This work is licensed under a Creative Commons Attribution-Noncommercial-Share Alike 2.0 UK: England & Wales License

1. Thermochemistry University of Lincoln presentation This work is licensed under a Creative Commons Attribution-Noncommercial-Share Alike 2.0 UK: England & Wales License

2. Thermochemistry Enthalpy changes in chemical reactions (+ video) Enthalpy Diagrams Thermochemical Equations Calorimetry and measuring enthalpy changes

3. Energy and Chemistry Petrol bombs What does this show? How to ensure your bonfire burns! Why does this happen?

4. Energy and Chemical Reactions • Formation of new substances • Redox reactions • Acid-base reactions • Precipitation reactions • Energy released/absorbed • Light (chemiluminescence) • Electrical energy (electrochemistry) • Heat (thermochemistry)

5. Reaction 1 Energy products>reactants so energy is absorbed from the surroundings Heat is lost from the surroundings so Temperature (surroundings) decreases - Endothermic e.g. Ba(OH)2.8H2O(s) and NH4Cl(s) (video) Heat Heat System Endothermic S u r r o u n d I n g s S u r r o u n d I n g s Reaction 2 Energy products<reactants so energy flows as heat from the system to the surroundings Temperature (surroundings) increases - Exothermic e.g. NaOH(aq) and HCl(aq). Heat Heat System Exothermic

6. Enthalpy level diagrams 2 mol Na(s) + 2 mol H2O(l) ΔH = -367.5 kJ (367.5 kJ of heat is released Enthalpy, H (kJ) 2 mol NaOH(aq) + 1 mol H2(g) The reaction between sodium metal and water – metal floats on water – effervescent reaction moves metal around- yellow flame above the metal- no solid residue For You To Do Draw diagrams for the reactions on the previous slide You will need to write balanced chemical equations first. For reaction1 assume that the products are NH3(g), H2O(l) and BaCl2(s) and that rH = +135 kJ mol-1 Reaction 2 is a straightforward neutralisation with a rH = -55 kJ mol-1

7. Thermochemical Equations Athermochemical equationis the chemical equation for a reaction (including state symbols) and the enthalpy of reaction for the molar amounts as given by the equation written directly after the equation.

8. Thermochemical Equations Why do we need state symbols? In athermochemical equationit is important to note state symbols because the enthalpy change depends on the physical state of the substances.

9. Two important rules Thermochemical Equations • When a thermochemical equation is multiplied by any factor, the value of H for the new equation is obtained by multiplying the DH in the original equation by that same factor. • When a chemical equation is reversed, the value of DH is reversed in sign.

10. What is the enthalpy change of reaction for the formation of 1 mole and 6 moles of water? -285.9 kJ mol-1; -1715.1 kJ mol-1 What is the enthalpy change for the splitting of 1 mole of water into hydrogen and oxygen gas? +285.9 kJ mol-1 Using Thermochemical Equations

11. Using Thermochemical Equations Consider the reaction of methane, CH4, burning in the presence of oxygen at constant pressure. Given the following equation, how much heat could be obtained by the combustion of 10.0 grams CH4? 1 g of methane would give Combustion of methane gives 55.6 kJ g-1 10 g of methane would give

12. Coffee-cup calorimeter Measuring enthalpy changes Thermometer HCl(aq) 2 polystyrene coffee cups NaOH (aq) • Measuring enthalpy changes is called calorimetry • Carry out the reaction in a calorimeter and measure the temperature change. • Calculate the energy transferred during the reaction from the temperature change. • Also require the mass of the substance and the specific heat capacity • Assumptions • All the energy change is transferred to the solution (water) • No losses of heat to the other surroundings

13. Need to knowSpecific Heat Capacity Definition The amount of energy required to raise the temperature of a specified mass of an object (substance) by 1 degree kelvin (K) units J g -1K-1 or J kg-1 K-1 Important example Water 4.184 J g-1 K-1

14. An Example: 25 cm3 of 2.00 mol dm-3 HCl(aq) is mixed with 25 cm3 of 2.00 mol dm-3 NaOH(aq). The temperature rises from 22.5 oC to 34.5 oC. Find the enthalpy change for the reaction 2508 J of heat is transferred from the reaction of 0.05 mol HCl with 0.05 mol NaOH

15. An Example continued For 1 mol of HCl and NaOH

16. Now Try This One 0.327 g of Zinc powder is added to 55 cm3 of aqueous copper sulfate solution at 22.8 oC. The copper sulfate is in excess of that needed to react all the zinc. The temperature rises to 32.3 oC. Calculate H for the following reaction:

17. Thermometer HCl(aq) 2 polystyrene coffee cups NaOH (aq) Limitations of this method ?? Coffee-cup calorimeter

18. A bomb calorimeter + Thermometer - Stirrer Current for ignition coil O2 Graphite sample Ignition coil How can we accurately measure enthalpy changes of combustion reactions? Needle Gas inlet Insulated jacket Steel bomb

19. Bomb Calorimetry- measurements + Thermometer - Stirrer Current for ignition coil O2 Graphite sample Ignition coil Some heat from reaction warms water qwater = mc∆T Needle Gas inlet Some heat from reaction warms “bomb” qbomb = heat capacity x ∆T Insulated jacket Steel bomb Total heat evolved, qtotal = qwater + qbomb Total heat from the reaction =qtotal

20. Calculate enthalpy of combustion of octane. C8H18(l) + 25/2 O2(g) 8CO2(g) + 9H2O(l) Burn 1.00 g of octane Temp rises from 25.00 to 33.20 oC Calorimeter contains 1200 g water Heat capacity of bomb = 837 J K-1 Calculating enthalpy changes from calorimetry data

21. Calculating enthalpy changes from calorimetry data Step 1: energy transferred from reaction to water. q = (4.184 J g-1K-1)(1200 g)(8.20 K) = 41170 J Step 2: energy transferred from reaction to bomb. q = (bomb heat capacity)(∆T) = (837 J K-1)(8.20 K) = 6860 J Step 3:Total energy transferred 41170 J + 6860 J = 48030 J Heat of combustion of 1.00 g of octane = - 48.0 kJ For 1 kg = -48 MJ kg-1 H=-48 kJ x 114 g mol-1=-5472 kJ mol-1

22. Video Click to link to “Thermochemistry” video

23. A case study- Self-heating cans Can Insert Quicklime Foil separator Button Water

24. The Chemistry CaO(s) + H2O(l) Ca(OH)2(s) quicklime slaked lime Water and quicklime packaged separately When mixed, exothermic reaction takes place and the temperature of the water increases Heat transferred to the drink rH = -65.1 kJ mol-1

25. How much quicklime is needed to heat up a coffee can? Think about what information you need to know for the calculation before doing the calculation – do some research and find approximate values Homework Draw an enthalpy level diagram for the reaction Do the calculation

26. FRS1027 Introductory Chemistry Hess’s law Standard enthalpy of formation Calculating enthalpy changes

27. Hess’s Law The enthalpy change on going from reactants to products is independent of the reaction path taken Can be used to calculate enthalpy changes

28. Hess’s Law & Energy Level Diagrams Reaction can be shown as a single step or in a two steps. ∆Htotal is the same no matter which path is followed. C(s) + O2(g) ΔH1 = -110.5 kJ CO(g) + ½ O2(g) ΔH3 = ΔH1 + ΔH2 = -393.5 kJ Energy ΔH2 = -283.0 kJ ∆H reaction path 1=∆H reaction path 2 CO2(g)

29. Standard enthalpy values (Ho) • Initial and final species are in their standard states • The standard state of a substance at a specified temperature is its pure form at 1 bar (100 kPa). T is usually 298 K (25 oC) but not always • rHo(298 K) is the standard enthalpy of reaction at 298 K • Some Physical States at 298 K • C = graphite; O2 = gas; CH4 = gas; H2O = liquid

30. Standard Enthalpy of Formation ∆fHo (298 K) = standard molar enthalpy of formation at 298 K enthalpy change when 1 mole of compound is formed from elements in their standard states at 298 K (These are available from data books) H2(g) + 1/2O2(g) H2O(g) ∆fHo (H2O, g) = -241.8 kJ mol-1 ∆fHo is zero for elements in their standard states.

31. Calculating Enthalpy Changes using standard enthalpies of formation Standard enthalpies of formation and Hess’s Law can be used to calculate unknown ∆rHo ∆rHo =  ∆Hfo(products) -  ∆Hfo(reactants) Enthalpy of reaction = sum of the enthalpies of formation of the products (correct molar amounts) – sum of the enthalpies of formation of the reactants (correct molar amounts) Why is this an application of Hess’s law?

32. Calculating Enthalpy Changes Calculate ∆cHo for methanol Standard state of methanol at 298 K is liquid CH3OH(l) + 3/2O2(g) CO2(g) + 2H2O(l) ∆cHo =  ∆Hfo(products) -  ∆Hfo(reactants) = {∆Hfo(CO2) + 2 ∆Hfo(H2O)} - {3/2 ∆Hfo(O2) + ∆Hfo(CH3OH)} = {(-393.5 kJ) + 2 (-285.8 kJ)} - {0 + (-238.9 kJ)} ∆cHo(298K)= -726.2 kJ mol-1 Now try the problems on the separate sheet

33. Video Link to “Hess’s Law” video

34. Some Other Problems to do

35. Describe the reaction

36. The enthalpy of reaction for black powder Black powder is a mixture of potassium nitrate (75%), charcoal (13%) and sulfur (12%). A simplified equation is: fHo values/kJ mol-1 KNO3(s) -494.6 K2S(s) -380.70 CO2(g) -393.51 Calculate the enthalpy of reaction in kJ mol-1 and kJ kg-1 of black powder.

37. The reaction of barium hydroxide with ammonium chloride Equation fHo values/ kJ mol-1 Ba(OH)2.8H2O(s) -3345 NH4Cl(s) -314 NH3(g) -46 H2O(l) -286 BaCl2(s) -859 Calculate rHo

38. Calculate the standard enthalpy of combustion of octane at 298 K fHo (octane)= -249.9 kJ mol-1

39. FRS1027 Introductory Chemistry Bond Dissociation Enthalpies

40. Definition The bond dissociation enthalpy (dissH) for an X-X diatomic molecule refers to the process: X2(g) 2X(g) at a given temperature (often 298 K) dissH = D(X-X) is the bond enthalpy for a specific process (a particular bond in a molecule) Breaking bonds is an endothermic process

41. D and D CH4(g) CH3(g) + H(g)H=436 kJ mol-1 CH3(g) CH2(g) + H(g)H=461 kJ mol-1 CH2(g) CH(g) + H(g)H=428 kJ mol-1 CH(g) C(g) + H(g)H=339 kJ mol-1 D depends on the bond D is an average value and is obtained from CH4(g) C(g) + H(g)H=1664 kJ mol-1 D (C-H) = 416 kJ mol-1

42. Enthalpy Changes in Chemical Reactions • Enthalpy difference between products and reactants because different chemical bonds are formed. • Enthalpy change can be estimated from the chemical bonds that are broken and made. • Breaking bonds is an endothermic process and making bonds is an exothermic process • Only an estimate because • Bond enthalpies are mean values • All species are in the gaseous state • A Hess’s law problem

43. Now have a go at the bond enthalpy problems and set up as an Excel spreadsheet. Can you set it up so that you only need to enter the number of C and H atoms to calculate the enthalpy change ???

44. Acknowledgements • JISC • HEA • Centre for Educational Research and Development • School of natural and applied sciences • School of Journalism • SirenFM • http://tango.freedesktop.org This work is licensed under a Creative Commons Attribution-Noncommercial-Share Alike 2.0 UK: England & Wales License