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Fall 2005 |
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| Exercise: Chemical Equilibrium |
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Example 1: Consider the gas phase reaction:
2 HI H2 + I 2
Keq = 2.06 x 10-2 @ 458°C
If both hydrogen and iodine are measured to have concentrations of 0.0135 M each at 458°C, what is the concentration hydrogen iodide?
First write out the equilibrium expression:
Keq = [H2] [I 2] / [HI]2 = 2.06 x 10-2
and [HI]2 = [H2] [I 2] / (2.06 x 10-2) = [0.0135] [0.0135] / (2.06 x 10-2)
[HI]2 = 0.8847
[HI] = 0.9406 M = 0.941 M
Example 2: Consider the gas phase reaction:
2 HBr
Keq = 1.5 x 10-5 @ 1400 K Calculate the concentrations of all species at equilibrium if we start with 0.15 moles each of hydrogen bromide and bromine in a 0.500 L container at 1400 K.
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2 HBr H2 + Br2 Before reaction 0.15/0.500 = 0.30 M 0 0.15/0.500 = 0.30 M From the equation some HBr will breakdown to give hydrogen and bromine: @ Equilibrium 0.30 - x x 0.30 + x .
At equilibrium then, Keq = [Br2] [H2] / [HBr]2
Substituting, Keq = (0.30 + x)(x) / (0.30 - x)2 = 1.5 x 10-5
Since Keq is small, assume x<<0.30. Can then simplify:
Keq = (0.30)(x) / (0.30)2 = 1.5 x 10-5
0.30x = (0.090)(1.5 x 10-5) = 1.35 x 10-6
and x = 4.5 x 10-6 <<0.30, Assumption OK!
At equilibrium:
HBr = 0.30 MH2 = 4.5 x 10-6 MBr2 = 0.30 M.
Example 3: Consider the gas phase dissociation of carbon dioxide to carbon monoxide and oxygen @ 1000 K.
If 0.200 moles of carbon dioxide is placed in a 1.00 L container at 1000 K calculate the concentrations of all species at equilibrium. Keq = 4.5 x 10-23 @ 1000 K.
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At equilibrium then, Keq = [CO]2 [H2] / [CO2]2
Substituting, Keq = (x)(2x)2 / (0.200 - 2x)2 = 4.5 x 10-23
Since Keq is small, assume x<<0.200. Can then simplify:
Keq = 4x3/ (0.200)2 = 4.5 x 10-23
4x3 = 0.0400(4.5 x 10-23)
x3 = 4.5 x 10-25
x = 7.7 x 10-9
and x = 7.7 x 10-9<<0.200, Assumption OK!
At equilibrium:
CO2 = 0.200 M
CO = 1.5 x 10-8M
Example 4: What is the concentration of hydrogen ion in a 0.20 M solution of acetic acid (CH3COOH) if Ka = 1.8 x 10-5 @ 25°C? ( Ka is the equilibrium constant for an acid dissociation.)
First need to write the equation for the dissociation of the acid:
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At equilibrium then, Ka = [CH3COO-][H+] / [CH3COOH]
Substituting, Ka = (x)(x) / (0.20 - x) = 1.8 x 10-5
Since Ka is small, assume x<<0.20. Can then simplify:
Ka = x2/ (0.20) = 1.8 x 10-5
x2 = 3.6 x 10-6
x = 1.9 x 10-3
and x = 1.9 x 10-3 is smaller than the error of significant figures, Assumption OK!
At equilibrium:
CH3COOH = 0.20 M
CH3COO- = 1.9 x 10-3 M
H+= 1.9 x 10-3 M
Example 5: Calculate the solubility of calcium carbonate. CaCO3 is used by organisms to make shells etc. Vast deposits of these shells make up limestone and (after heat and compression) marble deposits. An important consideration in determining whether shells can be formed, limestone deposited etc. is the solubility of calcium carbonate. Ksp = 8.7 x 10-9
When looking at solubility situations it is most common to look at the dissolution of the solid:
Ca2+ + CO32-
The equilibrium expression for this reaction can then be written as:
But recall that the activity (the effective or apparent concentration of a substance in a chemical system) is 1 for any solid (remember, a solid is a crystalline substance, and therefore a pure phase). We can thus write the simpler expression for the equilibrium and give a new symbol, Ksp, the solubility product = Keq for dissolution, for this expression:
We can now determine the solubility of calcium carbonate in water.
CaCO3(s) Ca2+ + CO32- Before reaction 1 0 0 [Ca2+] = [CO32-]; Let [Ca2+] = x @ Equilibrium 1 x x x2 = 8.7 x 10-9
Thus [Ca2+] = [CO32-] = (8.7 x 10-9)1/2 = 9.3 x 10-5
So the solubility of calcium carbonate = 9.3 x 10-5 M.
Example 6: Calculate the the Ksp for lead(II) chloride at 25°C if its solubility = 1.62 x 10-2 M.
Pb2+ + 2 Cl-
From the chemical equation we can see that twice as much chloride ion as lead ion results when lead chloride is dissolved,
PbCl2(s) Pb2+ + 2 Cl- Before reaction 1 0 0 Let [Pb2+] = x @ Equilibrium 1 x 2x thus a saturated solution of lead chloride will have:
Pb2+ = 1.62 x 10-2 M
Cl- = (1.62 x 10-2 M) = 3.24 x 10-2 M Substituting these values into the solubility product expression then gives:
Ksp = (1.62 x 10-2)(3.24 x 10-2)2 = 1.7 x 10-5
Example 7: What is the solubility of ferric hydroxide in a buffered aqueous solution where [OH-] = 1 x 10-6M. Ksp = 4 x 10-38
Fe(OH)3(s)
Ksp = [Fe3+] [OH-]3 = 4 x 10-38
For solubility note that for each Fe(OH)3(s) one [Fe3+] goes into solution, so [Fe3+] = solubility!
[Fe3+] = (4 x 10-38) / [OH-]3
but the buffer fixes [OH-] = 1 x 10-6M
so [Fe3+] = (4 x 10-38) / (10-6)3 = 4 x 10-20 M
© R A Paselk
Last modified 23 June 2006