2. Why do we need to study chemical energetics? It can lead us to understand a chemical change better. E.g What is the temperature for a reaction to occur. It will also tell us the energy barrier that reactants have to overcome before becoming products.
3. Energetics of chemical reactions Every reactant will have a discrete amount of energy. rearrangement of atoms that occurs in a chemical reaction is virtually always accompanied by the liberation or absorption of heat. Bonds are broken in reactants and bonds are formed in product
4. System and surroundings We will use 2 models to represent reactants and its environments around it. System – Reactants Surroundings – Environment around reactants
5. Exothermic Reaction Heat is given off As a result of this release of heat energy from the system (reactants), the surrounding will have a increase in temperature Energy of reactant < Energy of Surrounding
6. Exothermic Reaction So what happens after heat energy is being released? Products formed will have a energy level lower than reactants. Conservation of energy!
7. Exothermic Reaction Example of Exothermic Reaction Carbon combust under presence of oxygen to produce carbon dioxide. Neutralisation between an acid and alkali 3. Granular Zinc with Tin(II) Chloride. (Metal displacement)
8. Endothermic Reaction heat energy is taken in. System (reactants) take in energy from the surrounding. Surrounding loses heat to the reactants. Energy of reactant > Energy of Surrounding
9. Endothermic Reaction So what happen after heat energy is being absorbed? Product formed will have a higher energy than the reactants. Conservation of energy!
10. Endothermic Reaction Examples Thermal decomposition Hydrated copper (II) sulfate ----> Anhydrous Copper (II) sulfate CaCO3 --------> CaO + CO2 2) Photosynthesis 3)
11. So what do we name the difference in amount of energy between reactants and products? ENTHALPY CHANGE OF REACTION ∆H The Enthalpy change in a reaction has a symbol ∆H.
12. Enthalpy Enthalpy (H) is also known as heat content, is the total energy of a system, some of which is stored as chemical potential energy inside the chemical bonding. Absolute enthalpies for particular states cannot be measured, but the change in enthalpy that occurs during a reaction can be measured – Using a calorimetry
13. Enthalpy ∆H is positive (+) for endothermic reactions. Heat energy is gained by system from the surrounding. Hence the surrounding will a dip in temperature ∆H is negative (-) for exothermic reactions. Heat energy is lost to the surrounding and cause the temperature in the surrounding to rise.
14. Enthalpy For endothermic reaction, final products have a higher energy than the reactants. For Exothermic reaction, final products have a lower energy than the reactants.
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16. These covalent bonds have a specific amount of energy to hold the atoms together
19. How to calculate the enthalpy change of reaction? Step 3: Energy is absorbed by the reactants when bonds are broken. Hence it will be 436KJ + 243KJ = 679KJ Energy is released by product when new bonds are formed. 2 X 432KJ = 864KJ Step4: ∆H = (Energy absorbed when bond are broken in reactants) – (Energy released when bonds are formed in product)
20. How to calculate the enthalpy change of reaction? Step 5: ∆H = (Energy absorbed when bond are broken in reactants) – (Energy released when bonds are formed in product) ∆H = 679KJ – 864KJ = -185KJ
21. Assignment: Chemical Energetics Qn 3a) 2 H-O-O-H ------> O=O + 2 H-O-H ∆H= -206KJmol-1 Thought process: Enthalpy change of reaction already shows that there is a decrease in value as shown in the negative sign. This implies that it is an exothermic reaction. This will also mean that energy from bond breaking(reactants) is less than bond formation (products)
22. Assignment: Chemical Energetics Qn 3a) Answer: The amount of energy absorbed to break 4 O-H bonds and 2 O-O bonds is less than the amount of energy released in forming 1 O=O bond and 4 O-H bonds. b) ∆H = Energy from bond breaking – Energy from bond forming -206 KJ= 2(B.E O-O) + 4(B.E O-H) – [ B.E O=O + 4 B.E O-H] -206 KJ = 2(B.E O-O) + 4(463KJ) – [ 486 + 4(463)] B.E O-O = 140 KJ mol-1
23. Assignment: Chemical Energetics Qn 4: Though process: Strong bond means larger bond energy and weak bond means smaller bond energy. X-X + 3 Y-Y ----> 2XY3 ∆H = Energy absorbedbond-breaking – Energy releasedbond formation = [B.E x-x+ 3B.E Y-Y] – 6[B.E XY] Answer: Exothermic. Less Energy is needed to break the X-X and Y-Y bonds and more energy will be released when forming the X-Y bonds. Enthalpy change will be negative.