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Equilibrium Calculations Lesson: Keq Determination with Examples

Learn how to determine Keq in equilibrium calculations using practical examples and ICE method application. Understand how initial concentrations affect equilibrium concentrations. See step-by-step solutions for different scenarios in chemical equilibrium.

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Equilibrium Calculations Lesson: Keq Determination with Examples

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  1. Equilibrium Calculations Lesson 8

  2. 1. When 0.800 moles of SO2 and 0.800 moles of O2 a 2.00 L container and allowed to reach equilibrium, the equilibrium [SO3] is 0.300 M. Calculate the Keq. are placed into x1/2 x 2/2 Implies initial and notequilibrium concentrations - ICE 2SO2 (g) + O2 (g)⇋ 2SO3 (g) I 0.400 M 0.400M 0 -0.150 M +0.300 M C -0.300 M E 0.100 M 0.250 M 0.300 M Equilibrium concentrations go in the equilibrium equation! [SO3]2 (0.3)2 = = 36.0 Keq = [SO2]2[O2] (0.1)2(0.25)

  3. 2. 0.80 moles of H2 and 1.4 moles of S are initially put in a 4.0 L flask and allowed to reach equilibrium. Calculate the [H2] at equilibrium. H2(g) + S(s)⇄ H2S(g) Keq= 14 I0.20 M 0 Cx x E0.20 - x x [H2S] x = = 14 Keq = [H2] 0.2 - x

  4. x = 14 1 0.2 - x 1x = 14(0.2 - x) 1x = 2.8 - 14x 1x + 14x = 2.8 15x = 2.8 x = 0.19 M [H2] = 0.20 - x [H2] = 0.20 - 0.19 -note the lose of one sig fig! [H2] = 0.01 M

  5. 3. If 6.0 moles of HI are initially put in a 3.00 L vessel and allowed to reach equilibrium. What is the equilibrium concentration of H2? 2HI(g)⇄ H2(g) + I2(g) Keq = 0.0183 I 2.0 M 0 0 C -2x x x E 2.0 - 2x x x x2 [H2][I2] = = 0.0183 Keq = (2 - 2x)2 [HI]2

  6. x2 = 0.0183 (2 - 2x)2 take the square root of both sides x = 0.13528 (2 - 2x) 1 cross multiple 1x = 0.270555 -0.27055x 1.27055x = 0.270555 x = [H2] = 0.21 M

  7. 4. The same number of moles of I2 and Cl2are placed in a 10.0 L flask and allowed to reach equilibrium. If the equilibrium concentration of ICl is 0.40 M, calculate the initial number of moles of I2 and Cl2. I2 (g) + Cl2 (g)⇄ 2ICl (g)Keq = 10.0 Ix x 0 C0.20 M 0.20 M 0.40 M Ex - 0.20 x - 0.20 0.40 M [ICl]2 Keq= [I2][Cl2] 0.402 = = 10.0 (x - .20)2

  8. 0.402 = 10.0 (x - 0.20)2 0.40 = 3.16227 x - 0.20 1 0.4 = 3.16227x - 0.63246 1.03246 = 3.16227x x = [I2] = [Cl2] = 0.33 M 10.0 L x 0.33 mole = 3.3 mol L

  9. 5.Sketch the changes in concentrations of [O2] and [N2] as equilibrium is obtained. Calculate the value of the Keq. 2.40 M 4.00 M 1.20 M 2.80 M [N2O4] 40 20 Time (min) [N2][O2]2 Keq = [N2O4] [O2] Keq = [1.20][2.40]2 [N2] [2.80] = 2.47 N2O4(g)⇋ 2O2(g) + N2(g) I 4.00 0 0 C 1.20 2.40 1.20 E 2.80 2.40 1.20

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