The Equilibrium Constant, K P

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Marsha Anderson
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1 The Equilibrium Constant, K Consider the gas-phase reaction: aa + bb yy + zz The thermodynamic equilibrium constant K is Yy z Z K = aa b B (standard state is 1 bar pressure) And by the same argument used earlier, we find G = RT ln K

2 Concentration Equilibrium Constant, K C Consider again the gas-phase reaction: nrt V aa + bb yy + zz crt, where is now a concentration, with units 3 c mol/ dm. Define the standard equilibrium constant K C K Y Z A B (Usual standard state is 1 mol dm -3 ) y z o C a b G = RT ln K C Note that K = K C (RT) n n y z a b K C = K / (RT) n

3 ConcepTest 1 NO (g) N O 4 (g) 300K A. For the above reaction, find K C if K = 6.9 bar 1 RT=0.083 bar dm 3 K -1 mol -1 x 300 K = 4.9 bar dm 3 mol -1 K C = K /(RT) -1 A 17 dm 3 /mol B 0.8 dm 3 /mol C 0.8 mol/dm 3 D 17 mol/dm 3 B. If mol NO is introduced into an empty 1 dm 3 vessel, find the number of moles of [N O 4 ] at equilibrium. (still 300K) A. Approximately 0.0 D. Approximately 0.75 B. Approximately 0.5 E. Approximately 1.0 C. Approximately 0.5 Work in groups, and take 5-10 minutes to get an answer. I will then harrass someone to explain each answer. Let s now look at a solution:

4 Solution If 1.00 mol NO is injected into a 1 dm 3 vessel, find [N O 4 ] at equilibrium. At equilibrium, x K C = 17 dm 3 /mol and K = 6.9 bar 1 NO (g) N O 4 (g) 300K (1-x)/ moles of each Assuming gases are ideal, then NO n NO and N O 4 n N O 4 I will use K C to find [N O 4 ], but I could just as well have used K 1 x mol 3 3 Simplifying, dm 344x 1 x or dm x x 1 0 mol mol x 3 dm Using my quadratic equation solver, I find x = and x = Two solutions!! Are both right? Are both wrong? Did I make a mistake? Answer: an additional constraint: 0< x < 1 x= x NO = mol/dm 3 NO x N O 4 = (1-x)/ = mol/dm 3 N O 4

5 Effects of Changes on Equilibria The Le Chatlier rinciple results from consideration of the effects of temperature, pressure or quantity changes on an equilibrium constant. E.g., NO (g) N O 4 (g) The reaction shifts to the left when the # moles of N O 4 is increased. The reaction shifts to the right when the total pressure is increased. We now consider in a bit more detail pressure and temperature effects on equilibria. (Reasons that we might do this include having tests for equilibrium, and to have ways to measure H or S.)

6 Effects of ressure on Equilibria We know G = RT ln K ΔG is defined at = = 1 bar, and so it is independent of pressure. (ideal gases assumed) For the gas-phase reaction aa + bb yy + zz K Obviously, K must also be pressure independent. y z o Y Z a b A There are three ways that one could imagine changing the pressure: a) Add an inert gas to the reaction mixture b) Change the volume of the reaction mixture c) Change the quantity of one of the components (eg, A,B, Y or Z in above) Take a brief look at each of them. B

7 Effect of adding an ideal inert gas, M Add gas M aa + bb yy + zz K y A B M y z Y a Z A z o Y Z a b mm+ aa + bb yy + zz+ mm A B If there is no change in volume, then the partial pressures of each of the ideal gas components remains unchanged by the addition of M. No change y z m in any partial o Y Z M K pressures. a b m K is unchanged! If the reaction were taking place in an isolated system, M would almost surely also affect the final temperature G = RT ln K B b However, M might well affect the rate at which the process occurs.

8 Change the volume Changing the volume can have an effect if the number of moles of gas on the left and right sides of the equation are different. N O 4 NO Let be the fraction of the dimer that has dissociated. N(1-) N N in moles Total number of moles of gas = N(1 - ) + N = N(1 + ) We now write the mole fractions of each component: NO N NO N N 1 1 N

9 Change the volume () We use the mole fractions to obtain the partial pressures of each component: 1 NO 4 NO 1 1 Use these to write the pressure equilibrium constant Solve for : K NO 4 NO 1 4 K K 4 The degree of dissociation decreases when the pressure increases. Another manifestation of the Le Chatlier rinciple

10 Change quantity of one component NO (g) N O 4 (g) K NO NO The system will shift in a direction away from the component that was increased. This is a method that is often used to determine whether or not equilibrium has actually been attained. (Rather than simply having an equilibration time >> observation time) 4

General Chemistry revisited

General Chemistry revisited A(g) + B(g) C(g) + D(g) We said that G = H TS where, eg, H = f H(C) + f H(D) - f H(A) - f H(B) G < 0 implied spontaneous to right G > 0 implied spontaneous to left In a very