1. EQUILIBRIUM

2. Many chemical reactions are reversible: Reactants → Products or Reactants ← Products Reactants form Products Products form Reactants For example, under certain conditions, one mole of the colourless gas N2O4 will decompose to form two moles of brown NO2 gas: N2O4 2 NO2 Under other conditions, you can take 2 moles of brown NO2 gas and change it into one mole of N2O4 gas: N2O4 2 NO2

3. EQUILIBRIUM In other words, this reaction is as written can go forward or in reverse depending on the conditions If we were to put some N2O4 in a flask, the N2O4 molecules would collide with each other some of them would break apart to form NO2 This is called a FORWARD reaction N2O4 2 NO2

4. EQUILIBRIUM Once this has happened for awhile, there is a build up of NO2 molecules in the same flask Once in awhile, two NO2 molecules will collide with each other and join to form a molecule of N2O4 ! This is called a REVERSE reaction: N2O4 2 NO2

5. EQUILIBRIUM 2 things to notice: 1. as long as N2O4 present, the forward reaction will keep on happening 2. as long as there is NO2 present, the reverse reaction will keep on happening! At one particular time a molecule of N2O4 might be breaking up, and at the same time two molecules of NO2 might be joining to form another molecule of N2O4

6. EQUILIBRIUM In any reversible reaction, the forward reaction and the reverse reaction are going on at the same time N2O4 2 NO2 The double arrow means that both the forward and reverse reaction are happening at the same time.

7. RATE OF REACTION Rate of the forward reaction Rate of forward reaction will be fast at first Rate will slow down

8. RATE OF REACTION Rate of reverse reaction

9. RATE OF REACTION

11. The rate of the forward reaction = the rate of the reverse reaction N02 is being used up at the same rate that it is being formed N2O4 2 NO2 Rate of reaction does not change because both N202 and NO2 are constant

12. EQUILIBRIUM When this happens we say the system is at EQUILIBRIUM We have reached a state of DYNAMIC EQUILIBRIUM

13. KEY POINTS The reaction has not stopped The forward and reverse reactions are still taking place but their rates are equal so there are no changes in concentrations of the reactants and products It is happening at a microscopic level so we don’t see the individual molecules reacting (macroscopic)

14. KEY POINTS All observable properties are constant include the reactants, products, total pressure, temperature If no changes were made to the condition a system would stay at equilibrium forever

15. KEY POINTS Temperature must remain constant A system at equilibrium is a closed system Equilibrium can be approaches form the left (reactants) or from the right (products) https://www.youtube.com/watch?v=dUMmoPdwBy4 https://www.youtube.com/watch?v=wlD_ImYQAgQ

16. CALCULATIONS Equilibrium calculations We can figure out what the concentrations are for the products and reactants when they are at equilibrium We use: Keq It tells us the ratio of products: reactants when the reaction is at equilibrium

17. CALCULATIONS to figure out the concentration of the products and reactants at equilibrium we use: Keq = [products] [reactants] A + B C + D Keq = [C] [D] [A] [B]

18. CALCULATIONS Given the equilibrium system: PCl 5 (g) PCl 3 (g) + Cl 2 (g) The system is analyzed at a certain temperature and the equilibrium concentrations are as follows: [PCl 5 ] = 0.32 M, [PCl 3 ] = 0.40 M and the [Cl 2 ] = 0.40 M. Calculate the Keq for this reaction at the temperature this was carried out.

19. CALCULATIONS SOLUTION: Step 1 - Use the balanced equation to write the Keq expression: PCl5(g) PCl3(g) + Cl2(g) Keq = [PCl3] [Cl2] [PCl5] Step 2 - "Plug in" the values for the equilibrium Keq = [0.4] [0.4] [0.32] Step 3 - Calculate the value of Keq : Keq = 0.5

20. CALCULATIONS Given the equilibrium system: 3 PCl 5 (g) 2PCl 3 (g) + Cl 2 (g) [PCl 5 ] = 0.32 M, [PCl 3 ] = 0.40 M and the [Cl 2 ] = 0.40 M. We write it as: Keq = [PCl 3 ] 2 [Cl2] [PCl 5 ] 3 Raise to the power of the coefficient Keq = [0.4] 2 [0.4] [0.32] 3

21. CALCULATIONS At 200°C, the Keq for the reaction: N2(g) + 3H2(g) 2NH3(g) is 625 If the [N2] = 0.030 M, and the [NH3] = 0.12 M, at equilibrium, calculate the equilibrium [H2]. Solution: All concentrations given are at equilibrium so: Write out the Keq expression: Keq = [NH3] 2 [N2] [H2] 3 625 = [0.12] 2 [0.30] [H2] 3 (625) [H2] 3 (0.030) = (0.12) 2