1. Enthalpies of Reactions and Hess’ Law Advanced Chemistry Ms. Grobsky

2. THERMODYNAMICS - ENTHALPY CHANGES • Recall that: • Enthalpy is a measure of the heat that is released or absorbed in a chemical reaction at constant pressure • You cannot measure the actual enthalpy of a substance • You can measure an enthalpy CHANGE • Written as the symbol DH , “delta H ”

3. STANDARD ENTHALPY CHANGES • Why is there a standard? • Enthalpy values vary according to the conditions • A substance under these conditions is said to be in its standard state: • Pressure 1 atmosphere • Temperature 298K (25°C) • Concentration mol/L • As a guide, just think of how a substance would be under normal lab conditions • Assign the correct subscript [(g), (l) or (s) ] to indicate which state it is in • To tell if standard conditions are used we modify the symbol for DH: • Enthalpy Change Standard Enthalpy Change • (at 298K)

4. STANDARD HEATS (ENTHALPIES) OF FORMATION • Definition • The enthalpy change when ONE MOLE of a compound is formed in its standard state from its elements in their standard states • Symbol • DH°f • Values • Usually, but not exclusively, exothermic • Example(s) • C(graphite) + O2(g) ———> CO2(g) • H2(g)+½O2(g) ———>H2O(l) • 2C(graphite) + ½O2(g) + 3H2(g) ———>C2H5OH(l) • Note: • Only ONE MOLE of product on the right-hand side of the equation • ElementsIn their standard states have zero enthalpy of formation

5. STANDARD HEATS (ENTHALPIES) OF FORMATION • What good are they? • The heat of reaction (∆H) can be calculated by subtracting the heats of formation of the reactions FROM the products • Where can I find the values to use? • A thermochemical table!

6. STANDARD HEATS (ENTHALPIES) OF COMBUSTION • Definition • The enthalpy change when ONE MOLE of a substance undergoes complete combustion under standard conditions. All reactants and products are in their standard states • Symbol • DH°comb • Values • Always exothermic • Example(s) • C(graphite) + O2(g) ———> CO2(g) • H2(g)+½O2(g) ———>H2O(l) • C2H5OH(l) + 3O2(g) ———> 2CO2(g) + 3H2O(l) • Notes • Always only ONE MOLE of what you are burning on the LHS of the equation • To aid balancing the equation, remember: • You get one carbon dioxide molecule for every carbon atom in the original and onewater molecule for every two hydrogen atoms • When you have done this, go back and balance the oxygen

7. OTHER HEATS (ENTHALPIES) OF REACTION • Heat of Fusion • The heat absorbed by ONE MOLE of a substance in melting from a solid to a liquid • DH°fus • If ONE MOLE of a substance releases heat in freezing from a liquid to a solid, DH°fusis negative! • Heat of Neutralization • We will soon find out in lab! • Heat of Vaporization • The amount of heat necessary to vaporize ONE MOLE of a given liquid • Heat of Condensation • The amount of heat released when ONE MOLE of a vapor condenses • Heat of Solution • Heat change caused by dissolution of ONE MOLE of substance

8. Relation of Enthalpy to Bonds • Theory • Imagine that, during a reaction, all the bonds of reacting species are broken and the individual atoms join up again but in the form of products • Breaking a bond REQUIRES energy • ENDOTHERMIC • Forming a bond RELEASES energy • Exothermic • The overall energy change will depend on the difference between the energy required to break the bonds and that released as bonds are made • If energy released making bonds > energy used to break bonds • EXOTHERMIC • Products are lower in energy than the reactants • If energy used to break bonds > energy released making bonds • ENDOTHERMIC • Products are greater in energy than reactants

9. ENTHALPY REACTION CO-ORDINATE THERMODYNAMICS - ENTHALPY CHANGES • Turn to FRONT of page 165 • Enthalpy change (DH) = Enthalpy of products - Enthalpy of reactants • Enthalpy of reactants > products Enthalpy of reactants < products • DH = - DH = + • EXOTHERMICHeat given outENDOTHERMICHeat absorbed ENTHALPY REACTION CO-ORDINATE

10. Enthalpy and Hess’ Law

11. Enthalpy and Hess’ Law • Enthalpy is a state function • As such, DH for going from some initial state to some final state is pathway independent • Hess’ Law • DH for a process involving the transformation of reactants into products is not dependent on pathway • Therefore, we can pick any pathway to calculate DH for a reaction

12. Hess’s Law Start Finish • Both lines accomplished the same result, they went from start to finish • Net result = same

13. Campsite to Illustrate Altitude as a State Function

14. Hess’s Law • Hess’s law of heat summation states that for a chemical equation that can be written as the sum of two or more steps, the enthalpy change for the overall equation is the sum of the enthalpy changes for the individual steps.

15. A →B + C H = x B + C → D H = y H = ? A → D H = x + y

16. HESS’S LAW • “The enthalpy change is independent of the path taken” • How • The enthalpy change going from • A to B can be found by adding the • values of the enthalpy changes for • the reactions A to X, X to Y and Y to B. • DHr = DH1 + DH2 + DH3

17. HESS’S LAW • “The enthalpy change is independent of the path taken” • How • The enthalpy change going from • A to B can be found by adding the • values of the enthalpy changes for • the reactions A to X, X to Y and Y to B. • DHr = DH1 + DH2 + DH3 • If you go in the opposite direction of an arrow, you subtract the value of • the enthalpy change • DH2= -DH1+ DHr-DH3 • The values of DH1 andDH3 have been subtracted because the route involves going in the opposite direction to their definition

18. HESS’S LAW • “The enthalpy change is independent of the path taken” • Use • Applying Hess’s Law enables one to calculate enthalpy changes from • other data, such as: • Changes which cannot be measured directly • Lattice Enthalpy • Enthalpy change of reaction from bond enthalpy • Enthalpy change of reaction from DH°c • Enthalpy change of formation from DH°f

19. Enthalpy diagram illustrating Hess’s law

20. Could you use these data to obtain the enthalpy change for the following reaction? Hess’s Law • For example, suppose you are given the following data:

21. Hess’s Law • If we multiply the first equation by 2 and reverse the second equation, they will sum together to become the third.