Question

PART IV: We can model the equilibrium process using BB's to represent molecules and petri dishes to separate "reactant molecules" from "product molecules." 1) Consider the simple equilibrium A = B For this BI Кф [A] and we'll assign a value to Koq of 8. Keq = 8.0 Get 2 petri dishes and place 27 BB's in one of them. These represent 27 "A" molecules, while the empty dish shows that there are "B" molecules. no [B] value is 8 [A] Now move BB's from the "A" dish to the "B" dish until the At this point your system has reached "equilibrium." How many "A's" are there at equilibrium? How many "B's?" b) Now add to your "equilibrium" 18 new "B" molecules (18 BB's in the "B" petri dish, and a total of 45). Find the equilibrium values. Reminder: Keq doesn't change-the reaction shifts left or right. What are the equilibrium values of [A] and [B]? [A] [B с%3D D Kag 1/15 0.067 Now consider another new equilibrium 2) Begin with a total of 32 molecules (BB's) all representing "D." What are the equilibrium values of "C" and "D", assuming the total number of molecules (BB's) remains at 32 [CJ [DJ b) Keq for A B was greater than 1 and Ke for C =D was less than 1. What does the size of the Keq have to do with the relative amounts of reactants and products present at equilibrium? Occasionally one another. These equilibria cannot be rewritten into one equation since they really are separate processes that just happen to "overlap" one another. We experienced this in Part I II and III when the copper (II) ions were in equilibrium with hydroxide ions in one process and with ammonia in a separate process. Consider the following reactions occurring at the same time in the same container: two or more equilibria are sharing reactants or products and therefore affect 3) Keq 0.25 and F G Keq 5.0 [F] [G] 0.25 and 5.0 [F] [E] Get 3 petri dishes to represent E, F, and G molecules. Move BB's from E to F and from F to G until both equilibria are established. a) Begin with 30 BB's, all representing E molecules. What are the final amounts of b) Add 10 more molecules of "F" and find the new equilibrium values of [E], [F], and each type of molecule/BB? [G]. (40 total BBs) [EJ [E] [F] [F] G [GI

Answer 1

A 3 24 A B 8= 42 27 27-K E 3td 42-H IG [AJ 21-x 9x 216 A 24 42-y atequilibrium B 24 A 27-243 42-y 24t8y 9 newA 3t2= 5 - 42-240 C D Kea- 15 = 0 067 2. 32-7 D Kea 0.067 -0-067 32 32 1067x 32-a 0 32-30-2 30 with the sizel amount eaactants oybtarting umber equilin efeactartpaduet to do nothing has Ken equilibniwn Fnom and products at statiliy depen ding On wia each