1. Do Now: Calculate the mass of 4 mol of MgO
Calculate the amount (in moles) or 32 g of NaCl
Calculate the mass of 1.4 x 10-5 mol PbCl2
Calculate the mass of 2.46 x 10-3 mol CH4

2. Do Now: Copy and complete the table Solid
Type of particle (atom, ion, molecule)
Attractive forces
Melting point (high or low)
Aluminium Al
Hydrogen sulfide H2S
Graphite C
Calcium oxide CaO

3. Solid
Type of particle (atom, ion, molecule)
Attractive forces
Melting point (high or low)
Aluminium Al
Atom
Metallic bond
High
Hydrogen sulfide H2S
Molecule
Weak intermoelcular forces
low
Graphite C
Covalent bond
high
Calcium oxide CaO
ion
Ionic bond

4. Energy Changes

5. Kaupapa / Learning Outcome
At the end of todays lesson I will be able to identify exothermic and endothermic reactions using energy diagrams, ΔrH values and heat changes.

6. Enthalpy Heat content of substance Symbol ➜ H Unit ➜ Kilojoules (Kj)
Enthalpy change
(ΔrH) = Hproducts - Hreactants

7. Exothermic Vs Endothermic
video

8. Exothermic Reactants have more energy than products Releases energy
Temperature of surroundings increases
Feels warm
Negative ΔrH or in products
Forming bonds

9. Endothermic Reactants have less energy than products Absorbs energy
Temperature of surroundings decreases
Feels cold
Positive ΔrH or in reactants
Breaking bonds

10. Example
The sublimation of carbon dioxide is overall an endothermic process
CO2(s)  CO2(g)
There are no weak intermolecular bonds between CO2 gas molecules. No bonds formed no energy out
To break the weak intermolecular forces between CO2 molecules requires energy
Overall: Since energy in > energy out = endothermic so ∆H is positive

11. If a reaction is endothermic in one direction it will be exothermic in the opposite directions A + B ➜ C + D (endothermic) C + D ➜ A + B (exothermic)

12. Complete pages 61-62