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Equilibrium CalculationsPowerPoint Presentation

Equilibrium Calculations

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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

- 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

Answer

- K = [N2][H2]3
- [NH3]2
- K = [.18][.35]3
- [1.03]2
- K = .0073

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.

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]

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.

Answer

- K = [SO3] 2
- [SO2]2 [O2]
- 4.34 = [SO3]2
- [.28]2[.43]
- [SO3] = .38 M

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)

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.

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-]= .022M

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.

Problems

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

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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