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Equilibrium Calculations

Equilibrium Calculations. Law of chemical equilibrium. For an equilibrium a A + b B c C + d D K = [C] c [D] d [A] a [B] b K is the equilibrium constant for that reaction. The [ ] mean concentration in molarity. Problem. 2NH 3 (g) N 2 (g) + 3H 2 (g)

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Equilibrium Calculations

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  1. Equilibrium Calculations

  2. Law of chemical equilibrium • For an equilibrium • a A + b B c C + d D • K = [C]c[D]d • [A]a[B]b • K is the equilibrium constant for that reaction. • The [ ] mean concentration in molarity

  3. Problem • 2NH3 (g) N2 (g) + 3H2 (g) • Calculate the equilibrium constant for the above reaction if it comes to equilibrium with the following concentrations: N2 = .18 M, H2 =.35 M, and NH3 = 1.03 M

  4. Answer • K = [N2][H2]3 • [NH3]2 • K = [.18][.35]3 • [1.03]2 • K = .0073

  5. Equilibrium by phase • Equilibrium depends on the concentration of the reactants. • We can calculate the concentration of a gas or of anything dissolved (aqueous). • Insoluble solids or liquids won’t have a concentration. • They in essence are removed from the equilibrium.

  6. So using that • What would the equilibrium expression look like for the following reaction? • 2 H2O2(l) 2 H2O(l) + O2(g) • We ignore the liquids (and solids). • K = [O2]

  7. Another problem • 2 SO2 (g) +O2 (g) 2 SO3 (g) • K = 4.34, for the above reaction. Calculate the concentration of SO3 if the SO2 = .28 M and O2 = .43 M at equilibrium.

  8. Answer • K = [SO3] 2 • [SO2]2 [O2] • 4.34 = [SO3]2 • [.28]2[.43] • [SO3] = .38 M

  9. Solution Equilibrium • All dissociations we have done are equilibriums. • Before we simply stated something was soluble or insoluble. • Actually everything dissolves to some extent, and some dissolved substance fall out of solution. • Higher concentrations force more solute to fall out of solution. • So there is a maximum concentration of solute a solution can hold (saturation)

  10. Solution equilibrium • For Example: • Lead (II) Bromide • PbBr2 (s) Pb2+ (aq)+ 2 Br-(aq) • What would the equilibrium expression look like? • Ksp = [Pb2+ ][Br-]2 • Equilibrium constants for dissociations are called solubility products, and are denoted by Ksp.

  11. Cont. • The Ksp value for PbBr2 is 5.0 x 10-6. • PbBr2 (s) Pb2+ (aq)+ 2 Br-(aq) • I will assume 1 L, and that I start with x moles of PbBr2. • Using the balanced equation I have x mole/L of Pb2+ and 2x mole/L of Br- • So • 5.0 x 10-6= x (2x)2 • 5.0 x 10-6= 4x3 • x=[Pb2+ ] = .011 M 2x = [Br-]= .021M

  12. This means • A solution of PbBr2 would be saturated with [Pb2+ ] = .011 M and [Br-]= .021M • These are both very low concentrations so we say the is compound is insoluble.

  13. Problems • Calculate the saturation concentrations of solutes in aluminum hydroxide Ksp = 5.0 x10 -33, and Barium sulfate Ksp = 1.4 x10 -14

  14. Answer • Al(OH)3 Al3+ (aq)+ 3 OH-(aq) • Ksp = [Al3+ ][OH-]3 • 5.0x10-33 = x (3x)3 • x = [Al3+ ]= 3.7x10-9 M • 3x = [OH-] = 1.1 x10-8 M

  15. Answer • BaSO4 Ba2+ (aq)+ SO42-(aq) • Ksp = [Ba2+ ][SO42-] • 1.4x10-14 = x (x) • x = [Ba2+]= [SO42-] = 1.2 x10-7 M

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