Problem #1 6 mol of SO and 4 mol of O are plaed into a 1 L flask at temperature, T. The equilibrium onentration of SO is found to be 4 mol/l. Determine K. SO (g) + O (g) SO (g) K = [SO ] / [SO ] [O ] Answer: Easiest way to determine equilibrium onentrations is to set up a table as follows: SO + O SO initial on 6 4 0 hange -4 - +4 final (eq) on 4 note: hange of +4 for [SO ] is found by differene, hanges for [SO ] and [O ], -4, -, are found from the stoihiometry of the reation Now K = [SO ] / [SO ] [O ] = 4 / x = Problem # Here we start with 1.00 atm of N O 4 (g) and K p = 11 at 100 o C Calulate the partial pressures of eah gas at equilibrium Answer: N O 4 (g) NO (g) Initial 1.0 atm 0.0 atm hange -x +x equilibrium (1.0-x) x set up Equilibrium onstant expression : K p = 11 = PNO = (x) PN O 4 (1.0 -x) 4x +11x -11 = O using quadrati formula we get x = 0.78 therefore P N O 4 = 1.0 0.78 = 0. atm P NO = x= (0.78) = 1.56 atm at equilibrium 1
Problem #.0 moles of Br is plaed in a.0 L flask and it quikly dissoiates into Br radials under UV irradiation. Only 10 % of the Br dissoiates. Calulate K. Answer is K = 0.044 (ask me or Joe Antle for solution) 1.6 Fators that Alter the Composition of an Equilibrium Mixture. One a system is at equilibrium, we an hange the ratio of reagents to produts (i.e. disturb the equilbrium) by; 1) adding or removing one or more reagents or produts (hanging the on n or partial pressure of those substanes) ) ompressing or expanding the system (when dealing with gaseous reations) ) hanging the temperature We would like to know how to maximize a produt yield for a reation with a minimum of energy (and money) input. If a reation goes nearly to ompletion, then this isn t muh of a problem. However, most reations don t go to near ompletion (what does this say about their equilibrium onstants?) and so we must adjust experimental onditions so the reation proeeds as favourably as possible.
One the equilibrium has been disturbed, the reation will then inrease in rate either in a forward diretion or in the reverse diretion in order to re-establish equilibrium again. This fat was first stated by Le Chatelier in 1884 and is known as Le Chatelier s Priniple If a system at equilibrium is disturbed by a hange in onentration, pressure or temperature, the system will, if possible, shift so as to ounterat the hange (restore equilibrium onditions) 1.7 Altering an Equilibrium Mixture: Changes in Conentration N (g) + H (g) NH (g) K = 0.91 at 700 K. At first, the system is in equilibrium, with [N ] = 0.50 mol/l, [H ] =.00 mol/l, and [NH ] = 1.98 mol/l. At a point in time, the onentration of N is inreased to 1.50 mol/l by adding N to the reation system.
Le Chatelier s Priniple states the system will re-establish equilibrium by reating in suh a way as to derease the stress to the system. Sine we have added a reatant, the reation should proeed towards produts to minimize the amount of extra N in the system. N (g) + H (g) NH (g) This is refleted in the Figure where the reatant onentrations derease and the produt Conentration inreases until a new equilibrium mixture is reahed. Using the onentrations of the ammonia example, we saw before the introdution of more nitrogen that the reation quotient would be: NH (1.98) 0. K N H (0.50)(.00) Q 9 The system is at equilibrium. Obviously, if we add nitrogen to the system, the reation quotient will hange to: NH (1.98) 0. K N H (1.50)(.00) Q 0968 we see the reation quotient is now less than the equilibrium onstant, meaning the reation must move from left to right to reah the (new!) equilibrium mixture. At the new equilibrium, the onentrations are found to be [N ] = 1.1 mol/l, [H ] =.4mol/L, and [NH ] =.6 mol/l. The reation quotient equals the equilibrium onstant! NH (.6) 0. K N H (1.1)(.4) Q 96 4
Problem 1.16 Consider the equilibrium for the water-gas shift reation: CO (g) + H O (g) CO (g) + H (g) Use Le Chatelier s Priniple to predit how the onentration of H will hange when the equilibrium is disturbed by: a) Adding CO: Answer: More CO (a reatant) means the reation will shift to redue the amount of reatants by reating produts. Therefore the onentration of H will inrease. b) Adding CO: Answer: More CO (a produt) means the reation will shift to redue the amount of produts by reating reatants. Therefore the onentration of H will derease. Problem 1.16 (ontinued) CO (g) + H O (g) CO (g) + H (g) ) Removing H O: Answer: Less H O (a reatant) means the reation will shift to inrease the amount of reatants by onsuming produts. Therefore the onentration of H will derease. d) Removing CO ; also aount for the hange using the reation quotient Q : Answer: Less CO (a produt) means the reation will shift to inrease the amount of produts by onsuming reatants. Therefore the onentration of H will inrease. Reation quotient: Imagine the initial equilibrium onentrations of the hemials as written from left to right in the balaned equation are w, x, y, and z. The equilibrium onstant will be K = (yz) / (wx) If we remove, let s say, half of the CO (onentration of y/), the reation quotient will be Q = (yz)/(wx) whih will be less than K. This means the reation proeeds from left to right, meaning more hydrogen is formed. 5
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