables list heats of formation for compounds not elements - why? Think standard states The standard enthalpy of formation for all elements is zero

  18. Remember fH0 is the standard reaction enthalpy for the formation of one mole of substance from its elements in their most stable form at standard state. The units of fH0 are kJ mol-1 and the equation it refers to must have only one moleof product. Note: If 4C(s) + 6H2(g) + O2(g) 2C2H5OH (l) rH0 = -555 kJ Then f H0(C2H5OH, l) = -555  2 = - 277.5 kJ mol-1

  19. Exercise - Write equations for the reactions which have the following enthalpies of combustion: • cH0(H2, 25 oC) = -286 kJ mol-1 • cH0(CH3OH, 25 oC) = -726 kJ mol-1 H2(g) + ½ O2(g) H2O(l) ∆cH = -286kJmol-1 CH3OH(g) + 1 ½ O2(g) CO2(g) + 2H2O(l) ∆cH = -726kJmol-1

  20. Exercise 1 - Write balanced equations for the formation reactions from their elements • (a) C6H12O6(s) • 6C(s) + 6H2(g) + 3O2(g) C6H12O6 (s) • (b) CO(g) • C(s) + ½ O2(g) CO (g) • (c) HCl(g) Note : We use mole fractions where necessary to give one mole of the compound ½ H2 (g) + ½ Cl2(g) HCl (g)

  21. Starter cH0 means the e_______ of c__________ the 0 means s_______ c________ standard conditions - refers to ____ mole at ____degrees celsius and ___ atmosphere of p______

  22. ENERGY CALCULATIONS • Calculations involving thermochemical principles may include: • relating enthalpy changes to heat and mass, and use of specific heat capacity of water • application of ΣrH = ΣfH (products) - ΣfH (reactants) • application of Hess’s Law • use of average bond energies in enthalpy change calculations.

  23. Measuring enthalpies of reaction – calorimetry To measure enthalpy changes, the reaction is carried out in an insulated container (such as a polystyrene cup) and the temperature change is measured. Using this temperature change, ΔT, and the value of the specific heat capacity, c, the amount of energy transferred to the mass m of substance can be calculated using the expression E = m c ΔT If the reaction takes place in an aqueous solution, and the solutions are placed in an insulated container then we can assume: c =The specific heat capacity of the solutions is 4.18 J oC-1 g-1. The densities of these aqueous solutions can be taken as 1.0 g mL-1 so that 100 mL would have a mass of 100 g.

  24. Calorimetry

  25. Exercise 25.0 mL of 1.0 mol L-1 HCl is placed in a polystyrene cup and its temperature is measured as 21.0 oC. Following the addition of 25.0 mL of 1.0 mol L-1 NaOH solution, the mixture is stirred and the final temperature recorded as 27.8 oC. Calculate the molar enthalpy change, ΔrH in kJ mol-1,for this reaction. (Note: If the temperature increases then the reaction is exothermic and ΔrH will be negative) E = m c ΔT = 50g X 4.18 J oC-1 g-1 x 6.8 oC = 1421.2 J

  26. If the reaction takes place in aqueous solution, and the solutions are placed in an insulated container, the specific heat capacity of the solutions is 4.18 J oC-1 g-1. The densities of these aqueous solutions can be taken as 1.0 g mL-1 so that 100 mL would have a mass of 100 g.

  27. Energy per mole E = m c ΔT = 50g X 4.18 J oC-1 g-1 x 6.8 oC = 1421.2 J mol of NaOH (n) = 0.025L x 1molL-1 = 0.025 mol ΔrH

  28. Calculating Enthalpy From Heats of Formation Given the enthalpies of formation of all the compounds in a given reaction, we can calculate the enthalpy of reaction using the following formula ΣrH = ΣnfH (products) - ΣnfH (reactants) (the symbol Σ means sum of)

  29. Calculating Enthalpy From Heats of Formation • ΣrH = ΣnΔfH  (products) – ΣnΔfH (reactants) • These calculations are straight forward providing you remember the following: • The formula only works if the heats of formationfor all the compounds are used • Set your work out very carefully, paying particular attention to the + and - signs What does the “n” stand for? Always check your answers!

  30. Example • Using the standard heats of formation of CO2(g), H2O(l), and C6H12O6(s), calculate the standard enthalpy of combustion of glucose. • fHo(CO2, g) = -394 kJ mol-1 • fHo(H2O, l) = -286 kJ mol-1 • fHo(C6H12O6, s) = -1268 kJ mol-1

  31. fHo(CO2, g) = -394 kJ mol-1 fHo(H2O, l) = -286 kJ mol-1 fHo(C6H12O6, s) = -1268 kJ mol-1 Note - Start by writing an equation for the combustion of 1 mole of glucose. C6H12O6(s) + 6O2(g)  6CO2(g) + 6H2O(l) rHo =  nfHoproducts -  nfHoreactants rHo = ( 6 x -394 + 6 x -286) - ( -1268 + 0) = - 2812 kJ mol-1

  32. Ethanoic acid can be formed by the oxidation of ethanol. C2H5OH(l) + O2(g)  CH3COOH(l) + H2O(l) This reaction occurs when wine goes sour. By calculating the standard reaction enthalpy for this oxidation reaction, decide whether the reaction is exothermic or endothermic given the following data: fHo(H2O, l) = -286 kJ mol-1 fHo(CH3COOH, l) = -485 kJ mol-1 fHo(C2H5OH, l) = -278 kJ mol-1 rHo =  nfHoproducts -  nfHoreactants rHo = (- 485 + - 286) - (-278) = - 493 kJ mol-1 Therefore exothermic

  33. Exercises • 1. a) Using the standard heats of formation given calculate the enthalpy change for the following reaction. • 2SO2(g) + O2(g)  2SO3(g) • fHo(SO2, g) = -297 kJ mol-1 • fHo(SO3, g) = -396 kJ mol-1 rHo =  nfHoproducts -  nfHoreactants rHo = (2 x -396) - (2 x -297) rHo = (2 x -396) - (2 x -297) = - 198 kJ mol-1

  34. b) What is the enthalpy change when 100 g of SO2 burn to form SO3? 2SO2(g) + O2(g)  2SO3(g) fHo(SO2, g) = -297 kJ mol-1fHo(SO3, g) = -396 kJ mol-1 rHo =  nfHoproducts -  nfHoreactants rHo = (1.563 x -396) - (1.563 x -297) = - 154 .7 kJ = - 155 kJ(3SF)

  35. Calculate fHo(PCl5, s) from the following information. fHo(PCl3, l) = -320 kJ mol-1 PCl3(l) + Cl2(g)  PCl5(s) rH = -124 kJ mol-1 rHo =  nfHoproducts -  nfHoreactants This is the unknown (-124) = fHo - (-320) (-124) = fHo + 320 fHo = - 444 kJ mol-1

  36. Tricky enthalpy calculations 1. Calculate the final temperature when 50g of water at 70 C is mixed with 150g of water at 40 C ? Energy loss by hot water = Energy gained by cool water E lost = E gained (conservation of energy) ∆H(loss) = ∆H (gained) m c ∆T = m c ∆T m c ( 70 – T) = m c (T - 40) Hot water will cool Cool water will heat up

  37. Tricky enthalpy calculations m c ∆T = m c ∆T m c ( 70 – T) = m c (T - 40) Hot water will cool Cool water will heat up Solve for T

  38. Hess’s Law If an overall reaction can be broken down into a series of two or more steps, then the corresponding overall enthalpy of reaction is the sum of the enthalpies of the individual reaction steps. None of the steps need to be a reaction that can be carried out in the laboratory. Another way of saying this is that the energy difference depends only on the difference in energy between the reactants and products, not on the reaction path taken.

  39. The process of photosynthesis is an endothermic process in which energy from the sun is trapped and stored in the bonds of glucose. 6CO2(g) + 6H2O(l)  C6H12O6(aq) + 6O2(g) rH0 = +2808kJ mol-1 It is however easier to measure the enthalpy change for the reverse reaction, the combustion of glucose (i.e. the process of respiration). C6H12O6(aq) + 6O2(g)  6CO2(g) + 6H2O(l) rH0 = -2808kJ mol-1 Remember if you reverse the reaction reverse the sign of rH0

  40. reactants Hess’s Law illustrated Reaction 1, H1 Reaction 2, H2 Enthalpy, H the energy difference depends only on the difference in energy between the reactants and products, not on the reaction path taken. rH total Reaction 3, H3 products rH total = rH1 - rH2 + rH3

  41. As a result of this principle of conservation of energy we can say: Chemical reactions and their corresponding H values can be added , subtracted and multiplied as if they were algebraic equations

  42. Another illustration of Hess’s law is: + ½ O2 (g) CO (g) C(s) H o3 H o1 + ½ O2 H o2 + O2 H o 1=H o2 + H o3 CO2 (g) In practice it is impossible to measure the value of H o3 (the heat of formation of CO(g)) directly as some CO2 gas is always formed as a by product. However we can calculate it using Hess’s law. Using algebra and the following info find H o3 H o1 = -393kJmol-1 andH o2 = -282kJmol-1

  43. + ½ O2 (g) CO (g) C(s) H o3 H o1 + ½ O2 H o2 + O2 Answer H o 3 =H o1 - H o2 = -393 – (-282) = -111kJ mol-1 CO2 (g) How’d ya go?

  44. Solving Problems Using Hess’s Law • Write the data in the form of equations • Rewrite the equations to give the desired species on the correct side of the equation. If the reaction must be reversed (perhaps because we require a species to be a reactant and not a product) then the sign of the H must also be reversed. • Add the equations, and the H together • Check your working!!

  45. Hang on to your hats! Here we go………

  46. Hess’s Law Calculation • Calculate the heat of formation of CS2(l) given that the heats combustion of carbon, sulfur and carbon disulfide. Sulfur burns to SO2 in oxygen • -393, -297 and -1007 kJ mol -1 respectively • Write the data given in the form of equations • C (s) + O2(g) CO2 (g) H = -393 kJ mol -1 • S (s) + O2 (g) SO2 (g) H = -297 kJ mol -1 • CS2(s) + 3O2 (g) CO2 (g) + 2SO2 (g) H = -1007 kJ mol -1 • The equation we are looking for is C(s) + 2S(s) CS2 (l) • Now rewrite the equations to give the desired species on the correct side of the equation

  47. Hess’s Law Calculation The equation we are looking for is C(s) + 2S(s) CS2 (l) Rewrite the equations to give the desired species on the correct side of the equation • C (s) + O2(g) CO2 (g) H = -393 kJ mol -1 • 2S (s) + 2O2 (g) 2SO2 (g) H = ( 2 x -297) = - 594 kJ mol -1 • CO2 (g) + 2SO2 (g) CS2(s) + 3O2 (g)H = +1007 kJ mol -1 Now add equations and the H values together cancelling where appropriate • C (s) + O2(g) CO2 (g) H = -393 kJ mol -1 • 2S (s) + 2O2 (g) 2SO2 (g) H = -594 kJ mol -1 • CO2 (g) + 2SO2 (g) CS2(s) + 3O2 (g)H = +1007 kJ mol -1 Now add C(s) + 2S(s) CS2 (l) H = +20 kJ mol -1 Therefore the fH o =(CS2(l)) is +20 kJ mol -1

  48. Important • All booklets are to be handed in worksheets 1 (8),2 (9),3 (10), 4 (11) and 5 (12) Change • All worksheets must be completed and marked before next Monday as all booklets are handed in on day of test (Monday) • Read pathfinder thermochemistry page 54 to help understanding and bestchoice for extra practice

  49. Hess’s Law calculation

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