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## Endo/Exo (again)

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**Endo/Exo (again)**Unit 6 Topic 3**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(-)**H**• H represents the difference between the Hprod (heat of products) and the Hreact (heat of the reactants). • H = Hprod - Hreact**KJ**• Heat is expressed in units of KJ (kilojoules) for chemical reactions.**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.**Examples**• 2X + Y + Heat C (H +) • 2A + 3B + 300KJ 6AB (H = + 300KJ)**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.**Examples**• C + 3D 2A + Heat (H -) • 2Z X + Y + 150KJ (H = -150KJ)**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.**Exothermic synthesis of ammonia**• 1 N2 + 3 H2 2 NH3 + 160 KJ • H = -160 KJ**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**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**Try this one:**• So how much heat would you produce from reacting 0.25 moles of N2?**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**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.**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**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**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**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!**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.**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.**Celsius to Kelvin**• Freezing point of water, 0oC = 273 K • Boiling point of water, 100oC = 373 K • Therefore, 0 K = - 273oC**Converting Celsius to Kelvin**• To convert from Celsius to Kelvin: Add 273 • Determine the Kelvin equivalent of 25oC • 25 oC + 273 = 298 K**Converting Kelvin to Celsius**• To convert from Kelvin to Celsius: Subtract 273 • Determine the Celsius equivalent of 100 K • 100 K - 273 = - 173 oC**Try a few**• 35oC = ___________K • 297 K = _________ oC • -35oC = __________K • 45 K = ___________oC**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?**N2 + O2 + 189.6 KJ 2NO**• How much heat would you need to produce 10 grams of NO? • 31.60