Equilibrium

1. Equilibrium --another concept that the mentally unbalanced struggle with.

2. When you set up a reaction: A + B  C + D • The reactants in the flask start to make products. • [A] and [B] fall • The rate of the reaction begins to slow as reactants are used.

4. Almost any reaction is reversible C + D A + B • The rate of the reverse reaction starts at 0 M/s • This rate begins to increase as [C] and [D] increase • As C and D appear, they re-make the original reactants

6. Any reversible reaction: A + B C + D • Will come to a dynamic equilibrium. • When the rate of the forward reaction falls to the same rate that the reverse reaction rises to, both rates stabilize.

7. If you started with reactants: • The forward rate quits falling as the reverse reaction re-makes products fast enough • The reverse rate quits rising as the forward reaction quits increasing product concentrations • At equilibrium, concentrations become constant, unless the system is disturbed

9. Equilibrium means: • The rates of the forward and reverse reactions are equal. • Products and reactants are both being used and formed at the same rate • Concentrations are stable

10. Equilibrium does not mean: • The forward and reverse reactions stop. • Product and reactant molecules are stable • Product and reactant concentrations are equal Good foils for multiple choice!

11. This rate depends on…

12. This rate depends on… • Temperature • Concentrations (considered together) • Ea of forward and reverse reactions

13. The forward reaction

14. The reverse reaction Products Ea DG Reactants

15. PS • The reaction that starts off fastest is the one whose reactants were present • The reaction whose products end up more common (*) is the easier one to carry out (lower Ea) • An exothermic reaction has an endothermic reverse. The endothermic direction has a higher Ea

16. The Equilibrium Constant Expression • For a general reaction, aA + bB cC + dD …all substances being (aq) or (g), K= [C]c[D]d [A]a[B]b …is a constant at constant T

17. Write the equilibrium constant expression for … 1) the combustion of ammonia to form nitrogen monoxide and water vapor 2) the conversion of carbon monoxide and hydrogen to methanol 3) the association of ammonia and water to form ammonium and hydroxide ions

18. Rookie mistakes: • --putting reactants on top • --using coefficients inside the brackets • --adding instead of multiplying concentrations • --multiplying by coefficients, instead of raising to the power • --including liquids and solids. Avoid these errors!

19. What is the value of K? 1) For the combustion (in .30 M O2) of ammonia (.05 M) to form nitrogen monoxide (at .20 M) and water vapor (at .40 M)? 2) For the conversion of carbon monoxide (at.20 M) and hydrogen (at .20 M) to methanol (at .03 M) ? 3) For the association of ammonia (at .10 M) and water (at 55.5 M) to form ammonium (at .0013 M) and hydroxide ions (at .0013 M)?

20. Assuming these K’s are correct (at some temperature) 1) For the combustion (in .40 M O2) of ammonia (.06 M) to form nitrogen monoxide (at .15 M), what is the concentration of water vapor? 2) For the conversion of carbon monoxide (at what concentration?) and hydrogen (at .25 M) to methanol (at .07 M)? 3) For the association of ammonia (at ? M) and water (at 55.5 M) to form ammonium (at .0019 M) and hydroxide ions (at .0003 M)?

21. But that’s the easy part. • Given original concentrations, and K, what are the final concentrations? • Ex. For the synthesis of ammonia from its elements, K=.034 at some temperature. If you start with [N2]=[H2]=[NH3]=.50M, what are the final concentrations?

22. Use the ICE method. • Step 1: Write and balance • Step 2: Write eq. const. expression. • Step 3: Plug in initial values to find Q, the reaction quotient --if Q>K, the reaction shifts left --if Q<K, the reaction shifts right --if Q=K, (yeah, right. Like that’ll ever happen)

23. Step 4: Make an ICE table [A] [B] [C] • Initial [Ao] [Bo] [Co] • Change -ax -bx +cx • Equilibrium [Ao] -ax [Bo] -bx [Co] +cx (reverse signs on changes if Q>K) • Step 5: Plug the eq. values, solve for x.

24. Is this your card? • Step 1: N2(g) + 3H2(g) 2NH3(g) • Step 2: K= [NH3] 2 [N2] [H2]3 • Step 3:Q= .5M2=4 (greater than K) .5M x .5M3 …so the reaction will shift left More ammonia will decompose than will be made—but how much more?

25. Step 4 Make an ICE table [N2] [H2] [NH3] • Initial .5M .5M .5M • Change + +x +3x -2x • Equilibrium .5+x M .5+3x M .5-2x M • Step 5: Plug the equilibrium values into the expression, solve for x.

26. K=.034=[NH3] 2 = (.5-2x)2 [N2] [H2]3 (.5+x)(5+3x)3 • These are equal where x=.172 [N2] [H2] [NH3] • Equilibrium .5+xM .5+3xM .5-2xM =.672M =1.016M =.156M

27. You wouldn’t take my word for it, would you? • Check: • K= .156M2/.672M x 1.016M3=.035 (aka: “Close enough for government work.”)

28. What does the .172 say? • The reaction proceeds to the left .172 times through the reaction as written. • It proceeds .172 moles: --farther to the left than to the right --for a substance with a coefficient of one --at which point, the rates equalize.

29. What is x? What are the final concentrations? 1) [O2 ]=.40 M,[NH3 ]=.02 M,[NO] =.18 M, [H2O]=.20M 2) [CO ]=.03M,[H2]= .15 M,[CH3OH]=.02M 3) [NH3]=.12M,[NH4+]=.0019 M,[OH-]=.0005 M

30. What’s with this molarity for gasses? • Well, we could, but we usually measure pressure • Use PV=nRT to convert.

31. Come, let us reason together. • If N2(g) + 3H2(g) 2NH3(g) K= [NH3] 2 [N2] [H2]3 & M=n/V=P/RT Then: • K= (PNH3/RT)2= PNH32 (RT) 2 (PN2/RT)(PH2/RT)3 PN2PH23

32. Dealer’s Choice • Kc -- calculated by concentrations (M) • Kp -- calculated by pressures (atm.) Kp=Kc(RT)Dn Dn is the change in number of moles of gas

33. Kp=Kc(RT)Dn • ‘Bout says it all, doesn’t it? • No? • A pressure constant is related to a concentration constant only differing by the RT factor represented by the unequal moles of gas that don’t cancel. This reaction lost 2 moles of gas from reactants to products.

34. Totally gratuitous joke. • Oxygen gas can dissociate into individual atoms, O2 2 O • where K=[O]2/[O2], is a very small value at normal temperatures, larger at higher T

35. Totally gratuitous joke. • Oxygen gas can dissociate into individual atoms, O2 2 O • where K=[O]2/[O2], is a very small value at normal temperatures, larger at higher T OK?

36. What is the value of Kp? 1) For the combustion (in .30 M O2) of ammonia (.05 M) to form nitrogen monoxide (at .20 M) and water vapor (at .40 M)? 2) For the conversion of carbon monoxide (at.20M) and hydrogen (at .20 M) to methanol (at .03 M) ? 3) For the association of ammonia with water— Not a gas phase reaction. No Kp

37. Remember Dalton’s Law? • Dn = kDM (const. vol.) —used w/ Kc calculations • Dalton’s Law of partial pressures implies Dn = kDP —used w/ Kc calculations

38. What is x? What are the final concentrations? 1) PO2 =1.3 atm,PNH3=2.1 atm, PNO =1.8 atm, PH2O=2.0 atm 2) PCO =4.3 atm, PH2= 2.5 atm, PCH3OH=.80atm 3) PNH3=9.5 atm, PNH4+=1.1 atm, POH-=2.3 atm (Just joking on that last one—that really is in an aqueous environment—don’t try to use pressures)

39. Que? ? • If A + B C + D has an equilibrium constant, K1…. • --then C + DA + B has an equilibrium constant, K2=1/K1 (products and reactants are reversed) • --and 2A + 2B 2C + 2D has an equilibrium constant, K3=K12 (all of the concentrations are squared)

40. Heterogeneous Equilibria • If a solid or a liquid are part of a reversible reaction, they do not affect the equilibrium except by their absence. • If they are present—the system can be in equilibrium. If they are not present, the system is not in equilibrium

41. LeChatelier’s Principal Henri LeChatelier attended the Ecole Polytechique, in the Latin Quarter of Paris, where his principal, Monsieur Bouvier, always made a point of….

42. LeChatelier’s Principle If a system in equilibrium is subjected to a stress, the system will shift in the direction that will relieve that stress

43. Application of LeChatelier’s principle • “Shift right” --forward reaction is faster, --more of all products are formed --all reactants are used • “Shift left” --reverse reaction is faster, --more of all reactants are formed --all products are used

44. Application of LeChatelier’s principle • An aqueous or gas substance in the reaction added—shift away to use it up • Increasing pressure—shift toward side with fewer moles of gas to relieve pressure • Increasing temperature—shift in the endothermic direction to absorb heat

45. Application of LeChatelier’s principle • An aqueous or gas substance in the reaction removed—shift towards to replace it • Decreasing pressure—shift toward side with more moles of gas to fill the space • Decreasing temperature—shift in the exothermic direction to produce heat

46. Add N2(g) Add H2(g) Add NH3(g) Increase P (compress) Increase T Add a catalyst Remove N2(g) 7. Remove H2(g) 8. Remove NH3(g) 9. Decrease P (allow to expand) 10. Decrease T 11. Increase pressure by adding He (g) N2(g) + 3H2(g) 2NH3(g) + D Which way would the equilibrium shift if you:

47. Add CO2(g) Add CaO(s) Add CaCO3(s) Increase P (compress) Increase T Add a catalyst Remove CO2(g) 7. Remove CaO(s) 8. Remove CaCO3(s) 9. Decrease P (allow to expand) 10. Decrease T 11. Increase pressure by adding He (g) CO2(g) +CaO(s) CaCO3(s) + D Which way would the equilibrium shift if you:

48. Inappropriate Applications of LeChatelier’s Principle • An equilibrium will not shift if: --a solid or liquid is added or removed --a catalyst is used --gasses are compressed/allowed to expand and Dn=0 --forward/reverse rates are affected equally --a noble gas is added (Pt increases, partial pressures do not change)