1. The Equilibrium Law Section 17.2 (AHL)

2. Vocabulary Homogeneous equilibrium: all the reactants and products are in the same phase Heterogeneous equilibrium: when there are two or more phases Q c : reaction quotient, refers to a quotient obtained by applying the equilibrium law to initial concentrations (instead of equilibrium concentrations)

3. Using Q c The value of the reaction quotient can be used to determine how a reaction needs to shift in order to attain equilibrium Q c =K c equilibrium is established Q c < K c the system must shift to the right to establish equilibrium Q c > K c the system must shift to the left

4. Example Calculation #1 The acid-catalyzed hydrolysis of ethyl ethanoate can be achieved by mixing the ester with dilute HCl CH 3 COOC 2 H 5(l) + H 2 O (l) CH 3 COOH (l) + C 2 H 5 OH (l) The products are ethanoic acid and ethanol H+H+

5. Problem #1 Continued If 1.00 mole of ethyl ethanoate is mixed with 1.00 mole of water and the reaction allowed to reach equilibrium at a particular temperature, then 0.30 moles of ethanoic acid is found in the equilibrium mixture Calculate the value of K c at this temperature

6. Make an ICE table CH 3 COOC 2 H 5 H2OH2OCH 3 COOHC 2 H 5 OH Initial Amount in moles 1.00 0.00 Change in amount +0.30 Equilibrium amount in moles 0.30

7. Continued The rest of the table needs to be filled in to find K c at that temperature According to the stoichiometry, all the coefficients in the equation are all “1”, so if 0.30 mol of CH 3 COOH is produced, then 0.30 mol ofvC 2 H 5 OH is also produced

8. CH 3 COOC 2 H 5 H2OH2OCH 3 COOHC 2 H 5 OH Initial Amount in moles 1.00 0.00 Change in amount +0.30 Equilibrium amount in moles 0.30

9. Problem Continued If 0.30 mole are present in the equilibrium mixture, then they must have been produced from the reaction of 0.30 moles of the ester and the water. The amount of ester and water remaining at equilibrium must be 1.00-0.30 moles of each

10. CH 3 COOC 2 H 5 H2OH2OCH 3 COOHC 2 H 5 OH Initial Amount in moles 1.00 0.00 Change in amount -0.30 +0.30 Equilibrium amount in moles 0.70 0.30

11. Problem Continued We now have the number of moles, but not the volumes of each liquid. You can assume 1 dm 3 All molar values then become concentration values

12. Example Problem #2 (ex. on page 194) When a mixture initially containing 0.0200 mol dm -3 SO 2 and an equal concentration of O 2 is allowed to reach equilibrium in a closed container of fixed volume at 1000K, it is found that 80.0% of the SO 2 is converted to SO 3. Calculate the value of the equilibrium constant at that temperature.

13. Problem #2 Continued Write the balanced equation first 2SO 2(g) + O 2(g) 2SO 3(g) Make your table of ICE

14. SO 2 O2O2 SO 3 Initial Amount in moles 0.0200 0.00 Change Equilibrium amount in moles

15. Continued Equilibrium [SO 3 ] = (0.0200) (0.800) since 80% of the the SO 2 turns into SO 3 [SO 3 ] = 0.0160 mol dm -3 The stoichimetry says a 1:1 ratio between SO 3 and SO 2, so the [SO 2 ] is equal to 0.0200 minus 0.0160 which is 0.0040 mol dm -3

16. Continued According to the stoichiometry, each sulfur trioxide molecule requires only ½ an oxygen molecule so: [O 2 ] = 0.0200 – (1/2)(0.0160) = 0.0120 mol dm -3

17. SO 2 O2O2 SO 3 Initial Amount in moles 0.0200 0.00 Change -0.0160-0.0080+0.0160 Equilibrium amount in moles 0.00400.01200.0160

18. Calculate K c =1333 mol -1 dm 3 1330 mol -1 dm 3 (to 3 sig figs)

19. Another Fun Problem #3 SO 3(g) + NO (g) NO 2(g) + SO 2(g) K c for this reaction is 6.78 at a specified temperature If the initial concentrations of NO and SO 3 were both 0.03 mol dm -3, what would be the equilibrium concentration of each component?

20. Continued We will have to use variables for the exponents to keep track of the concentrations and to solve Let the change in concentration of the reactants both equal -x Let the change in concentration of the products both equal +x

21. SO 3 NONO 2 SO 2 Initial Amount in moles 0.03 0.00 Change in amount -x +x Equilibrium amount in moles 0.03-x xx

22. Continued

23. 2.60 (0.03 – x) = x 0.078 – 2.60x = x 0.078 = 3.60x 0.0217 = x (realize you are not finished when you calculate “x”)

24. SO 3 NONO 2 SO 2 Initial Amount in moles 0.03 0.00 Change in amount -x +x Equilibrium amount in moles 0.03-x xx

25. SO 3 NONO 2 SO 2 Initial Amount in moles 0.03 0.00 Change in amount -0.0217 +0.0217 Equilibrium amount in moles 0.0083 0.022

26. One more type of problem

27. N 2 (g)3H 2 (g)2NH 3 Initial Amount in moles 1.003.000.00 Change in amount -0.062÷2= 0.031 -(3*(0.062÷2))= 0.093 +0.062 Equilibrium amount in moles 0.9692.910.062

28. Finish K c = 1.61 x 10 -4