1 / 8

This problem can best be analyzed by an ICE table

redford
Download Presentation

This problem can best be analyzed by an ICE table

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Nitrosyl chloride decomposes by the following reaction:2 NOCl (g)  2 NO (g) + Cl2 (g) The equilibrium constant, Kc , at 50 C is 1.6 x 10 –7. An evacuated chamber is filled with nitrosyl chloride gas to a concentration of 2.00 M. The mixture is allowed to reach equilibrium at 50 C. What would be the concentration of each substance in the reaction equation. This problem can best be analyzed by an ICE table ICE stands for Initial, Change, Equilibrium The ICE table is based on the balanced equation and is used to analyze the concentrations of all substances in the reaction at three stages: initial, change, and equilibrium

  2. Nitrosyl chloride decomposes by the following reaction:2 NOCl (g)  2 NO (g) + Cl2 (g) The equilibrium constant, Kc , at 50 C is 1.6 x 10 –7 . An evacuated chamber is filled with nitrosyl chloride gas to a concentration of 2.00 M. The mixture is allowed to reach equilibrium at 50 C. What would be the concentration of each substance in the reaction equation. 2 NOCl (g)  2 NO (g) + Cl2 (g) Initial 2.00 0 0 The first section of the table includes all data on the initial concentrations of the substances in the reaction.

  3. Nitrosyl chloride decomposes by the following reaction:2 NOCl (g)  2 NO (g) + Cl2 (g) The equilibrium constant, Kc , at 50 C is 1.6 x 10 –7. An evacuated chamber is filled with nitrosyl chloride gas to a concentration of 2.00 M. The mixture is allowed to reach equilibrium at 50 C. What would be the concentration of each substance in the reaction equation. 2 NOCl (g)  2 NO (g) + Cl2 (g) Initial 2.00 0 0 Change – 2x 2x x The second section represents the change in the substances from the initial concentrations as the reaction moves to equilibrium. In this situation, we started with NOCl only; as the reaction proceeds, NOCl must go down and NO and Cl2 must go up. We do not know the exact value (hence it is represented by x), but we do know the correct ratios of the change.

  4. Nitrosyl chloride decomposes by the following reaction:2 NOCl (g)  2 NO (g) + Cl2 (g) The equilibrium constant, Kc , at 50 C is 1.6 x 10 –7. An evacuated chamber is filled with nitrosyl chloride gas to a concentration of 2.00 M. The mixture is allowed to reach equilibrium at 50 C. What would be the concentration of each substance in the reaction equation. 2 NOCl (g)  2 NO (g) + Cl2 (g) Initial 2.00 0 0 Change – 2x 2x x Equilibrium 2.00 - 2x 2x x The third section of the table represent the concentration of substances when the reaction reaches equilibrium. For each substance, it is the sum of the initial concentration and the change.

  5. Nitrosyl chloride decomposes by the following reaction:2 NOCl (g)  2 NO (g) + Cl2 (g) The equilibrium constant, Kc , at 50 C is 1.6 x 10 –7. An evacuated chamber is filled with nitrosyl chloride gas to a concentration of 2.00 M. The mixture is allowed to reach equilibrium at 50 C. What would be the concentration of each substance in the reaction equation. 2 NOCl (g)  2 NO (g) + Cl2 (g) Initial 2.00 0 0 Change – 2x 2x x Equilibrium 2.00 - 2x 2x x The equilibrium values can be then be applied to the equilibrium constant expression: 1.6 x 10 –7 = [NO] 2 [Cl2] = (2x)2 (x) [NOCl] 2 (2.00 – 2x) 2 4 x3 + 6.68 x 10 -7 x2 - 6.68 x 10 -7 = 0

  6. 2 NOCl (g)  2 NO (g) + Cl2 (g) Initial 2.00 0 0 Change – 2x 2x x Equilibrium 2.00 - 2x 2x x Since the equilibrium constant is very small (1.6 x 10 -7 ), the reaction is reactant favored and the change in products must be very small. Therefore x must be much smaller than 2.0. Subtracting even 2x from 2.0 should not result in a significant change from 2.0. The ICE table can therefore be simplified to: 2 NOCl (g)  2 NO (g) + Cl2 (g) Initial 2.00 0 0 Change – 2x 2x x Equilibrium 2.00 - 2x  2.00 2x x

  7. 2 NOCl (g)  2 NO (g) + Cl2 (g) Initial 2.00 0 0 Change – 2x 2x x Equilibrium 2.00 - 2x  2.00 2x x The equilibrium values can be then be applied to the equilibrium constant expression: 1.6 x 10 –7 = [NO] 2 [Cl2] = (2x)2 (x) [NOCl] 2 (2.00)2 1.6 x 10 -7 = x 3 x = 0.0054 At equilibrium: NOCl = 2.00 - .011 = 1.99 M NO = 0.011 M Cl2 = 0.0054 M Is the answer reasonable: Kc = (0.011)2 (0.0054) = 1.6 x 10 -7 (1.99)2

  8. Nitrosyl chloride decomposes by the following reaction:2 NOCl (g)  2 NO (g) + Cl2 (g) Kc = 5.21 x 10 –5 at 30 CIf an evacuated tank is filled with NOCl at 1.20 M, what would be [NO] at equilibrium and 30? 2 NOCl (g)  2 NO (g) + Cl2 (g) Initial 1.20 0 0 Change - 2x 2x x Equil 1.20 – 2x  1.20 2x x Kc = [NO] 2 [Cl2] 5.21 x 10 –5 = [2x] 2 [x] [NOCl] 2 [1.20] 2 4x3 = 7.5024 x 10 -5 x = 0.02657 [NO] = 2x = 2(0.02657) = 0.0531 M

More Related