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Equilibrium Calculations

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Presentation on theme: "Equilibrium Calculations"— Presentation transcript:

1 Equilibrium Calculations
Lesson 8

2 1. When moles of SO2 and moles of O a 2.00 L container and allowed to reach equilibrium, the equilibrium [SO3] is M. Calculate the Keq. are placed into Implies initial and not equilibrium concentrations - ICE 2SO2 (g) + O2 (g) ⇋ 2SO3 (g) I 0.400 M 0.400M x1/2 x 2/2 C M M M E 0.100 M 0.250 M 0.300 M Equilibrium concentrations go in the equilibrium equation! [SO3]2 (0.3)2 Keq = = = 36.0 [SO2]2[O2] (0.1)2(0.25)

3 2. 80 moles of H2 and 1. 4 moles of S are initially put in a 4
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 I 0.20 M 0 C x x E x x [H2S] x Keq = = = [H2] x

4 x x = 1 1x = 14( x) 1x = x 1x x = 15x = x = 0.19 M [H2] = x [H2] = -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
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 = I 2.0 M 0 0 C -2x x x E x x x [H2][I2] x2 Keq = = = ( x)2 [HI]2

6 x2 ( x)2 = take the square root of both sides = 1 x ( x) cross multiple 1x = x x = x = [H2] = 0.21 M

7 4. The same number of moles of I2 and Cl2 are placed in a 10
4. The same number of moles of I2 and Cl2 are 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) ⇄ ICl (g) Keq = I x x 0 C M M M E x x M [ICl]2 Keq = [I2][Cl2] 0.402 = = 10.0 (x )2

8 0.402 (x )2 = 10.0 = x 0.4 = x = x x = [I2] = [Cl2] = 0.33 M 10.0 L x mole = mol L

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

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