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# Chemical Equilibrium – Part 2b GD: Chpt 7 (7.2, 17.2); CHANG: Chpt 14 Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction.

## Presentation on theme: "Chemical Equilibrium – Part 2b GD: Chpt 7 (7.2, 17.2); CHANG: Chpt 14 Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction."— Presentation transcript:

Chemical Equilibrium – Part 2b GD: Chpt 7 (7.2, 17.2); CHANG: Chpt 14 Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Calculating Equilibrium Constants 1.Use the balanced chemical equation to write the equilibrium constant expression in terms of the equilibrium concentrations. 2.Input the equilibrium values for the concentration of all involved species. (Sometimes these are given, sometimes you have to calculate these ) 3.Solve for K c

The equilibrium concentrations for the reaction between carbon monoxide and molecular chlorine to form COCl 2 (g) at 74 0 C are [CO] = 0.012 M, [Cl 2 ] = 0.054 M, and [COCl 2 ] = 0.14 M. Calculate the equilibrium constant K c. 14.2 Calculating Equilibrium Constants – Sample 1

The reaction between nitrogen monoxide and oxygen to form the brown nitrogen dioxide gas has been studied at 230°C. In one experiment the concentrations of the reacting species at equilibrium are found to be [NO] = 0.0542 M, [O 2 ] = 0.127 M, and [NO 2 ] = 15.5 M. Calculate the equilibrium constant K c. 14.2 Calculating Equilibrium Constants – Sample 2 – C14.2

How to… Calculate Equilibrium Concentrations given K c and Initial Concentration Values 1.Express the equilibrium concentrations of all species in terms of the initial concentrations and a single unknown x, which represents the change in concentration. 2.Write the equilibrium constant expression in terms of the equilibrium concentrations. Knowing the value of the equilibrium constant, solve for x. 3.Having solved for x, calculate the equilibrium concentrations of all species. 14.4

Sample Problem # 3 - C14.9 A mixture of 0.500 mol H 2 and 0.500 mol I 2 was placed in a 1.00 L stainless-steel flask at 430°C. The equilibrium constant Kc for the reaction of hydrogen and iodine to form hydrogen iodide is 54.3 at this temperature. Calculate the concentrations of H 2, I 2, and HI at equilibrium. H 2 (g) + I 2 (g) 2HI (g) Let x be the change in concentration of H 2 Initial (M) Change (M) Equilibrium (M) 14.4

Taking the square root of both sides, we get x = 0.393 At equilibrium, [H 2 ] = 0.500 – 0.393 = 0.170 M 14.4 K c = (2x) 2 (0.500 – x) 2 = 54.3 K c = 2x 0.500 – x = 7.37 At equilibrium, [I 2 ] = 0.500 – 0.393 = 0.170 M At equilibrium, [H I ] = 0.000 + (2 x 0.393) = 0.786 M [HI] 2 [H 2 ] [I 2 ] K c = Solve for x

Sample Problem # 4 At 1280 0 C the equilibrium constant (K c ) for the reaction Is 1.1 x 10 -3. If the initial concentrations are [Br 2 ] = 0.063 M and [Br] = 0.012 M, calculate the concentrations of these species at equilibrium. Br 2 (g) 2Br (g) Let x be the change in concentration of Br 2 Initial (M) Change (M) Equilibrium (M) 14.4

K c = Solve for x 14.4 4x 2 + 0.048x + 0.000144 = 0.0000693 – 0.0011x 4x 2 + 0.0491x + 0.0000747 = 0 ax 2 + bx + c =0 -b ± b 2 – 4ac  2a2a x = x = -0.00178x = -0.0105 =

Br 2 (g) 2Br (g) Initial (M) Change (M) Equilibrium (M) 0.0630.012 -x-x+2x 0.063 - x0.012 + 2x 14.4 Now insert the x values back into the ICE set up to solve. At equilibrium, [Br] = 0.012 + 2x = -0.009 Mor 0.00844 M At equilibrium, [Br 2 ] = 0.062 – x = 0.0648 M

When a mixture initially containing 0.0200 mol dm -3 sulfur dioxide and an equal concentration of oxygen is allowed to reach equilibrium in a container of fixed volume at 1000K, it is found that 80.0% of the sulfur dioxide is converted to sulfur trioxide. Calculate the value of the equilibrium constant at that temperature. SO 2 (g) + O 2 (g) 2 SO 3 (g) 14.2 Calculating Equilibrium Constants – Sample 5 – IB p 194 Assume the fixed volume is 1 dm 3 in order to determine [ ] Set up ICE… (next slide)

The change in concentration of SO 2 is expressed as 80% Initial (M) Change (M) Equilibrium (M) 2 SO 2 (g) + O 2 (g) 2 SO 3 (g) Kc =Kc = [SO 3 ] 2 [SO 2 ] 2 [O 2 ] =

In the gas phase at 730 K, the equilibrium constant for the reaction of hydrogen and iodine to form hydrogen iodide has a value of 490. If the initial concentration of iodine is 0.0200 mol dm -3, what concentration of hydrogen is required for 90.0% of the iodine to be converted to hydrogen iodide? 14.2 Calculating Equilibrium Constants – Sample 6 – IB – p. 195 The change in concentration of I 2 is expressed as 90% I 2 (g) + H 2 (g) 2 HI (g) Initial (M) Change (M) Equilibrium (M)

Kc =Kc = = = expression [ ]valuesK c value Solve for x:

Homework Reread Section 7.2 pp 184-185 Read Section 17.2 pp 194-195 Do Ex 17.2 pp 195-196 # 1, 3, 4, 5, 6

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