1. Endo/Exo (again) Unit 6 Topic 3

2. Changes in Heat oChemical reactions are accompanied by changes in heat, H. oReactions that are endothermic have a positive H (+), oreactions which are exothermic have a negative H (-)

3. HH oH represents the difference between the Hprod (heat of products) and the Hreact (heat of the reactants). oH = Hprod - Hreact

4. KJ o Heat is expressed in units of KJ (kilojoules) for chemical reactions.

5. Endothermic In endothermic reactions heat is absorbed therefore heat (KJ) is a reactant. For endothermic reactions the heat term (KJ) must be placed on the left-hand (reactants) side of the equation.

6. Examples o2X + Y + Heat  C (H +) o2A + 3B + 300KJ  6AB (H = + 300KJ)

7. Exothermic In exothermic reactions heat is released therefore heat (KJ) is a product. For exothermic reactions the heat term (KJ) must be placed on the right-hand (products) side of the equation.

8. Examples oC + 3D  2A + Heat (H -) o2Z  X + Y + 150KJ (H = -150KJ)

9. Specific Heat When the specific amount of heat (in KJ) is known, then the amount of heat required (endothermic) or produced (exothermic) in a chemical reaction can be calculated from the mass or number of moles of any substance in the reaction.

10. Exothermic synthesis of ammonia o1 N 2 + 3 H 2  2 NH 3 + 160 KJ oH = -160 KJ

11. Exothermic synthesis of ammonia o1 N 2 + 3 H 2  2 NH 3 + 160 KJ oFrom the balanced equation we know the following: o1 mole N 2 = 160 KJ of heat o3 moles of H 2 = 160 KJ of heat

12. Problem oSo how much heat would you produce from reacting 1.00 moles of H2? X KJ = 160 KJ 1.00 moles H 2 3 moles H 2 X = 53.3 KJ

13. Try this one: oSo how much heat would you produce from reacting 0.25 moles of N2?

14. Try this one: oSo how much heat would you produce from reacting 0.25 moles of N2? o X KJ = 160 KJ 0.25 mole N 2 1 moles N 2 oX = 40.0 KJ

15. Mass In the lab we cannot measure out moles we measure out mass. Therefore you need to be able to determine the heat produced from a given mass of a reactant.

16. Harder Problem o1 N 2 + 3 H 2  2 NH 3 + 160 KJ H = -160 KJ oHow much heat would be evolved when 5.00g of hydrogen is reacted completely? oStep 1: Change grams to mole oStep 2: Use Heat Ratio

17. STEP 1: 5.00 g H 2 = ? moles o5.00 grams = 2 grams x moles 1 mole 2X = 5 2X/2 = 5/2 X = 2.50 moles

18. Step 2: Heat Ratio o1 N 2 + 3 H 2  2 NH 3 + 160 KJ H = -160 KJ o3 moles H 2 = 160 KJ oHow much heat is produced if there are 2.50 moles? o X KJ = 160 KJ 2.5 moles 3 mole 3X = 2.5 * 160 X =133.33 KJ

19. Change in Temperature Changes in heat result in a change in temperature. This is why we monitor heat changes by measuring temperature. We cannot measure heat directly!

20. Endo/Exo changes oFor endothermic reactions, heat is absorbed by the reaction (taken away from the thermometer) so the temperature decreases. oFor exothermic reactions, heat is released (to the thermometer) so the temperature increases.

21. Temperature Temperature measures the average kinetic energy (motion) of the particles of matter. In chemistry we use two different temperature scales, Celsius (o C) and Kelvin (K). It is important to know when to use each and how to convert between them.

22. Celsius to Kelvin oFreezing point of water, 0 o C = 273 K oBoiling point of water, 100 o C = 373 K o Therefore, 0 K = - 273 o C

23. Converting Celsius to Kelvin oTo convert from Celsius to Kelvin: Add 273 oDetermine the Kelvin equivalent of 25 o C o25 o C + 273 = 298 K

24. Converting Kelvin to Celsius oTo convert from Kelvin to Celsius: Subtract 273 oDetermine the Celsius equivalent of 100 K o100 K - 273 = - 173 o C

25. Try a few o35 o C = ___________K o297 K = _________ o C o-35 o C = __________K o45 K = ___________ o C

26. Try some more oN 2 + O 2 + 189.6 KJ  2NO oH = ? KJ (endo or exo?) oH =189.6 KJ oHow much heat would you need to produce 10 grams of NO?

27. oN 2 + O 2 + 189.6 KJ  2NO oHow much heat would you need to produce 10 grams of NO? o31.60