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Endo/Exo (again)

Endo/Exo (again)

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Endo/Exo (again)

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

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

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

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

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

  6. Examples • 2X + Y + Heat  C (H +) • 2A + 3B + 300KJ  6AB (H = + 300KJ)

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

  8. Examples • C + 3D  2A + Heat (H -) • 2Z  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 • 1 N2 + 3 H2 2 NH3 + 160 KJ • H = -160 KJ

  11. Exothermic synthesis of ammonia • 1 N2 + 3 H2 2 NH3 + 160 KJ • From the balanced equation we know the following: • 1 mole N2 = 160 KJ of heat • 3 moles of H2 = 160 KJ of heat

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

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

  14. Try this one: • So how much heat would you produce from reacting 0.25 moles of N2? • X KJ = 160 KJ 0.25 mole N2 1 moles N2 • X = 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 • 1 N2 + 3 H2 2 NH3 + 160 KJ H = -160 KJ • How much heat would be evolved when 5.00g of hydrogen is reacted completely? • Step 1:Change grams to mole • Step 2:Use Heat Ratio

  17. STEP 1: 5.00 g H2 = ? moles • 5.00 grams = 2 grams x moles 1 mole 2X = 5 2X/2 = 5/2 X = 2.50 moles

  18. Step 2: Heat Ratio • 1 N2 + 3 H2 2 NH3 + 160 KJ H = -160 KJ • 3 moles H2 = 160 KJ • How much heat is produced if there are 2.50 moles? • 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 • For endothermic reactions, heat is absorbed by the reaction (taken away from the thermometer) so the temperature decreases. • For 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 • Freezing point of water, 0oC = 273 K • Boiling point of water, 100oC = 373 K • Therefore, 0 K = - 273oC

  23. Converting Celsius to Kelvin • To convert from Celsius to Kelvin: Add 273 • Determine the Kelvin equivalent of 25oC • 25 oC + 273 = 298 K

  24. Converting Kelvin to Celsius • To convert from Kelvin to Celsius: Subtract 273 • Determine the Celsius equivalent of 100 K • 100 K - 273 = - 173 oC

  25. Try a few • 35oC = ___________K • 297 K = _________ oC • -35oC = __________K • 45 K = ___________oC

  26. Try some more • N2 + O2 + 189.6 KJ 2NO • H = ? KJ (endo or exo?) • H =189.6 KJ • How much heat would you need to produce 10 grams of NO?

  27. N2 + O2 + 189.6 KJ 2NO • How much heat would you need to produce 10 grams of NO? • 31.60