Question

3. The scenes below show a reaction involving the atoms X (orange) and Y(black). The molecular reaction is: X+Y2 → XY+Y. Assume each molecule is 0.1 mol and the volume is 1 Liter. 1 2 3 A) If K=4.5, which mixture is at equilibrium? Show your work. B) Rank the mixtures from most positive AG to most negative AG. Explain why. C) What is the difference between AG and AGº? When does AG = AGº? IF

Answer 1

хо у — ху x+y, Hху +7 , k = 4.5 ke Сх7] Су] (x] (77) 1 2 3 ха 0.1 mal x — оЗmol х О. 2 ma) Ча 0.2 mo) 7 - 0:2 тд) у — 0.3 mg] ХУ - о: 4 то) ху - о: 2то) 24 7 0.3 md) О-І то | 9, - 7, 6:2 mo) y, Po-1 me) К. К. k Схүј Су] Сх] (72) Сх7] Ту] Сх] (77) Сх7] Су] Сх] (72) (оч) (0:2) 10 (о: 2) (0- 2) (o-3) (62) ka - (03) (0:3) (o'2) (от, (0:0) (0-7) 5 = 8 Kc 2. с О: k= 4-5 cet equilibrium (9) sol mixture 3 is dhe

(3) Kc an will means on on= - RT enka increasing the decrease. so higher k = more reasing crcler mixture (2) < mixture (3) < mixturec3) negative on so Kc incore so order of on from most positive to most rejective mixture (3) mixture (2) < mixture (3) < mixture most positve 2 must negative

ct (C) OG > Standard standard gibbs free energy change conditions. like cat as c I catm AG - at Gribbs free energy change any condition ou = Ari + RT lnke k = equilibrium consstant so for og هون Rilnke こ which is only possible at ke=1 so when then we can Kc=1 Ari < 07\w > say ou ou