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2-01 Equilibrium We have hinted that reactions can proceed in the forward or reverse direction • CH3-CH(OH)-CH3 with an H2SO4 catalyst CH3-CH=CH2 • But CH3-CH=CH2 with an H2SO4 catalyst CH3-CH(OH)-CH3 • This makes sense from our PE diagrams • The forward and reverse reactions just have different activation energies and opposite ∆H values A reversible reaction is said to be at equilibrium when the rate of the forward reaction is equal to the rate of the reverse reaction • If the products are able to escape from the reaction vessel they won’t be available to reform reactants o The reaction must be in a closed system Closed system • Nothing can enter or leave Assign 1-2 • The temperature affects equilibrium o A new equilibrium is attained at a new temperature • The same equilibrium will exist regardless of whether we start with excess reactants or products • When a system is at equilibrium no macroscopic changes occur Dynamic equilibrium (chemistry) • Equilibrium in which microscopic changes but not macroscopic changes occur • Constant microscopic back and forth between reactants and products but the rates cancel out and nothing is visible on the large scale • http://www.mhhe.com/physsci/chemistry/animations/chang_ 7e_esp/kim2s2_5.swf reversible-reactions.jar Static equilibrium (Physics) • No changes occur at all Assign 3-5 2-02 Characteristics of Equilibrium Since equilibrium reactions are reversible there is both a forward and reverse rate. REACTANTS ↔ PRODUCTS The rate of a reaction increases when the [reactants] increases • The rate must be proportional to the [reactants] • Rateforward = kforward [reactants] • Ratereverse = kreverse [products] • K is some constant for each reaction…a number Try 6 and 7 on graph paper Should have found that at equilibrium: • Rate of consumption of reactants = rate of production of reactants • [reactants] is not = [products] in general • [reactants] is constant and [products] is constant • forward and reverse rates do not change • a system which is not at equilibrium will tend to move towards equilibrium once equilibrium is reached, no further change occurs reversible-reactions.jnlp For H2 + Cl2 <--> 2HCl H2 + Cl2 = HCl The large boxes represent the total amounts at some equilibrium • the number of moles of each gas present is not evident just from the equation • the ratio of [product]/[reactant] is a constant at equilibrium more later… The small boxes are the amounts that react every second at equilibrium • total reactants changing = total products changing back= no net change • 1H2 and 1Cl2 react to make 2HCl while at the same time 2HCl react to make 1H2 and 1Cl2 • The rate that something is reacted is equal to the rate that it is produced Assign 8-13 2-03 Predicting if a reaction is spontaneous or not Spontaneous change • Occurs by itself without outside assistance • Some conditions allow while other forbid spontaneous reactions Exothermic reactions should be spontaneous provided the activation energy is not too high. PE Reaction proceeds If you give the reaction a little nudge it goes to completion • Match to paper • Ignite explosives… But endothermic reactions require constant energy input to proceed and are not expected to be spontaneous unless sufficient heat is available. • Molecules on the downhill side would be expected to stay there unless energy is supplied to them. • The side having minimum energy (minimum enthalpy) is favored • lazy There is more to spontaneity • Some endothermic reactions are spontaneous o Ammonium nitrate and water react spontaneously and absorb energy from the surroundings • Some exothermic reactions don’t go to completion o Equilibrium is established as the reverse endothermic reaction occurs Randomness is a second factor that comes into play for spontaneity • Most arrangements in space are disordered • Only a few special arrangements are ordered o Throw a deck of cards in the air • There are a few situations where order could develop • Card house • Back in sequence • All Spades together • Most situations from that random event are random (disordered) • A disordered result from a random event is more probable than an ordered result. As a crystal of salt dissolves water molecules strike the ions and they bounce out of the lattice • To get back in another molecule of water will have to hit it perfectly to go back into its ordered position o There are many other possible ways the ion could be hit that won’t take it back to its crystal o Disorder is favored and the crystal dissolves Entropy is the amount of randomness in a system. • Entropy tends to a maximum value • Things tend to get jumbled up over time • Slob Ba2+(aq) + 2SCN-(aq) 2NH3(g) + 10H2O(l) 2NH4SCN (s) + Ba(OH)2.8H2O(s) The reactants are highly ordered crystals and the products are very random ions so entropy favors the reaction even if enthalpy does not. The reactants absorb energy and then become so random that they can’t reorganize to reform. • The reaction is spontaneous and can drop the surrounding temperature by 55o C All reactions are driven towards maximum randomness (entropy) and minimum energy (enthalpy) • The universe favors the sloppy and the lazy Tendency to minimum enthalpy Exothermic AB + heat Endothermic C + heat D The arrow points to the tendency for minimum enthalpy • the down hill side • the side with heat as a reactant Tendency to maximum entropy • look at the relative randomness of phases • gases >> solutions > liquids >> solids H2O (s) ↔ H2O(l) • the liquid is more random and the products are favored CaC(s) + 2H2O(l) ↔ C2H2(g) + Ca(OH)2(aq) • the most random phase present is gas and the products are favored A(g) + B(s) ↔ 2C(g) + D(s) • if the reactants and products are in the same phase then the side with more molecules is more random (more possible disordered arrangements) Consider • C2H2(g) + 2Cl2(g) C2H2Cl4(l) + 386kJ o Enthalpy favors the products o Entropy favors the reactants o Opposite tendencies lead to equilibrium • CH4(g) + O2(g) CO2(g) + 2H2O(g) + 394 kJ o Enthalpy favors the products o Entropy favors the products o Reaction goes “100%” • 4Au(s) + 3O2(g) + 162kJ 2Au2O3(s) o enthalpy favors the reactants o entropy favors the reactants o the reaction goes “0%” no reaction o gold doesn’t oxidize! Try 14-16 Notes 2-04 Le Chatelier’s Principle Le Chatelier’s Principle • If a closed system at equilibrium is subjected to a change, processes will occur that tend to counteract that change • Whatever we do, nature tries to undo This principle is crucial to quickly predicting what will happen in equilibrium systems and will be used throughout the course. Fe3+ + SCN- FeSCN2+ • If we increase [Fe3+] or [SCN-] then the [FeSCN2+] increases. • If we decrease [Fe3+] or [SCN-] then the [FeSCN2+] decreases. Think of the water in a bathtub representing a system at equilibrium with the reactants at one side of the tub and the products at the other. • If you dump some water at one end of the tub the water flows towards the other end of the tub to reestablish equilibrium • If you scoop out some water at one end of the tub the water flows back towards the “hole” to reestablish equilibrium. The effect of temperature changes: • If we decrease the temperature in a reaction like: 2NO(g) + Cl2 (g) ↔ 2NOCl(g) + 76kJ • We removed energy (a product) • It’s like we scooped out some water on the product side • The reaction shifts to the right to compensate • More reactants become products Cl2 NO […] NOCl Time at which temp changed [Cl2] only changes half as much as [NO] since they react in a 1:2 ratio Time The effect of concentration changes: • If we increase the [Cl2] in a reaction like: 2NO(g) + Cl2 (g) ↔ 2NOCl(g) + 76kJ • We added a reactant • It’s like we dumped some water on the reactant side • The reaction shifts to the right to compensate • More reactants become products Cl2 Time at which Cl2 was injected quickly NO Not all the Cl2 added is used up… the change can’t be cancelled completely because the other parts changed […] NOCl Time The effect of pressure changes: • A decrease in volume increases the pressure and the concentration of all the gases. 2NO(g) + Cl2 (g) ↔ 2NOCl(g) + 76kJ • Figure out which side has the most gases (3 moles reactants and 2 moles products) • It’s like we dumped some water on the high gas side (reactants) • The reaction shifts to the right to compensate (want fewer gases at high pressure) • More reactants become products Cl2 Time at which pressure changed NO All gases increase in concentration at first and then the equilibrium shifts to make products […] NOCl Time The effect of adding a catalyst • Since a catalyst speeds up the forward and reverse rate the reaction remains at equilibrium • Just gets there faster if it wasn’t already there. Graphing notes: • Temperature changes • Concentrations change slowly to a new value • Changing the concentration of a species • One species suddenly jumps up or down then all adjust slowly to compensate • Changing the pressure • The concentration of all gasses simultaneously jump up or down then slowly adjust. Assign 17-28 PS It helps to draw a little mound or hole over the substance that increases or decreases to visualize which way the equilibrium will shift. Notes 02-05 Equilibrium in industry In WWll Germany couldn’t get nitrates from the main global source, Chile. Nitrates are used to make Trinitrotoluene (TNT). Fritz Haber figured out how to make ammonia on a huge scale, which can easily be oxidized to nitrates. The Haber Process for making ammonia (important to remember) • N2(g) + 3H2(g) ↔ 2NH3(g) + 92kJ Figure out what conditions will maximize the yield of ammonia • Pressure…high • Temperature? o Reaction rate = high vs yield = low • Other…iron oxide catalyst http://www.youtube.com/watch?v=c4BmmcuXMu8 The Making of Cement from Limestone • CaCO3(s) + 175kJ ↔ CaO(s) + CO2(g) • Best conditions are o Pressure….low o Temperature…high Is there a conflict between rate and yield?…no Notes 2-06 The equilibrium expression and the equilibrium constant Pull out the two graphs we drew for questions 6 and 7. At the first equilibrium there was 5 times as much B as A. At the second equilibrium there was 5 times as much B as A. • The constant ratio is not a fluke! For the equilibrium equation: A+B⇔C+D it has been found that [C ]× [D] = a constant [A]× [B] the expression is called the equilibrium expression and Keq is a number called the equilibrium constant. K eq = Examples • H2O(g) + CO(g) ↔ H2(g) + CO2(g) o K eq = [H2(g) ]× [CO2(g)] [H2O(g) ]× [CO(g)] • PCl5(g) ↔ PCl3(g) + Cl2(g) o K eq = [PCl3(g) ]× [Cl2(g)] [PCl5(g) ] • H2(g) + F2(g) ↔ HF(g) + HF(g) or H2(g) + F2(g) ↔ 2HF(g) 2 o K eq = [HF(g) ]× [HF(g)] = [HF(g) ] [H2(g) ]× [F2(g) ] [H2(g) ]× [F2(g) ] o The coefficient becomes the exponent In general: • aA + bB + cC +… ↔ pP + qQ + rR +… gives p q r [ P ] × [Q] × [R ] ... • K eq = a or [A] × [B]B × [C ]c ... • K eq = [Products ] [Reactants] But there is a little wrinkle: • for CaF2(s) ↔Ca2+(aq) + 2F-(aq) • [Ca = 2+ ]× [F − ] 2 = 8.4 ×10 −13 (measured experimentally) [CaF2(s)] • but for a pure solid the concentration is a constant because its density is a constant o so [CaF2(s)] = 40.7M from 3.18 g/L and 78.1 g/mol o if the concentration of the solid never changes the expression is really: K eq ( aq ) K eq ( aq ) [Ca = 2+ ( aq ) ]× [F − ] 2 ( aq ) 40.7 = 8.4 ×10 −13 Why bother having 2 constants in the same expression? Combine the constants to get K eq ( new) [Ca = 2+ ( aq ) ]× [F − ] 2 ( aq ) = 3.4 ×10 −11 Instead of having to calculate the concentration of a solid every time, just combine it into the Keq. • Eliminate any concentrations that have a constant value in the equilibrium expression o Solids…the concentration comes from the density which doesn’t change o Pure liquids…the concentration can come from density Only counts if it is the only liquid that exists in the entire equilibrium expression If there is more than one then dilution occurs and concentration will vary 2 [ HBr ( g ) ] • Br2(l) + H2(g) ↔ 2HBr(g) gives K eq = [H2(g)] • CH3COCH3(l) +Cl2(g) ↔ CH3COCH2Cl(l) + HCl(g) gives • Cl2(g) + 8H2O(l) ↔ Cl2.8H2O(s) gives Adding a reactant or a product which is a solid or pure liquid will have no effect on the equilibrium Assign 31-35 02-07 Le Chatelier’s Principle and the Equilibrium Constant When the concentration pressure or temperature are changed the reaction tends to counteract these changes. • One shot change The system regains equilibrium • Continuous change You have an open system Does not reach equilibrium When temperature is decreased and held 2NO(g) + Cl2(g) ↔ 2NOCl(g) + 76kJ shifts to the product side. This results in an increase in [products] and a decrease in [reactants] so that increases. To figure out the direction of the change in Keq draw an arrow in the Keq expression that shows the shift. • An upward arrow indicates a higher Keq and vice versa • Only changes in temperature change Keq Large Keq means that [products] > [reactants] • lots of products at equilibrium Small Keq means that [products] < [reactants] • lots of reactants at equilibrium Assign 36-46 02-08 Equilibrium Calculations To determine the value of Keq for any reaction you need to accurately know the concentrations of all of the species present. • Must be experimentally determined Example A For 2NO(g) + O2(g) ↔2NO2(g) A 2.0 L bulb contains 6.00 mol of NO2(g), 3.0 mol of NO(g) and 0.20 mol of O2(g) at equilibrium. What is the Keq? First determine the equilibrium expression: Use the data to calculate each of the concentrations: Substitute the values into the expression: [NO 2 ]2 [3.0]2 = = 40 K eq = [NO]2 [O 2 ] [1.5]2 [0.10] Example B 4.00 mol of NO2 is introduced into a 2.00L bulb. After a while equilibrium is attained. At equilibrium 0.50 mol of NO is found. What is Keq? The question implies that time had to pass for equilibrium to pass. The initial concentration is not an equilibrium concentration. ∴ you need to figure out the equilibrium [ ]’s. [NO 2 ]2 The Keq expression is K eq = [NO]2 [O 2 ] Calculate the concentrations from the questions: 4.00mol = 2.0M 2.00L Set up an “ice” box [NO 2 ]start = 2NO(g) [NO]eq = 0.500mol = 0.250M 2.00L O2(g) ↔ 2NO2(g) Initial 0 0 2.00 Change Equilibrium 0.250 Units are omitted but they must all be the same (usually all M but it can work with mol, contrary to what your book says) Fill in the rest of the box. The values in the change row obey the rules of stoichiometry. 2:1:2 ratio ∆ NO = + 0.250 (a shift left) ∆ O2 = ½ ∆ NO = + 0.125 (the shift added to the reactants) ∆ΝΟ2 = − ∆ NO = - 0.250 (the shift took away NO2) Initial Change Equilibrium 2NO(g) O2(g) 0 +0.250 0.250 0 +0.125 0.125 ↔ 2NO2(g) 2.00 -0.250 1.75 Substitute the equilibrium concentrations into the equilibrium expression: [NO 2 ]2 [1.75]2 K eq = = = 392 [NO]2 [O 2 ] [0.250]2 [0.125] Example C An unknown amount of NO2 was introduced to a 5.00L bulb. When equilibrium was attained according to the equation 2NO(g) + O2(g) ↔2NO2(g) the concentration of NO was 0.800M. If Keq is 24.0, how many moles of NO2 were originally put in the bulb? There is a start and then time passing before equilibrium. We know [NO]eq = 0.800M and we want [NO2]start = x. Set up the ice box Initial Change Equilibrium 2NO(g) O2(g) 0 0 ↔ 2NO2(g) x 0.800 [NO] went from 0 to 0.800 so the shift was left. Fill in the box with changes that match the stoichiometry of 2:1:2. ∆ NO = + 0.800 ∆ O2 = ½ ∆ NO = + 0.400 ∆ΝΟ2 = − ∆ NO = - 0.800 Initial Change Equilibrium 2NO(g) O2(g) 0 +0.800 0.800 0 +0.400 0.400 ↔ 2NO2(g) x -0.800 x-0.800 Substitute the values into the equilibrium expression: [NO 2 ]2 [x - 0.800]2 = = 24.0 K eq = [NO]2 [O 2 ] [0.800]2 [0.400] You won’t have to use the quadratic equation in chem. 12. There will always be a way around it. (x-0.800)2=(24.0)(0.800)2(0.400)=6.144 take the square root of both sides x-0.800=2.479 x=3.279M = [NO2]start We needed moles at the start so: mol 3.279 × 5.00L = 16.4mol # moles NO2 = L Example D Keq= 49 for 2NO(g) + O2(g) ↔2NO2(g). if 2.0mol of NO , 0.2mol O2 and 0.40mol of NO2 are put into a 2.0L bulb, which way will the reaction shift in order to reach equilibrium? This question is asking for a decision not a numerical answer. You need to compare two numbers [NO 2 ]2 K eq = = Q , where Q is a trial value for Keq [NO]2 [O 2 ] if Q = Keq then the system is at equilibrium if Q < Keq then [products] is too small and the reaction will shift to [reactants] products to raise Keq if Q > Keq then [products] is too big and the reaction will shift to [reactants] reactants to lower Keq First find the equilibrium expression: K eq = [products] [NO 2 ]2 = [reactants] [NO]2 [O 2 ] Calculate the starting concentrations: [NO]start = 2.0mol = 1.0M 2.0L [NO 2 ]start = 0.40mol = 0.20M 2.0L Calculate the reaction quotient ∴ the reaction must shift right to the products Example E Keq = 3.5 for SO2(g) + NO2(g) ↔ SO3(g) + NO(g). If 4.0 mol of SO2(g) and 4.0 mol of NO2(g) are placed in a 5.0 L bulb and allowed to come to equilibrium. What will be the equilibrium concentrations of all the species? First write the equilibrium expression K eq = [SO 3 ][NO] [SO 2 ][NO 2 ] Set up an ice box Initial Change Equilibrium SO2(g) 0.80 NO2(g) 0.80 SO3(g) 0 NO(g) 0 We will need to enter variables for the rest of the spaces. It’s easiest to assign x as the change for one of the species: SO2(g) NO2(g) SO3(g) NO(g) Initial 0.80 0.80 0 0 Change -x -x +x +x Equilibrium 0.80-x 0.80-x x x Substitute the equilibrium concentrations into the Keq expression: [SO 3 ][NO] x2 K eq = = = 3.5 take the square root of both [SO 2 ][NO 2 ] (0.80 - x) 2 sides ∴[SO3]=[NO]= 0.52M; [SO2]=[NO2]=0.28M Example F A 1.0 L reaction vessel contained 1.0 mol of SO2, 4.0 mol of NO2, 4.0 mol of SO3 and 4.0 mol of NO at equilibrium according to SO2(g) + NO2(g) ↔ SO3(g) + NO(g). If 3.0 mol of SO2 is added to the mixture, what will the new concentration of NO be when equilibrium is re attained? You have all the information needed to calculate Keq. Set up an ice box (any additions or removals are put in the initial line) SO2(g) NO2(g) SO3(g) NO(g) Initial 1.0 +3.0 4.0 4.0 4.0 Change -x -x +x +x Equilibrium 4.0-x 4.0-x 4.0+x 4.0+x Take the root of both sides ∴ [NO] = 4.0 + x = 5.3M Assign 47-65 odd and 66.