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Dario Bressanini1 G spontaneous Extent of Reaction, equilibrium Reaction Equilibrium for A B (reaction Gibbs energy)

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Dario Bressanini2

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3 1 B A Even though B has a lower value of the standard molar Gibbs energy the system can still achieve a lower Gibbs energy by having some A present in equilibrium with B.

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Dario Bressanini4 1 B A Generalizing, we can say that no chemical reaction of gases goes to completion; however it may be difficult to detect reactants at equilibrium if the products have a very much lower Gibbs energy. equilibrium

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Dario Bressanini5 Equilibrium Constants at equilibrium, at equilibrium, K p = K c (RT) n aA + bB cC + dD

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Dario Bressanini6 Values for the standard state are included in the equilibrium constant, but we usually do not write them down explicitly. Since their numerical value is unity, they really only serve to make the units work out. In the limit that solutes approach ideal behavior the activity coefficients approach unity. In terms of molarity

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Dario Bressanini7 Given that the standard Gibbs energy of reaction at 2000 K for the decomposition of water is 135.2 kJ/mol. Suppose steam at a pressure of 2.00 bar is passed through a furnace tube at that temperature. Calculate (a) the equilibrium constant K at 298 K. (b) degree of dissociation, and (c) mole fraction of O 2 present in the output gas stream. K 1 Reaction favors reactants. In other words very little water dissociates to hydrogen and oxygen. H 2 O(g) H 2 (g) + 1/2 O 2 (g)

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Dario Bressanini8 (b) To simplify notation let p j = p j /p , and since the magnitude of p equals unity, it will be dropped from the equations. H 2 O H 2 O 2 Initial n 0 0 Final (equilibrium) (1 ) n n 1/2 n Mole Fraction p = total gas pressure = 2 bar

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Dario Bressanini9 (b) continued Question: Does higher pressure favor reactants or products?

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Dario Bressanini10 (c) Calculate the mole fraction of O 2 (g) in the output stream.

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Dario Bressanini11 The Real K for Gas Reactions From slide two, K = (a C c a D d ) / (a A a a B b ) From slide two, K = (a C c a D d ) / (a A a a B b ) For gases the activity may be replaced by the fugacity. For gases the activity may be replaced by the fugacity. If the gas is ideal, the fugacity is the same as the pressure. Otherwise, the fugacity coefficient must be used and K = [( C c D d )/( A a B b )] [(f C c f D d )/(f A a f B b )] This means that K p is probably closer the true thermodynamic equilibrium constant that K c is. This means that K p is probably closer the true thermodynamic equilibrium constant that K c is. But ONLY b/c the standard states of gases are usually defined in terms of pressure rather than concentration.

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Dario Bressanini12 Ks for Liquid Solutions The activity of solute J in solution, in terms of its molality, is given by The activity of solute J in solution, in terms of its molality, is given by a J = J b J /b o Here b is the molality. It is divided by b o, the standard molality (usually exactly one molal) in order to avoid having any units for the activity. In terms of mole fraction the activity is given by In terms of mole fraction the activity is given by a J = J x J This can be used for both gases and liquid solutions.

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Dario Bressanini13 Equilibria with Pure Liquid or Pure Solid concentrations (or partial pressures) are actually ratios of the actual concentration to the concentration in the standard state. concentrations (or partial pressures) are actually ratios of the actual concentration to the concentration in the standard state.ACTIVITIES 2HgO(s) 2Hg( ) + O 2 (g)

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Dario Bressanini14 Activity BY CONVENTION: for solution standard state = 1 M; activity = concentration. for solution standard state = 1 M; activity = concentration. for gas, standard state = 1 atmos, activity = pressure for gas, standard state = 1 atmos, activity = pressure for solids & liquids: standard state = pure solid or liquid, whose concentration does not change; activity = 1 always for solids & liquids: standard state = pure solid or liquid, whose concentration does not change; activity = 1 always the concentrations of liquids and solids do not appear in the expressions for K c, K p

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Dario Bressanini15 K c = [O 2 ]; K p = p O2 independent of amount of solid or liquid present Example: evaporation of water Example: evaporation of water Vapour pressure in sealed vessel in presence of liquid is always the same (at given temperature) irrespective of how much liquid water is present cf textbook fig 17.4 Hence for 2HgO(s) 2Hg( ) + O 2 (g) : H 2 O( ) H 2 O(g)

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Dario Bressanini16 Le Chateliers Principle A stress applied to a system at equilibrium will respond in such a way as to relieve that stress. 1 moles of product compared to 1 mole of reactant. An increase in pressure at constant temperature favors the formation of water vapor H 2 O (g). An increase in temperature (thermal energy) at constant pressure favors the breakup of water to form hydrogen and oxygen gas (reverse reaction). An increase in temperature at constant pressure for an endothermic reaction drives the reaction towards products. In this case the formation of ethene and hydrogen gas.

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Dario Bressanini17 A.Adding or subtracting a product or reactant if species added is a solid or liquid, there is no effect on a gaseous equilibrium. if species added or subtracted is a gas then: a.adding species shifts direction away from the species added. b.subtracting a species shifts direction towards the species removed. Add product -> more reactant Remove product > more product When a reactant is added to a reaction mixture at equilibrium the reaction tends to form products When a reactant is removed, more reactant tends to form

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Dario Bressanini18 B. Change in volume 1.if volume is decreased, the reaction proceeds towards the side with least moles of gas. 2.if volume is increased, the reaction proceeds towards side with most moles of gas. 3.if in the balanced equation there are the same number of moles of gas on both sides, a volume change will not affect the equilibrium. Expansion -> drives reaction in direction that increases # of gas molecules Compression -> drives reaction in direction that reduces # of gas molecules

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Dario Bressanini19 Changing the volume of the container. Increase in the volume. Rxn will shift toward the side with the most moles of gas. Decreasing the volume. Rxn will shift toward the side with the fewest moles of gas. Changing the volume will make no difference if there are the same number of moles of gas on each side.

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Dario Bressanini20 C. Change in pressure 1. increase in pressure shifts equilibrium in direction of decrease in the number of moles of gas 2. decrease in pressure shifts equilibrium in direction of increase in number of moles of gas. 3. if in the balanced equation there are the same number of moles of gas on both sides, a pressure change will not affect the equilibrium. Increasing the pressure by adding an inert gas has no effect on the equilibrium concentration

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Dario Bressanini21 Changing the pressure on the container. Increase in the pressure. Rxn will shift toward the side with the fewest moles of gas. -Decreasing the pressure. Rxn will shift toward the side with the most moles of gas. Changing the pressure will make no difference if there are the same number of moles of gas on each side.

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Dario Bressanini22 H 2 (g) + I 2 (g) 2HI(g). In this example the number of moles of gas on both sides of the balanced chemical reaction is the same, so increasing or decreasing the total pressure will have little or no effect. The equilibrium system cannot by reaction change the number of moles of gas present. N 2 + 3H 2 2NH 3. The number of moles of gas changes in this reaction, from 4 to 2 as written. The equilibrium system can by reaction change the number of moles of gas present, so changing the total pressure will have a significant effect. Increasing the total pressure will cause the reaction to proceed to the right, decreasing total moles of gas, while increasing the total pressure will cause the reaction to proceed to the left (in reverse), increasing the total moles of gas.

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Dario Bressanini23 N 2 O 4 2NO 2. The number of moles of gas changes in this reaction, from 1 to 2 as written. The equilibrium system can by reaction change the number of moles of gas present, so changing the total pressure will have a significant effect. Increasing the total pressure will cause the reaction to proceed to the left (in reverse), decreasing the total moles of gas, while decreasing the total pressure will cause the reaction to proceed to the right, increasing the total moles of gas.

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Dario Bressanini24 Effect of Pressure on Dissociation The pressure depen-dence of the degree of dissociation, , at equi- librium for the reaction A(g) 2B(g) for different values of the equilibrium constant K. The pressure depen-dence of the degree of dissociation, , at equi- librium for the reaction A(g) 2B(g) for different values of the equilibrium constant K. The value = 0 corresponds to pure A = 1 corresponds to pure B. 1 =. (1 + 4p/K) =. (1 + 4p/K)

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Dario Bressanini25 Response of Equilibria to Temperature ln K 1/T Reaction enthalpies can be determined based on knowledge of equilibrium concentrations and temperature ln K = H o /R (1/T)

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Dario Bressanini26 Determination of K at different temperatures. Assuming that the reaction enthalpy does not vary much over the temperature range of interest we get the following result.

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Dario Bressanini27 A container holds a mixture of the isomeric compounds cis-2-butene and trans-2-butene at 400 K. What is the ratio of trans to cis isomer? cis trans r G can be found from the thermodynamic tables in your book. r G at 298 K

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Dario Bressanini28 r H can be found from the thermodynamic tables in your book. Now calculate K(T 2 ).

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Dario Bressanini29 Optimising Reaction Conditions CONCENTRATION CONCENTRATION remove the product or increase the reactants, ie make Q c smaller and push the reaction toward products eg liquefy the ammonia and recycle the N 2 and H 2 PRESSURE increasing the total pressure (for this reaction!) makes Q p smaller since there are 4 moles of gaseous reactants to 2 moles gaseous product; reaction is shifted toward products CATALYSTS No effect on value of K c or position of equilibrium but increases rate at which equilibrium is attained. N 2 (g) + 3H 2 (g) 2NH 3 (g) Run at high T, high pressure with V2O5 catalyst Run at high T, high pressure with V2O5 catalyst Gives compromise between rate and yield (ie, how fast reaction goes to equilibrium vs optimal concentrations at equilibrium Gives compromise between rate and yield (ie, how fast reaction goes to equilibrium vs optimal concentrations at equilibrium

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Dario Bressanini30 The Haber Process N2 + 3H2 2NH3 forwards reaction is exothermic Conditions for best yield of ammonia: a) Temperature - high temperature would favour the endothermic process, so if it's too high, this could decompose too much of the ammonia. If it's too low, the production would be too slow, or not enough ammonia would be produced - a compromise must be reached. b) Pressure - increased pressure will produce more ammonia, but it is expensive in machinery and energy and it can also be dangerous. At 1 atmosphere pressure the yield is very low, and so high pressure is necessary. c) Concentration - if the ammonia could be continuously removed, the equilibrium would shift to the right hand side. If extra nitrogen or hydrogen were added, the equilibrium would again shift to the right hand side. d) Catalyst - the addition of a catalyst doesn't increase the yield in any way, however, it does make the overall reaction proceed much quicker. Solution: A temperature of 500C, a pressure of 200 atmospheres, and an iron catalyst are used. The ammonium is removed by liquifying (condensing) and the unreacted nitrogen and hydrogen is continuously recycled. A yield of 15% is achieved, but it is quick and continuous and there is no wastage and so it costs relatively little. Equilibrium is never achieved.