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
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
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 1.000 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:
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 17 344x x 1 0 mol mol x 3 dm Using my quadratic equation solver, I find x = 0.0548 and x =-0.0554 Two solutions!! Are both right? Are both wrong? Did I make a mistake? Answer: an additional constraint: 0< x < 1 x= x NO = 0.0548 mol/dm 3 NO x N O 4 = (1-x)/ = 0.474 mol/dm 3 N O 4
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.)
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
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.
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 1 1 1 1 4
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
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