1 / 27

270 likes | 456 Views

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.

Download Presentation
## Endo/Exo (again)

**An Image/Link below is provided (as is) to download presentation**
Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.
Content is provided to you AS IS for your information and personal use only.
Download presentation by click this link.
While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.
During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

**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

More Related