Equilibrium calculations
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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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Equilibrium calculations

Equilibrium Calculations


Law of chemical equilibrium

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

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

Answer

  • K = [N2][H2]3

  • [NH3]2

  • K = [.18][.35]3

  • [1.03]2

  • K = .0073


Equilibrium by phase

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

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

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.


Answer1

Answer

  • K = [SO3] 2

  • [SO2]2 [O2]

  • 4.34 = [SO3]2

  • [.28]2[.43]

  • [SO3] = .38 M


Solution equilibrium

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 equilibrium1

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.


Equilibrium calculations

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

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

Problems

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


Answer2

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


Answer3

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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