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CHEMICAL EQUILIBRIUM

CHEMICAL EQUILIBRIUM. Calculation of the Equilibrium Constant. 1.00 mol SO 2 and 1.00 mol O 2 are confined at 1000 K in a 1.00-liter container. At equilibrium, 0.925 mol SO 3 has been formed. Calculate Kc for the reaction 2SO 2 (g) + O 2 (g)  2SO 3 (g) at 1000 K. Solution:

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CHEMICAL EQUILIBRIUM

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  1. CHEMICAL EQUILIBRIUM joma

  2. Calculation of the Equilibrium Constant • 1.00 mol SO2 and 1.00 mol O2 are confined at 1000 K in a 1.00-liter container. At equilibrium, 0.925 mol SO3 has been formed. Calculate Kc for the reaction 2SO2(g) + O2(g)  2SO3(g) at 1000 K. Solution: 2SO2(g) + O2(g)  2SO3(g) Initial moles 1.00 1.00 0 Change - 2y - y + 2y Moles at equil. 1.00 – 2y 1.00 – y 0 + 2y = 0.925 mol n SO3 = 2y = 0.925 mol n SO2 = 1.00 – 2y, 1.00 – 0.925 = 0.075 mol n O2 = 1.00 – y, 1.00 – 0.46 = 0.54 mol joma

  3. 2SO2(g) + O2(g)  2SO3(g) • K = [SO3]2/[SO2]2 [O2] • K = [0.925]2/[0.075]2 [0.54] • K = 282 joma

  4. ２) 3.00 moles of pure SO3 are introduced into an 8.00-liter container at 1105 K. At equilibrium 0.58 mole of O2 has been formed. Calculate Kc for the reaction:2 SO3(g)  2 SO2(g) + O2(g) at 1105 K. • Answer: 2 SO3(g)  2 SO2(g) + O2(g) Initial: 3.00 0 0 Change: - 2x + 2x + 1x Equilibrium: 3.00-2x 0+2x 0+1x= 0.58 mole joma

  5. n O2 = 0 + x = 0.58 mole n SO3 = 3 – 2x, 3 – 1.16 = 1.84 mole n SO2 = 0 + 2x = 1.16 mole [O2] = 0.58/8 = 0.0725 mole/liter [SO3] = 1.84/8 = 0.23 M [SO2] = 1.16/8 = 0.145 M K = [SO2]2 [O2] /[SO3]2 K = [0.145]2 [0.0725] /[0.23]2 K = 2.88 X10-2 joma

  6. 3) A 1-liter container is filled with 0.50 mol of HI at 448oC. The value of the equilibrium constant for the reaction: H2(g) + I2(g)  2 HI(g) at this temperature is 50.5. What are the concentrations of H2, I2 and HI in the vessel at equilibrium? Answer: Given: K = 50.5 H2(g) + I2(g)  2 HI(g) Initial: 0 0 0.50 mol Change: x x - 2x Equilibrium: 0+x 0+x 0.50-2x joma

  7. K = [HI]2/[H2] [I2] • 50.5 = [0.5 – 2x]2/[x] [x] • (x = - 0.0979 mol) or (x = 0.0549 mol) • x = 0.0549 mol H2(g) + I2(g)  2 HI(g) Equilibrium: 0+x 0+x 0.50-2x • [H2] and [I2] = 0.0549 M [HI] = 0.39 M joma

  8. Checking: H2(g) + I2(g)  2 HI(g) Equilibrium: 0+x 0+x 0.50-2x • [H2] and [I2] = 0.0549 M [HI] = 0.39 M • K = [HI]2/[H2] [I2] • K = [0.39]2/[0.0549] [0.0549] • K = 50.5 joma

  9. 4) For the reaction: H2(g) + I2(g)  2 HI(g) ; the equilibrium constant at 420oC is 54.3. Suppose that the initial concentrations of H2 and I2 forming HI are 0.00623 M, 0.00414 M, and 0.0224 M respectively, what are the concentrations of these species at equilibrium? Answer: H2(g) + I2(g)  2 HI(g) Initial: 0.00623 0.00414 0.0224 Change: - x - x + 2x Equil.: 0.00623-x 0.00414-x 0.0224+2x joma

  10. Solution: 54.3 = [0.0224 + 2x]2 [0.00623 – x] [0.00414 –x] x = 0.0114 M or x = 0.00156 M x = 0.00156 M [H2] = 0.00467 [I2] = 0.00258 M [HI] = 0.0255 M Checking: K = [HI]2 = [0.0255]2 [H2] [I2] [0.00467] [0.00258] K = 53.97 K = 54 joma

  11. 5) At 1600 K, the equilibrium constant (Kc) for the reaction: Br2(g)  2 Br(g) is 1.04 X10-3. The amount of 0.10 mol Br2 is confined in a 1.0-liter container and heated to 1600 K. Calculate (a) the concentration of Br atoms present at equilibrium and (b) the percentage of the initial Br2 that is dissociated into atoms. • % dissociation = quantity dissociatedX 100 original quantity joma

  12. Solution: K = 1.04 X10-3 Br2(g)  2 Br(g) Initial: 0.1 0 Change: - y + 2y Equili.: 0.1 – y 0 + 2y 1.04 X10-3 = [2y]2 [0.1-y] y = 4.97 x10-3 M (a) 2y = 9.94 x10-3 M (b) % dissociation = 4.97 x10-3 MX 100 0.10 M = 4.97% or 5% joma

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