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# More Heat Calculations

More Heat Calculations. What have we done?. What have we done?. We can figure out heat values and then put them into kJ / mole. 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!.

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## More Heat Calculations

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1. More Heat Calculations What have we done?

2. What have we done? • 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 CH4(g) + 2O2(g) ==> CO2(g) + 2H2O (g) H = ? Find the Standard Heat of Formation for each substance. Hºf

7. Hºf • Hºf  how it exists at 1 atm and 25ºC.

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

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

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

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

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

13. Hºf • Remember! Hºf is the change in enthalpy that accompanies the formation of 1 mole of a compound from its elements. Therefore: In order to find the Hºf of a compound, you need to break it up into individual elements! Ca(s) + S(s) + 2O2(g)  CaSO4(s)

14. Hºf CH4(g) + 2O2(g) ==> CO2(g) + 2H2O(g) H = ? Look up values. Make sure you have correct states of matter.

15. Hºf CH4(g) + 2O2(g) ==> C02(g) + 2H20(l) H = ? ((-75 kJ) + 2(0kJ)) ==> ((-393.5kJ) + 2(-242kJ))

16. Hºf CH4(g) + 2O2(g) ==> C02(g) + 2H20(l) H = ? ((-75 kJ) + 2(0kJ)) ==> ((-393.5kJ) + 2(-242kJ)) -75 kJ ==>-393.5 kJ + -484kJ -877.5 kJ

17. Hºf CH4(g) + 2O2(g) ==> C02(g) + 2H20g) 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. Hºf Problems Calculate the standard change in enthalpy for the following reaction. • 2 Al(s) + Fe203(s)  Al203(s) + 2 Fe(s)

19. One More • 2) Students tend not to like this format! a) Write the combustion reaction for methanol. • 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 CO2 (g) = -393.5 kJ/mole H°f for H2O (g) = -242 kJ/mole

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