2. The collision Model Chemists believe that for a chemical reaction to occur, the reactants must collide with enough force to break bonds, allowing the reactants to rearrange themselves to form products. For example: 2BrNO (l) 2NO (g) +Br 2(g)

3. Molecules not correctly orientated, so no reaction Molecules are correctly orientated, so reaction occurs

4. So for a chemical reaction to occur the following must happen: Reactants collide Collide at the right orientation Collide with enough violence What factors affect the rate of a chemical reaction? Temperature 2NO 2 N 2 O 4 Surface area (Grain explosion) Concentration (aq)Mg + HCl H 2 + Cl 2 Catalysts catalase in animal liver 2H 2 O 2 2H 2 O + O 2 Volume (gases)

5. Activation Energy (E a ) This is the minimum energy required for a reaction to occur

7. An Endothermic Reaction

8. Catalyst This is a substance that speeds up a reaction without being consumed. Enzymes are catalysts that are found in our bodies. They allow our body to speed up the rate of reactions that would be too slow at normal body temperature. A good example is the breakdown of ozone, which is catalyzed by chlorine. Cl + O 3  ClO +O 2 O + ClO  Cl +O 2 Sum: Cl + O 3 + O + ClO  ClO +O 2 + Cl + O 2 O 3 + O  2O 2

11. Equilibrium means balance or steadiness In chemistry, an equilibrium reaction is when there are two opposite reactions that are balanced. We tend to think that all reactions go in only one direction. What is called the forward reaction. CaCO 3 (aq) + HCl (aq)  CaCl 2 (aq) + CO 2 (g) + H 2 O (l) In this reaction the reactants react to form products. It goes to completion, or until one of the reactants runs out But not reactions are like this!

12. Equilibrium Reactions: This is reaction that has two directions, a forward and reverse reaction. An equilibrium reaction is reversible H 2 O (g) + CO (g) H 2 (g) + CO 2 (g) Forward reaction Reverse reaction

13. Dynamic Equilibrium: If kept in a closed system, then both reactions will occur. If this reaction is left then it will reach a point of equilibrium where the rate of the forward reaction equals the rate of the reverse reaction. So the concentrations of the reactants doesn’t change. The equilibrium is dynamic, because even though the concentrations stay the same and nothing appears to be happening, both the forward and reverse reactions are continuing to occur.

14. Changes in reaction rates of the forward and reverse reactions for: H 2 + CO H 2 +CO 2 Rate of forward reaction decreases while reverse increases till the concentrations reach a level at which the rate of the forward and reverse reactions is the same. The system has reached equilibrium.

15. [ ] of A decreases while [ ] of B increases till equilibrium is reached. Equilibrium is reached when rate of forward reaction is the same as the reverse reaction.

16. The Equilibrium Constant: When the system is at equilibrium, the concentrations remain constant. So the ratio of the concentrations also remains constant. E.g. aA + bB  cC + dDA, B, C, D = chemical species a, b, c, d = coefficients [ ] = molarity in mol L -1 K eq = equilibrium constant E.g. 4NH 3 (g) + 7O 2 (g)  4NO 2 (g) + 6H 2 O (g)

17. Write equilibrium expressions for these reactions: a) 2O 3 (g)  3O 2 (g) b) H 2 (g) + F 2 (g)  2HF (g) The equilibrium constant means that for a given reaction at a given temperature the ratio of the concentration of reactants to products is always be constant. You should always make sure that you write balanced equations that include the physical states of the reactants and products.

18. Calculating Equilibrium Constants: The actual value of K eq is found experimentally. The individual concentrations of all the reactants is calculated, and the temperature recorded. This is because the value of K eq will change with temperature. What can K eq tell you?  Which direction is favored If the value of K eq is one then …. If the value of K eq is less than one …… If the value of K eq is greater than one then ………

19. Results for three experiments for the reaction: N 2 (g) + 3H 2 (g) 2NH 3 (g) at 500  C

20. Calculating K eq for a reaction: E.g. 4SO 2 (g) + O 2 (g)  2SO 3(g) Experiment 1 Initial Equilibrium [SO 2 ] = 2.00M [SO 2 ] = 1.50M [O 2 ] = 1.50M [O 2 ] = 1.25M [SO 3 ] = 3.00M[SO 3 ] = 3.50M Experiment 2 InitialEquilibrium [SO 2 ] = 0.500M [SO 2 ] = 0.590M [O 2 ] = 0.00M [O 2 ] = 0.045M [SO 2 ] = 0.350M[SO 3 ] = 0.260M Equilibrium constant for Experiment 1 = Equilibrium constant for Experiment 2 =

21. Heterogeneous Equilibria: A homogenous reaction is one in which all the substances are in the same state. A heterogeneous reaction is one in which all the substances are not in the same state. CaCO 3 (s) CO 2 (g) + CaO (s) Calcium carbonatecarbon dioxidelime When writing equilibrium constant expressions for Heterogeneous equilibria, you don’t include pure solids or pure liquids. Their concentrations don’t change K eq = [CO 2 ][ CaO ] K eq = [CO 2 ] [ CaCO 3 ] Try writing an K eq expression for this reaction: 2H 2 O (l) 2H 2 (g) +O 2(g)