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# Equilibrium Calculations - PowerPoint PPT Presentation

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

• 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

• 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

• K = [N2][H2]3

• [NH3]2

• K = [.18][.35]3

• [1.03]2

• K = .0073

• 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.

• 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]

• 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.

• K = [SO3] 2

• [SO2]2 [O2]

• 4.34 = [SO3]2

• [.28]2[.43]

• [SO3] = .38 M

• 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)

• For Example:

• 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.

• 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-]= .022M

• 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.

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

• 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

• BaSO4 Ba2+ (aq)+ SO42-(aq)

• Ksp = [Ba2+ ][SO42-]

• 1.4x10-14 = x (x)

• x = [Ba2+]= [SO42-] = 1.2 x10-7 M