1. Chemical Equilibrium Advanced Higher Chemistry Unit 2b

2. Revision Dynamic But to observe you would see no change Rate of forward reaction = rate of reverse reaction Concentrations of reactants and products are constant, but not (necessarily) equal

3. 2Hl(g) H 2 (g) + I 2 (g) To which side does the equilibrium lie? How would the graph differ if you started with H 2 and I 2 ?

4. Calculating compositions To determine equilibrium composition of a mixture, you need equilibrium amount of only ONE of the species Example: CO(g) + 3H 2 (g) CH 4 (g) + H 2 O(g) You place 1.000 mol CO and 3.000 mol H 2 in a reaction vessel at 1200 K and allow the reaction to come to equilibrium. The mixture is found to contain 0.387 mol H 2 O. What is the molar composition of the equilibrium mixture?

5. Calculation strategy You have starting amounts These will change over time They will remain constant at equilibrium Amount (mol) CO3H 2 CH 4 H2OH2O Starting Change Equilibrium 1.000 -x 1.000-x 3.000 -3x 3.000-3x 0 +x x 0 +x x = 0.387 Therefore: 0.613 mol CO, 1.839 mol H 2, 0.387 mol CH 4, 0.387 mol H 2 O CO(g) + 3H 2 (g) CH 4 (g) + H 2 O(g)

6. Practice 1. You place 1.50 mol of dinitrogen trioxide into a flask, where it decomposes at 25ºC and 1.00 atm: N 2 O 3 (g) NO 2 (g) + NO(g) What is the composition of the reaction mixture at equilibrium if it contains 0.45 mol of nitrogen dioxide? 2. Nitric oxide, NO, reacts with bromine to give nitrosyl bromide, NOBr 2NO(g) + Br 2 (g) 2NOBr(g) A sample of 0.0655 mol NO with 0.0328 mol Br 2 gives an equilibrium mixture containing 0.0389 mol NOBr. What is the composition of the equilibrium mixture?

7. The Equilibrium Constant, K c For the reaction: aA + bB cC + dD [C] c [D] d [A] a [B] b K c = The value of the equilibrium constant is constant for a particular reaction at a given temperature, whatever the initial starting concentrations

8. Practice Write the equilibrium constant expression for: 1.The synthesis of ammonia by the Haber process 2.The catalytic hydration of methane

9. Obtaining values for K c Consider the reaction: CO(g) + 3H 2 (g) CH 4 (g) + H 2 O(g) The equilibrium composition is: 0.613 mol CO, 1.839 mol H 2, 0.387 mol CH 4, 0.387 mol H 2 O The volume of the reaction vessel is 10.00 l. So the equilibrium concentration of CO is 0.613/10 = 0.0613 What is the value of the equilibrium constant?

10. Practice Suppose for the same reaction the equilibrium compositions in a 10 l vessel are: 1.522 mol CO, 1.566 mol H 2, 0.478 mol CH 4, 0.478 mol H 2 O What is the vale of K c ?

11. Homo- / Heterogeneous Equilibria Consider the reaction: 3Fe(s) + 4H 2 O(g) Fe 3 O 4 (s) + 4H 2 (g) Is this hetero- or homogeneous? For heterogeneous equilibria, pure solids and liquids are omitted from the K c Write an expression for K c for the above reaction This means that equilibrium is not affected by amounts of these substances

12. The Equilibrium Constant, K p For the reaction: CO(g) + 3H 2 (g) CH 4 (g) + H 2 O(g) The equilibrium constant can be expressed in terms of partial pressures: P CH 4 P H 2 O P CO P H 2 3 K p = In general, for a particular reaction the value of K p will be different from K c

13. Interpretation of K Does a large K mean that reactants or products are favoured? CH 4 (g) + 2O 2 (g) CO 2 (g) + 2H 2 O(g) K c = 1 x 10 140 at 25ºC N 2 (g) + O 2 (g) 2NO(g) K c = 4.6 x 10 -31 at 25ºC If K is around 1, neither reactants or products are favoured

14. Changing the reaction conditions Conditions can be changed to maximise product: –Concentration: adding reactants, removing products –Pressure: change the volume of the vessel –Temperture NOT a catalyst

15. Concentration change Example: CO(g) + 3H 2 (g) CH 4 (g) + H 2 O(g) If cooled, water can be condensed and removed Moles Stage of processCOH2H2 CH 4 H2OH2O Original mixture0.6131.8390.387 After removing water (before equilibrium readjusts) 0.6131.8390.3870 Equilibrium re-established0.4911.4730.5090.122 Assuming a 1 l vessel, what is K c for the original equilibrium and for the re-established equilibrium?

16. Pressure change How can you change the pressure of a system? What is meant by partial pressure?

17. Changing the volume of a reactant container changes the concentration of gaseous reactants and therefore their partial pressures Equilibrium position will therefore move The value of K c or K p does NOT change Changing pressure by adding more of an inert gas has no effect of the equilibrium position - No effect on partial pressures Pressure change

18. Temperature change Like concentration and pressure, temperature affects the position of equilibrium Unlike concentration and pressure, temperature, temperature does affect the size of the equilibrium constant Consider the following data: Temperature (K)KcKc 298 800 1000 1200 4.9 x 10 27 1.38 x 10 5 2.54 x 10 2 3.92 Is this reaction exothermic or endothermic?

19. Summary K is temperature dependant, but is unaffected by changes in concentrations or partial pressures Change Equilibrium position Value of K Concentration Pressure Temperature Catalyst Changes No change Changes No change