Question

please answer all

5. A 50.0 L reaction vessel contains 1.00 mol N., 3 moim vessel contains 1.00 mol N3 mol H, and 0.500 mol NH Will more ammonia ed or will it be removed when the mixture goes to equilibrium at 400°C? The reaction N2 (9) + 3H2 (9) - 2 NH (9). The value of K is 0.521 at 400°C. 6. At a certain temperature, K. is 4.13 x 10-2 for the equilibrium: 2 Br (g) 1 (9) + Bry (9) Assume that equilibrium is established at the above temperature by adding only Br (9) to the reaction flask. What are the concentrations of I, (q) and Br, (g) in equilibrium with 0.0124 moles/liter of IBr(9) ? Ans: 2.52 x 103 M What was the initial concentration of Br before equilibrium was established? Ans: 0.0174 7. The equilibrium constant for the reaction: 2 NO (9) N, (g) + O, (g) is 2.60 x 10 at 1100 °C. If 0.820 mole of NO (9) and 0.223 mole each of N, (g) and o, (g) are mixed in a 1.00 liter container at 1100 °C, what are the concentrations of NO (g), N (9), and O. (g) at equilibrium? Ans: [NO]= 1.15 M [N] = 0.059 M (0.1= 0.059 M

Answer 1

Tuesday NOVEMBER 2 Brig) ř Iz (g) + Brisg ke=4.38 102 Since Let the the reaction Initial started by moles of taking only Ibr by Ibr. c C-2x k= [born] [[₂] Given I Br] at Gm = 0.024 [Ibs] 2 4.13 X 102 - ² (0-0124)2 = 140.0124)-(4-13x10-2) X = 0.0025232.52x10-3M Conch of Iz = br=2.52 x 103 M Initial conch of IBr = (-2x = 0.0124 3) c = 0.0124 +240-00259). Ic = 0.0174 M I 200 19 = N(g) + O₂(g) K= 2.6*10-3 0.820 0.223 0.223 Qc = (0.223) (0.223) 20.073 (0.8202 Since value of Q7K 8x4 stufts in backward dish to attain DECEMBER 0.0124 10.820+ 2x 0 223-2 0. 223-2

Wk 48/331-034 TUBER 2012 TUVLIVIBER 1 2 15 16 3 4 5 6 7 8 9 10 11 12 13 17 18 19 20 21 22 23 24 25 26 27 5. 2017 N₂ +3th 29 30 31 SMTWTFS 20th reaction Qc = Quotient Kc = [NH3)2 [12] [112]3 K = 0.521 at 400°C (0.5) ² (5072 (3) 3 = 23.14 since the value of shifts in buckward this NH will be equilibrium. Qc > ko, this the reaction direction to acheive equilibrium. decreases as the reaction approach

Saturday NOVEMBER 23 24 2017 - 2.68103 [ Or] [Ne) [NO]2 = 2.68 10 3 (0.223 - 2C) (0.223-x) (0.820 +221) > x = 0.164 & [NO] - 0.820 + 20 2 1.148 M ų 1.15 M 2 [02] [N2) = 0.223 – X = 0.059 M Sunday DECEMBER