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Slide 1 ___________________________________ Chemical Equilibrium ___________________________________ Chemistry in Two Directions ___________________________________ ___________________________________ Text 692019 and Questions to 37607 1 ___________________________________ ___________________________________ ___________________________________ Slide 2 ___________________________________ Chemical Reactions Up until now, we have talked about reactions as though they proceed in one direction: from left to right, as written in a balanced equation: A(aq) + 2 B(aq) →6 C ___________________________________ ___________________________________ (aq) In fact, this is rarely true. ___________________________________ Text 692019 and Questions to 37607 2 ___________________________________ ___________________________________ ___________________________________ Slide 3 If you can go left, you can turn around and go right ___________________________________ If you consider the energetics of a reaction, it is usually represented by a graph of Energy vs. Reaction Coordinate ___________________________________ (Reaction Coordinate is complicated, but you can think of it as the lowest energy path between chemical species) ___________________________________ ___________________________________ Text 692019 and Questions to 37607 3 ___________________________________ ___________________________________ ___________________________________ Slide 4 ___________________________________ Typical Reaction Energy Diagram ___________________________________ ___________________________________ ___________________________________ Text 692019 and Questions to 37607 4 ___________________________________ ___________________________________ ___________________________________ Slide 5 ___________________________________ What if you just flip it around? ___________________________________ ___________________________________ ___________________________________ Text 692019 and Questions to 37607 5 ___________________________________ ___________________________________ ___________________________________ Slide 6 ___________________________________ Everything is simply reversed: The Ea is different, the ΔH is inverted ___________________________________ ___________________________________ ___________________________________ Text 692019 and Questions to 37607 6 ___________________________________ ___________________________________ ___________________________________ Slide 7 ___________________________________ Reactions can go both ways The hill is just higher going one way than the other. If it is exothermic going one way, it is endothermic going the other way. The world is reversible! (Well, except for time ) ___________________________________ ___________________________________ ___________________________________ Text 692019 and Questions to 37607 7 ___________________________________ ___________________________________ ___________________________________ Slide 8 ___________________________________ Implications for Chemical Reactions The reversibility of chemical reactions means that rather than proceed from reactants to products, most reactions reach a state where there is no further change. ___________________________________ ___________________________________ THIS DOES NOT MEAN NOTHING IS HAPPENING! ___________________________________ Text 692019 and Questions to 37607 8 ___________________________________ ___________________________________ ___________________________________ Slide 9 ___________________________________ DYNAMIC EQUILIBRIUM Chemical Equilibrium is a Dynamic Equilibrium. ___________________________________ It is not that there is no reaction occurring, it is that the forward reaction and the reverse reaction are occurring at equal rates. ___________________________________ ___________________________________ Text 692019 and Questions to 37607 9 ___________________________________ ___________________________________ ___________________________________ Slide 10 A(aq) + 2 B(aq) ↔6 C (aq) ___________________________________ 6C ___________________________________ (aq) → A(aq) + 2 B(aq) (C is disappearing) AND A(aq) + 2 B(aq) → 6 C (aq) ___________________________________ (C is being created) Both reactions occur simultaneously with the Rate of destruction = rate of creation Text 692019 and Questions to 37607 ___________________________________ 10 ___________________________________ ___________________________________ ___________________________________ Slide 11 ___________________________________ Stand ↔ Sit This is r c ion…er, sort of. ___________________________________ 1st L ’s ssum i is order in both directions – do sn’ m r, jus m k s i sy. ___________________________________ What is the rate law for the forward reaction? ___________________________________ Text 692019 and Questions to 37607 11 ___________________________________ ___________________________________ ___________________________________ Slide 12 ___________________________________ Stand ↔ Sit L ’s ssum i is 1st order in both directions – do sn’ m r, jus m k s i sy. What is the rate law for the forward reaction? A. Rate = k[Stand] B. Rate = k[Sit] C. ln[Stand]t =-kt+ln[Stand]0 D. A and C E. I c n’ ll wi hou mor inform ion. Text 692019 and Questions to 37607 ___________________________________ ___________________________________ ___________________________________ 12 ___________________________________ ___________________________________ ___________________________________ Slide 13 ___________________________________ Stand ↔ Sit L ’s ssum i is 1st order in both directions – do sn’ m r, jus m k s i sy. What is the rate law for the Reverse reaction? A. Rate = k[Stand] B. Rate = k[Sit] C. ln[Stand]t =-kt+ln[Stand]0 D. A and C E. I c n’ ll wi hou mor inform ion. Text 692019 and Questions to 37607 ___________________________________ ___________________________________ ___________________________________ 13 ___________________________________ ___________________________________ ___________________________________ Slide 14 ___________________________________ Let’s do the reaction!!! Forward: Rate = ½ [Stand] Reverse: Rate = ¼ [Sit] ___________________________________ Th r ’s 160 of you – all sitting. ___________________________________ ___________________________________ Text 692019 and Questions to 37607 14 ___________________________________ ___________________________________ ___________________________________ Slide 15 ___________________________________ Time Stand Sit 0 0 160 =1/2 [Stand] =1/4[Sit] 0-0+40=40 160-40+0=120 40 120 ½[Stand]=20 ¼[Sit]=30 40-20+30=50 120-30+20=110 50 110 ½[50]=25 ¼[110]=27.5 50+28-25=53 110-28+25 = 107 53 107 ½[53]=27 ¼[107]=27 1 2 3 Text 692019 and Questions to 37607 ___________________________________ ___________________________________ ___________________________________ 15 ___________________________________ ___________________________________ ___________________________________ Slide 16 ___________________________________ Equilibrium is Balance The forward and reverse reactions are balanced The concentrations of all species (reactants and products) become stable The equilibrium position is not the same for all reactions ___________________________________ ___________________________________ ___________________________________ Text 692019 and Questions to 37607 16 ___________________________________ ___________________________________ ___________________________________ Slide 17 ___________________________________ The Equilibrium Constant The balance between forward and reverse reaction is summ d up in h Equilibrium Cons n (“Keq”) ___________________________________ If K is large, most of the reactants form products. ___________________________________ If K is small, most of the reactants stay as reactants Text 692019 and Questions to 37607 ___________________________________ 17 ___________________________________ ___________________________________ ___________________________________ Slide 18 ___________________________________ Writing Equilibrium Constant Expressions I is v ry simpl o wri h xpr ssion for “Keq” ___________________________________ You simply take the concentration (molarity) of all gases, and solutions and raise them to their stoichiometric coefficient ___________________________________ You ignore pure liquids and solids – h y don’ h v “conc n r ion” nd h ir conc n r ion is considered constant (a unitless 1) and so it do sn’ con ribu ___________________________________ Text 692019 and Questions to 37607 18 ___________________________________ ___________________________________ ___________________________________ Slide 19 ___________________________________ A(aq) + 2 B(aq) ↔ 6 C (aq) ___________________________________ ___________________________________ ___________________________________ Text 692019 and Questions to 37607 19 ___________________________________ ___________________________________ ___________________________________ Slide 20 ___________________________________ 2 H2 (g) + O2 (g) ↔ 2 H2O (g) ___________________________________ ___________________________________ ___________________________________ Text 692019 and Questions to 37607 20 ___________________________________ ___________________________________ ___________________________________ Slide 21 ___________________________________ 2 H2 (g) + O2 (g) ↔ 2 H2O (l) ___________________________________ ___________________________________ Notice that the pure liquid is not included – i do sn’ r lly h v conc n r ion nd so i b com s “1” ___________________________________ Text 692019 and Questions to 37607 21 ___________________________________ ___________________________________ ___________________________________ Slide 22 ___________________________________ 2 Na(s) + Cl2 (g) ↔ 2 NaCl (s) ___________________________________ Keq = 1 [Cl2] ___________________________________ Notice how the 2 solids do not contribute to the equilibrium constant ___________________________________ Text 692019 and Questions to 37607 22 ___________________________________ ___________________________________ ___________________________________ Slide 23 ___________________________________ H2S (g) + 2 O2 (g) ↔ H2SO4 (g) ___________________________________ Keq = [H2SO4] [O2]2[H2S] ___________________________________ ___________________________________ Text 692019 and Questions to 37607 23 ___________________________________ ___________________________________ ___________________________________ Slide 24 ___________________________________ Write the equilibrium constant expression: 2 HCl(aq) + CaCO3(s)CaCl2(aq) + H2CO3(aq) ___________________________________ A. K = [H2CO3][CaCl2] [HCl]2[CaCO3] B. K = [H2CO3][CaCl2] [HCl][CaCO3] ___________________________________ C. K = [H2CO3][CaCl2] [HCl]2 D. K = [HCl]2[CaCO3] E. K = [HCl]2 [CaCl2][H2CO3] [CaCl2][H2CO3] ___________________________________ Text 692019 and Questions to 37607 24 ___________________________________ ___________________________________ ___________________________________ Slide 25 ___________________________________ Different kinds of “K” No matter what the subscript, K IS K IS K IS K IS K! They are all just equilibrium constants and they all get written and used the same way. ___________________________________ ___________________________________ Kc Kp Ka Kb Ksp Kf Kw ___________________________________ Text 692019 and Questions to 37607 25 ___________________________________ ___________________________________ ___________________________________ Slide 26 ___________________________________ Kc vs Kp When a reaction occurs in the gas phase, you can use the partial pressure of the gas instead of the concentration. ___________________________________ To separate the 2 different expressions, they are written differently: Kc = equilibrium constant with concentrations of species Kp = equilibrium constant with partial pressures of the species ___________________________________ ___________________________________ Text 692019 and Questions to 37607 26 ___________________________________ ___________________________________ ___________________________________ Slide 27 What is the preferred unit of concentration for a gas? ___________________________________ AM B kg C atm DL E Christian brothers ___________________________________ ___________________________________ ___________________________________ Text 692019 and Questions to 37607 27 ___________________________________ ___________________________________ ___________________________________ Slide 28 ___________________________________ Why is n/V of a gas only “sort of M”? A. B. C. D. E. F. G. H. I. Christian brothers Sisters of St. Joseph V is not in L V is not the solvent V is a horrible TV show V is victory – Sorry Team Canada Church of Scientology-sorry Tom Cruise Church of Heidi Klum – Sorry, sorry Seal Your mother – sorry to your mother Text 692019 and Questions to 37607 ___________________________________ ___________________________________ ___________________________________ 28 ___________________________________ ___________________________________ ___________________________________ Slide 29 ___________________________________ “Thermodynamic” K Keq = PO2 [H2] ___________________________________ ___________________________________ ___________________________________ Text 692019 and Questions to 37607 29 ___________________________________ ___________________________________ ___________________________________ Slide 30 ___________________________________ 2Na(aq) + Cl2 (g) = 2 NaCl (aq) ___________________________________ ___________________________________ No Kp ___________________________________ Text 692019 and Questions to 37607 30 ___________________________________ ___________________________________ ___________________________________ Slide 31 ___________________________________ If everything is a gas, you can use Kp H2(g) + Cl2(g) 2 HCl(g) ___________________________________ Kc = [HCl]2 [H2][Cl2] ___________________________________ P2 Kp = HCl = Keq PH2PCl2 Text 692019 and Questions to 37607 ___________________________________ 31 ___________________________________ ___________________________________ ___________________________________ Slide 32 ___________________________________ K IS K IS K IS K IS K The subscripts – and we will see more of them – just tell you what specific type of reaction you are talking about. ___________________________________ Everything else about them is the same. ___________________________________ ___________________________________ Text 692019 and Questions to 37607 32 ___________________________________ ___________________________________ ___________________________________ Slide 33 ___________________________________ Equilibrium Problems There are two main types of problems: You know the concentrations and you are calculating the equilibrium constant You know the equilibrium constant and you are calculating the concentrations Of course, there are nuances to these problems. ___________________________________ ___________________________________ ___________________________________ Text 692019 and Questions to 37607 33 ___________________________________ ___________________________________ ___________________________________ Slide 34 ___________________________________ Calculating the Equilibrium Constant Th only “ rick” o c lcul ing n quilibrium constant is that you must know the concentrations AT EQUILIBRIUM. ___________________________________ You cannot simply take any old concentrations at any old time. The reaction must have reached equilibrium ___________________________________ ___________________________________ Text 692019 and Questions to 37607 34 ___________________________________ ___________________________________ ___________________________________ Slide 35 ___________________________________ A simple example Hydrogen and oxygen gas will react to form steam (gaseous water). Hydrogen and oxygen gases were mixed in a 2 L flask at 450 C. After equilibrium was established, it was determined that there were 3 moles of water, 1.2 moles of hydrogen and 0.8 moles of oxygen in the flask. What is the equilibrium constant for this reaction at 450 C? ___________________________________ ___________________________________ ___________________________________ Text 692019 and Questions to 37607 35 ___________________________________ ___________________________________ ___________________________________ Slide 36 ___________________________________ A simple solution 1st you need a balanced equation: 2 H2 (g) + O2 (g) 2 H2O (g) ___________________________________ This allows us to immediately write the equilibrium constant expression: ___________________________________ All we need to know are the concentrations! But the must be the EQUILIBRIUM concentrations. Text 692019 and Questions to 37607 ___________________________________ 36 ___________________________________ ___________________________________ ___________________________________ Slide 37 ___________________________________ A simple example Hydrogen and oxygen gas will react to form steam (gaseous water). Hydrogen and oxygen gases were mixed in a 2 L flask at 450 C. After equilibrium was established, it was determined that there were 3 moles of water, 1.2 moles of hydrogen and 0.8 moles of oxygen in the flask. What is the equilibrium constant for this reaction at 450 C? ___________________________________ ___________________________________ ___________________________________ Text 692019 and Questions to 37607 37 ___________________________________ ___________________________________ ___________________________________ Slide 38 ___________________________________ A simple solution Since we know the equilibrium concentrations: ___________________________________ ___________________________________ ___________________________________ Kc =15.63 Note: You could also calculate Kp by using the ideal gas law - more on that later. Text 692019 and Questions to 37607 38 ___________________________________ ___________________________________ ___________________________________ Slide 39 ___________________________________ You canNOT use MOLES in K ___________________________________ ___________________________________ ___________________________________ Text 692019 and Questions to 37607 39 ___________________________________ ___________________________________ ___________________________________ Slide 40 ___________________________________ A Note about Units Notice that I wrote the equilibrium constant without units even though the concentrations have units: ___________________________________ Kc =15.63 ___________________________________ Since the units of K will depend on stoichiometry, it do sn’ h v consis n uni s. I is lw ys written as a UNITLESS QUANTITY! Text 692019 and Questions to 37607 ___________________________________ 40 ___________________________________ ___________________________________ ___________________________________ Slide 41 ___________________________________ Clicker H2 and O2 gas will react to form steam. H2 and O2 gases were mixed in a 3 L flask at 450 C. After equilibrium was established, it was determined that there were 1.5 moles of water, 2 moles of H2 and 0.8 moles of O2 in the flask. What is Kc at 450 C? A. K=2.81 D. K=0.938 B. K=1.42 E. K=0.703 C. K=2.11 Text 692019 and Questions to 37607 ___________________________________ ___________________________________ ___________________________________ 41 ___________________________________ ___________________________________ ___________________________________ Slide 42 ___________________________________ A more complicated problem Hydrogen and oxygen gas will react to form steam (gaseous water). 4.36 g of hydrogen and 28.6 g of oxygen were mixed in a 2 L flask at 250 C. After equilibrium was established, it was determined that there was 6.6 g of water What is the equilibrium constant for this reaction at 250 C? ___________________________________ ___________________________________ ___________________________________ Text 692019 and Questions to 37607 42 ___________________________________ ___________________________________ ___________________________________ Slide 43 ___________________________________ A series of simple calculations 1st you need a balanced equation: 2 H2 (g) + O2 (g) 2 H2O (g) ___________________________________ This allows us to immediately write the equilibrium constant expression: ___________________________________ The question is: what are the equilibrium concentrations of all of the species? Text 692019 and Questions to 37607 ___________________________________ 43 ___________________________________ ___________________________________ ___________________________________ Slide 44 ___________________________________ Determining the concentrations ICE - ICE - BABY - ICE – ICE ___________________________________ The easiest way to solve this problem is by using an I-C-E ch r (“ic ch r ”) wh r I = initial concentration, C= change in concentration, and E = the equilibrium concentration. ___________________________________ ___________________________________ Text 692019 and Questions to 37607 44 ___________________________________ ___________________________________ ___________________________________ Slide 45 ___________________________________ An ICE Chart 2 H2 (g) + O2 (g) 2 H2O ___________________________________ (g) Initial ___________________________________ Change Equilibrium I+C I+C I+C ___________________________________ Text 692019 and Questions to 37607 45 ___________________________________ ___________________________________ ___________________________________ Slide 46 ___________________________________ What do you know? 4.36 g hydrogen * 1 mol H2 = 2.16 mol H2 2.016 g H2 (this is the INITIAL amount) ___________________________________ 28.6 g oxygen * 1 mol O2 = 0.894 mol O2 32.0 g O2 (this is the INITIAL amount) ___________________________________ 6.6 g H2O * 1 mol H2O = 0.366 mol H2O 18.02 g H2O (this is the EQUILIBRIUM AMOUNT) Text 692019 and Questions to 37607 ___________________________________ 46 ___________________________________ ___________________________________ ___________________________________ Slide 47 ___________________________________ UNITS! UNITS! UNITS! An ICE chart can use EITHER moles or concentration (molarity) or even pressure (atm), but you must use only one of these in any single ICE chart. ___________________________________ Kc uses molarity, so it is usually easiest to use concentration I will do the problem both ways! Text 692019 and Questions to 37607 ___________________________________ ___________________________________ 47 ___________________________________ ___________________________________ ___________________________________ Slide 48 ___________________________________ An ICE Chart 2 H2 Initial Change (g) + O2 (g) 2 H2O ___________________________________ (g) 2.16 mol 0.894 mol 0 mol ????? ????? ?????? ???? ????? 0.366 mol ___________________________________ Equilibrium ___________________________________ What is the change in quantities? Text 692019 and Questions to 37607 48 ___________________________________ ___________________________________ ___________________________________ Slide 49 ___________________________________ The “change” is all about stoichiometry! 2 H2 Initial Change (g) + O2 (g) 2 H2O ___________________________________ (g) 2.16 mol 0.894 mol 0 mol -2x -x +2x ???? ????? 0.366 mol ___________________________________ Equilibrium ___________________________________ Now it is easy to finish filling in the ICE chart! Text 692019 and Questions to 37607 49 ___________________________________ ___________________________________ ___________________________________ Slide 50 ___________________________________ An ICE chart is really just “accounting for moles” 2 H2 Initial Change Equilibrium (g) + O2 (g) 2 H2O ___________________________________ (g) 2.16 mol 0.894 mol 0 mol -2x -x +2x 2.16 – 2 x=? 0.894 – x=? 2 x = 0.366 mol ___________________________________ It is often helpful to use an ICE chart for other types of problems, it is a great way to keep track of what is going on. Text 692019 and Questions to 37607 ___________________________________ 50 ___________________________________ ___________________________________ ___________________________________ Slide 51 ___________________________________ Determining x allows me to fill in the rest of the chart 2 H2 (g) + O2 (g) 2 H2O ___________________________________ (g) 2.16 mol 0.894 mol 0 mol Change -2x -x +2x Equilibrium 2.16 – 2 x 0.894 – x 2 x = 0.366 mol Initial 2 x = 0.366 mol x = 0.183 mol ___________________________________ ___________________________________ Text 692019 and Questions to 37607 51 ___________________________________ ___________________________________ ___________________________________ Slide 52 ___________________________________ Determining x allows me to fill in the rest of the chart 2 H2 (g) + O2 (g) 2 H2O ___________________________________ (g) 2.16 mol 0.894 mol 0 mol -2 (0.183 mol) -0.183 + 2 (0.183 mol) 2.16 – 2 (0.183) 1.794 mol 0.894 – 0.183 0.711 mol 0.366 mol Initial Change Equilibrium mol ___________________________________ ___________________________________ Text 692019 and Questions to 37607 52 ___________________________________ ___________________________________ ___________________________________ Slide 53 ___________________________________ Now we need to calculate the concentrations and put them into Kc ___________________________________ ___________________________________ ___________________________________ Kc = 0.117 Text 692019 and Questions to 37607 53 ___________________________________ ___________________________________ ___________________________________ Slide 54 ___________________________________ We could do the same calculation using concentrations directly in the ICE chart ___________________________________ (this is the INITIAL concentration) ___________________________________ (this is the INITIAL concentration) ___________________________________ (this is the EQUILIBRIUM concentration) Text 692019 and Questions to 37607 54 ___________________________________ ___________________________________ ___________________________________ Slide 55 ___________________________________ An ICE Chart 2 H2 Initial Change + O2 (g) (g) 2 H2O ___________________________________ (g) 1.08 M 0.447 M 0M -2x -x +2x 1.08 – 2x 0.447 - x 2x = 0.183 M ___________________________________ Equilibrium 2x = 0.183 M x = 0.0915 M ___________________________________ Text 692019 and Questions to 37607 55 ___________________________________ ___________________________________ ___________________________________ Slide 56 ___________________________________ An ICE Chart 2 H2 (g) + O2 (g) 2 H2O 1.08 M 0.497 M 0M -2x -x +2x 1.08 – 2(0.0915) 0.447 – 0.0915 0.183 M 0.897 M 0.356 M ___________________________________ (g) Initial Change Equilibrium ___________________________________ ___________________________________ Text 692019 and Questions to 37607 56 ___________________________________ ___________________________________ ___________________________________ Slide 57 ___________________________________ Now we simply put the concentrations into Kc ___________________________________ ___________________________________ Kc = 0.117 ___________________________________ Same answer as before! Text 692019 and Questions to 37607 57 ___________________________________ ___________________________________ ___________________________________ Slide 58 ___________________________________ How to use Kc What if you already know the equilibrium constant? How do you use it? ___________________________________ If you know the equilibrium constant, you can use it to determine the equilibrium concentrations of all of the reacting species! ___________________________________ ___________________________________ Text 692019 and Questions to 37607 58 ___________________________________ ___________________________________ ___________________________________ Slide 59 ___________________________________ An ICE Chart I+ C=E 2 H2 (g) + O2 (g) 2 H2O ___________________________________ (g) Initial Change -2x -x +2x Equilibrium I+C I+C I+C ___________________________________ ___________________________________ Text 692019 and Questions to 37607 59 ___________________________________ ___________________________________ ___________________________________ Slide 60 ___________________________________ Another Simple Problem The Kc value for the formation of water from hydrogen and oxygen at 850 C is 4x10-6. If I mix 5.0 grams of hydrogen and 5.0 grams of oxygen in a 3 L flask at 850 C, what is the equilibrium concentration of the water? ___________________________________ ___________________________________ ___________________________________ Text 692019 and Questions to 37607 60 ___________________________________ ___________________________________ ___________________________________ Slide 61 ___________________________________ Another simple solution 1st you need a balanced equation: 2 H2 (g) + O2 (g) 2 H2O ___________________________________ (g) This allows us to immediately write the equilibrium constant expression: ___________________________________ ___________________________________ Text 692019 and Questions to 37607 61 ___________________________________ ___________________________________ ___________________________________ Slide 62 ___________________________________ Again, the Power of ICE 2 H2 (g) + O2 (g) 2 H2O ___________________________________ (g) Initial Change -2x -x ___________________________________ +2x Equilibrium ___________________________________ Th “Ch ng ” lin is lw ys jus s oichiom ry Text 692019 and Questions to 37607 62 ___________________________________ ___________________________________ ___________________________________ Slide 63 ___________________________________ We already know a couple of things 5.0 g hydrogen * 1 mol H2 = 2.48 mol H2 2.016 g H2 2.48 mol H2 = 0.827 M 3L 5.0 g oxygen * 1 mol O2 = 0.156 mol O2 32.0 g O2 0.156 mol O2 = 0.0533 M 3L Text 692019 and Questions to 37607 ___________________________________ ___________________________________ ___________________________________ 63 ___________________________________ ___________________________________ ___________________________________ Slide 64 ___________________________________ Again, the Power of ICE 2 H2 Initial Change (g) + O2 (g) 2 H2O ___________________________________ (g) 0.827 M 0.0533 M 0M -2x -x +2x 0.827 – 2 x 0.0533 – x ___________________________________ Equilibrium 2x ___________________________________ Now, we know everything – well, sort of. Text 692019 and Questions to 37607 64 ___________________________________ ___________________________________ ___________________________________ Slide 65 We have all of the equilibrium concentrations in terms of x… ___________________________________ …w c n us Kc to solve for x ___________________________________ ___________________________________ ___________________________________ Text 692019 and Questions to 37607 65 ___________________________________ ___________________________________ ___________________________________ Slide 66 ___________________________________ It looks like a mess… … nd i sor of is ( l hough your c lcul or can probably solve it) ___________________________________ BUT you can simplify it with a helpful assumption: ___________________________________ ASSUME x<<0.0533 ___________________________________ Text 692019 and Questions to 37607 66 ___________________________________ ___________________________________ ___________________________________ Slide 67 ___________________________________ If we assume x is small ___________________________________ ___________________________________ .0533-x ≈ 0.0533 0.827 – 2x ≈ 0.827 ___________________________________ Text 692019 and Questions to 37607 67 ___________________________________ ___________________________________ ___________________________________ Slide 68 ___________________________________ I’m NOT assuming x=0 I ’s qu s ion of sc l . I’m ssuming x is so sm ll h sub r c ing (or dding) i do sn’ ch ng h v lu . ___________________________________ ___________________________________ ___________________________________ Text 692019 and Questions to 37607 68 ___________________________________ ___________________________________ ___________________________________ Slide 69 ___________________________________ I’m NOT assuming x=0 Bill Gates is worth $52.5 billion. If he gives you $100, what is his net worth? A. B. C. D. $52.5 billion $52, 499,999,900 $52, 500,000,100 Who c r s, i ’s bu Brick. ___________________________________ ___________________________________ lo d of mon y, Text 692019 and Questions to 37607 ___________________________________ 69 ___________________________________ ___________________________________ ___________________________________ Slide 70 IT’S ALL ABOUT THE SIG FIGS, BENJAMIN! ___________________________________ 52,500,000,000 100 $52.5 billion to the right number of sig figs ___________________________________ ___________________________________ ___________________________________ Text 692019 and Questions to 37607 70 ___________________________________ ___________________________________ ___________________________________ Slide 71 Don’t forget to square EVERYTHING! ___________________________________ ___________________________________ ___________________________________ ___________________________________ Text 692019 and Questions to 37607 71 ___________________________________ ___________________________________ ___________________________________ Slide 72 Is x=1.91x10-4 really “much smaller than “0.0533” A. B. C. D. E. Yes No Maybe, it depends Yo, wh r ’d you g Yo, momma ___________________________________ ___________________________________ h gr n d , Brick? ___________________________________ ___________________________________ Text 692019 and Questions to 37607 72 ___________________________________ ___________________________________ ___________________________________ Slide 73 ___________________________________ Was the assumption good? We started by assuming x<<0.0533 ___________________________________ W now “know” h , wi h his ssump ion, x is 1.91x10-4 ___________________________________ Is 1.91x10-4 << 0.0533? 0.0533-0.000191=0.0531 Text 692019 and Questions to 37607 ___________________________________ 73 ___________________________________ ___________________________________ ___________________________________ Slide 74 ___________________________________ Critical Judgment How small is small depends on how accurate an answer you need: If you need 1 sig fig, than any number that is a factor of 10-20 smaller is insignificant. If you need 2 sig figs, then it must be about 100 times smaller. If you need 3 sig figs it must be about 1000 times smaller. ___________________________________ ___________________________________ A good general rule for our purposes is that if a number is <5% of another number, it is significantly smaller. This maintains 2 sig figs in all the concentrations – usually enough. ___________________________________ Text 692019 and Questions to 37607 74 ___________________________________ ___________________________________ ___________________________________ Slide 75 ___________________________________ 0.0533-x : x=1.91x10-4 ___________________________________ Compare 1.91x10-4 to 1/20th of 0.0533 IF 1.91x10-4<0.00267, I c ll i “good ssump ion”. If 1.91x10-4>0.00267, I c ll i “b d ss” ___________________________________ ___________________________________ Text 692019 and Questions to 37607 75 ___________________________________ ___________________________________ ___________________________________ Slide 76 ___________________________________ We put x back into the ICE chart 2 H2 Initial Change (g) + O2 (g) 2 H2O ___________________________________ (g) 0.827 M 0.0533 M 0M -2 (1.91x10-4) - 1.91x10-4 + 2 (1.91x10-4) 0.827 – 2 (1.91x10-4) 0.0533 – 1.91x10-4 = 0.827 = 0.0531 ___________________________________ Equilibrium 3.8x10-4 ___________________________________ And we have our answer. Text 692019 and Questions to 37607 76 ___________________________________ ___________________________________ ___________________________________ Slide 77 ___________________________________ Clickers! The Kc value for the formation of H2O from H2 and O2 at 500 C is 3.2x10-4. I put 5.0 mol of H2 and 5.0 mol of O2 in a 2 L flask at 500 C, what is the [H2O] at equilibrium? A. 0.035 M B. 0.050 M C. 0.001 M ___________________________________ ___________________________________ D. 0.070 M E. 0.100 M ___________________________________ Text 692019 and Questions to 37607 77 ___________________________________ ___________________________________ ___________________________________ Slide 78 ___________________________________ First Exam Yes, to graphing calculators – as long as there is no cheating. Yes, to cell phone calculators – as long as there is no cheating. ___________________________________ ___________________________________ You get 1 (ONE) 8-1/2 x 11 inch “formul sh ” – no restrictions. ___________________________________ Did I mention – NO CHEATING! Text 692019 and Questions to 37607 78 ___________________________________ ___________________________________ ___________________________________ Slide 79 ___________________________________ Topics on 1st exam 1. 2. 3. 4. 5. 6. 7. Intermolecular forces – bp, mp, phase changes Colligative properties – bp, mp, vp, osmotic pressure Phase Diagrams Heating/Cooling curves Rate laws Integrated Rate laws Arrhenius Equation Text 692019 and Questions to 37607 ___________________________________ ___________________________________ ___________________________________ 79 ___________________________________ ___________________________________ ___________________________________ Slide 80 ___________________________________ Exam Review Homework Online now. ___________________________________ Due at EIGHT 8 EIGHT 8 EIGHT p.m. Thursday. ___________________________________ Answer key will be available at 9 p.m on Thursday. ___________________________________ Text 692019 and Questions to 37607 80 ___________________________________ ___________________________________ ___________________________________