This document discusses various topics in thermochemistry including: - Enthalpy changes in chemical reactions and how they are measured using calorimetry. Exothermic and endothermic reactions are explained. - Hess's law, which states that the enthalpy change of a reaction is independent of the reaction pathway. It can be used to calculate enthalpy changes. - Standard enthalpies of formation and how they allow calculation of enthalpy changes using Hess's law and bond dissociation enthalpies. - Measuring enthalpy changes using bomb calorimetry and coffee cup calorimetry. Limitations of each method are discussed.

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

5. 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). 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 .8H 2 O(s) and NH 4 Cl(s) (video) Endothermic Exothermic S u r r o u n d I n g s S u r r o u n d I n g s System System Heat Heat Heat Heat

6. Enthalpy level diagrams 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 NH 3 (g), H 2 O(l) and BaCl 2 (s) and that  r H = +135 kJ mol -1 Reaction 2 is a straightforward neutralisation with a  r H = -55 kJ mol -1 The reaction between sodium metal and water – metal floats on water – effervescent reaction moves metal around- yellow flame above the metal- no solid residue 2 mol Na(s) + 2 mol H 2 O(l) Enthalpy, H (kJ) 2 mol NaOH(aq) + 1 mol H 2 (g) Δ H = -367.5 kJ (367.5 kJ of heat is released

14. An Example: 25 cm 3 of 2.00 mol dm -3 HCl(aq) is mixed with 25 cm 3 of 2.00 mol dm -3 NaOH(aq). The temperature rises from 22.5 o C to 34.5 o C. 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 cm 3 of aqueous copper sulfate solution at 22.8 o C. The copper sulfate is in excess of that needed to react all the zinc. The temperature rises to 32.3 o C. Calculate  H for the following reaction:

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

18. A bomb calorimeter How can we accurately measure enthalpy changes of combustion reactions? Needle Gas inlet Insulated jacket Steel bomb + - O 2 Thermometer Current for ignition coil Stirrer Ignition coil Graphite sample

19. Bomb Calorimetry- measurements Some heat from reaction warms “bomb” q bomb = heat capacity x ∆T Total heat evolved, q total = q water + q bomb Total heat from the reaction =qtotal Needle Gas inlet Steel bomb Some heat from reaction warms water q water = mc∆T Insulated jacket + - O 2 Thermometer Current for ignition coil Stirrer Ignition coil Graphite sample

21. Step 1 : energy transferred from reaction to water. q = (4.184 J g -1 K -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 Calculating enthalpy changes from calorimetry data

22. Video Click to link to “Thermochemistry” video

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

30. Standard Enthalpy of Formation ∆ f H o (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) H 2 (g) + 1/2O 2 (g) H 2 O(g) ∆ f H o (H 2 O, g) = -241.8 kJ mol -1 ∆ f H o is zero for elements in their standard states.

31. Standard enthalpies of formation and Hess’s Law can be used to calculate unknown ∆ r H o ∆ r H o =  ∆H f o (products) -  ∆H f o (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? Calculating Enthalpy Changes using standard enthalpies of formation

32. Calculate ∆ c H o for methanol Standard state of methanol at 298 K is liquid CH 3 OH(l) + 3/2O 2 (g) CO 2 (g) + 2H 2 O(l) ∆ c H o =  ∆H f o (products) -  ∆H f o (reactants) = {∆H f o (CO 2 ) + 2 ∆H f o (H 2 O)} - {3/2 ∆H f o (O 2 ) + ∆H f o (CH 3 OH)} = {(-393.5 kJ) + 2 (-285.8 kJ)} - {0 + (-238.9 kJ)} ∆ c H o (298K)= -726.2 kJ mol -1 Now try the problems on the separate sheet Calculating Enthalpy Changes

33. Video Link to “Hess’s Law” video

34. Some Other Problems to do

35. Describe the reaction

39. FRS1027 Introductory Chemistry Bond Dissociation Enthalpies

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 ??? -818 Tota enthalpy change of combustion kJ mol -1 -3466 Total energy out/kJ mol -1 2648 Total energy in/kJ mol -1 -1856 4 0 0 464 O-H 0 0 996 2 498 O=O -1610 2 0 0 805 C=O 0 0 1652 4 413 C-H 0 0 0 0 347 C-C Energy out KJ mol -1 Number made Energy in KJ mol -1 Number broken Bond energy Bond type CH 4 Compound Enthalpy changes of combustion for hydrocarbons