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Most chemical reactions are reversible. Chemical equilibrium can only occur in a closed system. Equilibrium is when the rates are equal AND the [ ]s remains constant. K c can be used to determine the equilibrium position.

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K c and Equilibrium Problems (4 types)

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CAN YOU? / HAVE YOU? Make equilibrium constant (K) calculations with various missing values. Use ICE tables to successfully complete Equilibrium calculations. *Always make sure the equation is balanced, FIRST*

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1. Plugand solve

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At 225°C, a 2.0 L container holds moles of N 2, 0.15 moles of H 2 and 0.50 moles of NH 3. If the system is at equilibrium, calculate K C. 1. Change all into concentrations - mol/L 2. Write the equilibrium law for the reaction N 2(g) + 3 H 2(g) 2 NH 3(g) mol N L = M N 2 = M H 2 = 0.25 M NH 3

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3. Substitute the concentrations and calculate K. (Note: no units for K)

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The equilibrium concentrations of N 2 and NO are 1.40 mol/L and 5.20 mol/L respectively. Calculate the K C. According to Eq stoichiometry – the same amount of O 2 will be produced as N 2. 2 NO (g) N 2(g) + O 2(g) 1.40 M 5.20 M 1.40 M K c = [N 2 ][O 2 ] [NO] 2 K c = [1.40][1.40] [5.20] 2 =0.0725

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and solve 2. Rearrange

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At 210°C, the K c is 64.0 The equilibrium concentrations of N 2 and O 2 are 0.40 mol/L and 0.60 mol/L, respectively. Calculate the equilibrium concentration of NO. 1. Write out the Eq Law. 2. Rearrange for [NO]. N 2(g) + O 2(g) 2 NO (g)

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3. Substitute concentrations then solve.

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At 10.0°C, the K c is The equilibrium concentrations of SO 2 is 9.40 mol/L. Calculate the [S] and [O 2 ] at equilibrium. SO 2(g) S (g) + O 2(g) Values should be the same amount for S and O 2 – assign the unknown “x” x M 9.40 M x M K c = [S][O 2 ] [SO 2 ] = K c [S][O 2 ] [SO 2 ]

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[x][x] = 215 [9.40] = K c [S][O 2 ] [SO 2 ] [x] 2 = 2021 √√ x = 45.0 M [S] eq and [O 2 ] eq = 45.0 M

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3. Simple I.C.E. Tables

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1.00 mole of hydrogen and 1.00 mole of fluorine are sealed in a 1.00 L flask at 150.0°C and allowed to react. At equilibrium, 1.32 moles of HF are present. Calculate K C. H 2(g) + F 2(g) 2 HF (g) [Eqlbm] [Initial] 1.00 mol/L 0 [Change] Write out the Eq Law – for the equation as written.

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[Eq] H 2(g) + F 2(g) 2 HF (g)

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Initially 2.0 mol of SO 2, 1.0 mol of O 2 are mixed in a 3.0 L reaction container. At equilibrium, 0.20 mol of O 2 are found to remain. Calculate the K c. 2 SO 3 (g) 2 SO 2 (g) + O 2 (g) [E][E] [I][I] mol/L0.33 mol/L [C][C] K c = [SO 2 ] 2 [O 2 ] [SO 3 ] 2

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K c = K c = 5.6 x [E][E] K c = [SO 2 ] 2 [O 2 ] [SO 3 ] 2 2 SO 3 (g) 2 SO 2 (g) + O 2 (g) K c = [0.15] 2 [0.067] [0.52] 2

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4. HARD I.C.E. Tables

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[E][E] x 2x x 6.0 moles of N 2 and O 2 gases are placed in a 1.0 L container, what are all the concentrations at equilibrium? The K c is N 2(g) + O 2(g) 2 NO (g) [I][I] 6.0 mol/L 0 [C][C]- x + 2x

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2. Get rid of the square by taking the square root of both sides. 1. Substitute known values.

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3. Isolate, and solve for x. [I][I] [C][C] [E][E] 6.0 mol/L (3.4) x2x6.0 - x - x + 2x

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H 2 (g) + I 2 (g) 2 HI (g) 2.0 moles of HI were placed in a 1.0 L flask at 430ºC. (K c = 54.3) Calculate the equilibrium concentrations. [I][I] [C][C] [E][E] x - 2x x+ x K c = [HI] 2 [H 2 ][I 2 ]

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54.3 = [ x] 2 [+ x][+ x] K c = [HI] 2 [H 2 ][I 2 ] 7.37 = 2.0 – 2x x 7.37 x = 2.0 – 2x √√ 7.37 x + 2x = x = 2.0 x = 0.21 [I][I] [C][C] [E][E] x - 2x x+ x

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The K c for the reaction 5.3 x at 0ºC. Initially, 2.5 mol each particle was injected into a 1 L reaction vessel. Find the Eq concentrations x x -2x + x 2.5 mol/L [E][E] [C][C] [I][I] H 2 (g) + Cl 2 (g) 2 HCl (g) K c = [HCl] 2 [H 2 ][Cl 2 ]

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5.3 x = [ x] 2 [2.5 + x][2.5 + x] K c = [HCl] 2 [H 2 ][Cl 2 ] √ 5.3 x = [ x] 2 [2.5+x] 2 √ √ 0.23 = 2.5 – 2x 2.5+ x

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x = 0.86 mol/L x = x = x 2.5 mol/L [E][E] [C][C] [I][I] (0.86) 0.8 mol/L 3.4 mol/L 2.23 x = 1.92

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CAN YOU? / HAVE YOU? Make equilibrium constant (K) calculations with various missing values. Use ICE tables to successfully complete Equilibrium calculations. *Always make sure the equation is balanced, FIRST*

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