1. More Heat Calculations What have we done?

2. We can figure out heat values and then put them into kJ / mole

3. What have we done? We can figure out heat values and then put them into kJ / mole We can put heat on the correct side of the equation and then do STOICH!

4. What have we done? We can figure out heat values and then put them into kJ / mole We can put heat on the correct side of the equation and then do STOICH! Hess’s Law problems

5. What have we done? We can figure out heat values and then put them into kJ / mole We can put heat on the correct side of the equation and then do STOICH! Hess’s Law problems Now, let’s calculate heat by just having the equation

6. Heat of formation CH 4 (g) + 2O 2 (g) ==> CO 2 (g) + 2H 2 O (g)  H = ? Find the Standard Heat of Formation for each substance. Hº f

7. Hº f Hº f  how it exists at 1 atm and 25ºC.

8. Hº f Hº f --> how it exist at 1 atm and 25 °C. Get used to seeing °

9. Hº f Hº f  how it exist at 1 atm and 25ºC. Get used to seeing ° For elements and diatomics molecules Hº f = 0

10. Hº f Hº f --> how it exists at 1 atm and 25ºC. Get used to seeing ° For elements and diatomics molecules Hº f =0 The rest - Look at Appendix

11. Hº f Hº f --> how it exists at 1 atm and 25 ºC. Get used to seeing ° For elements and diatomics molecules Hº f = 0 The rest - Look at Appendix Coefficients act as multipliers. WHY? Look at units.

12. Hº f Hº f --> how it exists at 1 atm and 25 ºC. Get used to seeing ° For elements and diatomics molecules Hº f = 0 The rest - Look at Appendix 4 page A21 Coefficients act as multipliers. WHY? Look at units. Always!  H = ∑P - ∑R ∑=sum

13. Hº f Remember! Hº f is the change in enthalpy that accompanies the formation of 1 mole of a compound from its elements. Therefore: In order to find the Hº f of a compound, you need to break it up into individual elements! Ca(s) + S(s) + 2O 2 (g)  CaSO 4 (s)

14. Hº f CH 4 (g) + 2O 2 (g) ==> CO 2 (g) + 2H 2 O(g)  H = ? Look up values. Make sure you have correct states of matter.

15. Hº f CH 4 (g) + 2O 2 (g) ==> C0 2 (g) + 2H 2 0(l)  H = ? ((-75 kJ) + 2(0kJ)) ==> ((-393.5kJ) + 2(-242kJ))

16. Hº f CH 4 (g) + 2O 2 (g) ==> C0 2 (g) + 2H 2 0(l)  H = ? ((-75 kJ) + 2(0kJ)) ==> ((-393.5kJ) + 2(-242kJ)) -75 kJ ==> -393.5 kJ + -484kJ -877.5 kJ

17. Hº f CH 4 (g) + 2O 2 (g) ==> C0 2 (g) + 2H 2 0g)  H = ? ((-75 kJ) + 2(0kJ)) ==> ((-393.5kJ) + 2(-242kJ)) -75 kJ ==> -393.5 kJ + -484 kJ -877.5 kJ  H = -877.5 kJ - (-75 kJ)  H = -802.5 kJ

18. Hº f Problems Calculate the standard change in enthalpy for the following reaction. 2 Al(s) + Fe 2 0 3 (s)  Al 2 0 3 (s) + 2 Fe(s)

19. One More 2) Students tend not to like this format! a) Write the combustion reaction for methanol. b)For the combustion of methanol,  H, is equal to -1454 kJ. Find the heat of formation of methanol given only the information below. H° f for CO 2 (g) = -393.5 kJ/mole H° f for H 2 O (g) = -242 kJ/mole