1. Topic 3.6 Energy

2. HH 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 represents the difference between the H prod (heat of products) and the H react (heat of the reactants). Heat is expressed in units of KJ (kilojoules) for chemical reactions.

3. Endothermic Reactions In endothermic reactions heat is absorbed therefore heat (KJ) is a reactant. Reactants are placed on the left-hand side of the arrow. For endothermic reactions the heat term (KJ) must also be placed on the left-hand (reactants) side of the equation. 2X + Y + Heat  C Endothermic (  H +) or 2A + 3B + 300KJ  2D + M Endothermic (  H = + 300KJ)

4. Exothermic Reactions In exothermic reactions heat is released therefore heat (KJ) is a product. Products are placed on the right-hand side of the arrow. For exothermic reactions the heat term (KJ) must be placed on the right-hand (products) side of the equation C + 3D  2A + Heat Exothermic (  H -) 2Z  X + Y + 150KJ Exothermic (  H = 150KJ)

5. 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. For the exothermic synthesis of ammonia: 1 N 2 + 3 H 2  2 NH 3 + 160 KJ  H = -160 KJ

6. 1 N 2 + 3 H 2  2 NH 3 + 160 KJ  H = -160 KJ From the balanced equation we know the following: 1 mole N 2 produces 160 KJ of heat or 1 mole N 2 = 160 KJ of heat 3 moles of H 2 produces 160 KJ of heat

7. So how much heat would you produce from reacting 1.00 moles of H 2 ? 160 KJ = X KJX = 53.3 KJ 3 moles H 2 1.00 moles H 2 So how much heat would you produce from reacting 0.25 moles of N 2 ? 160 KJ = X KJX = 40.0 KJ 1 mole N 2 0.25 moles N 2 *When setting up these proportions, make sure the information from the equation is on one side and the given information is on the other. Make sure the units match!

8. 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. –Freezing point of water, H 2 O 0 o C = 273 K –Boiling point of water, H 2 O 100 o C = 373 K

9. To convert from Celsius to Kelvin: Add 273 Determine the Kelvin equivalent of 25 o C 25 o C + 273 = 298 K To convert from Kelvin to Celsius: Subtract 273 Determine the Celsius equivalent of 100 K 100 K - 273 = - 173 o C