the equilibrium law section 17.2 (ahl). vocabulary homogeneous equilibrium: all the reactants and...
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The Equilibrium Law
Section 17.2 (AHL)
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Vocabulary
Homogeneous equilibrium: all the reactants and products are in the same phase
Heterogeneous equilibrium: when there are two or more phases
Qc: reaction quotient, refers to a quotient obtained
by applying the equilibrium law to initial concentrations (instead of equilibrium concentrations)
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Using Qc
The value of the reaction quotient can be used to determine how a reaction needs to shift in order to attain equilibrium
Qc=K
c equilibrium is established
Qc< K
c the system must shift to the right to
establish equilibrium Q
c> K
c the system must shift to the left
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Example Calculation #1
The acid-catalyzed hydrolysis of ethyl ethanoate can be achieved by mixing the ester with dilute HCl
CH3COOC
2H
5(l) + H
2O
(l) CH
3COOH
(l) + C
2H
5OH
(l)
The products are ethanoic acid and ethanol
H+
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Problem #1 Continued
If 1.00 mole of ethyl ethanoate is mixed with 1.00 mole of water and the reaction allowed to reach equilibrium at a particular temperature, then 0.30 moles of ethanoic acid is found in the equilibrium mixture
Calculate the value of Kc at this temperature
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Make an ICE table
CH3COOC2H5 H2O CH3COOH C2H5OH
Initial Amount in moles
1.00 1.00 0.00 0.00
Change in amount +0.30
Equilibrium amount in moles
0.30
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Continued
The rest of the table needs to be filled in to find K
c at that temperature
According to the stoichiometry, all the coefficients in the equation are all “1”, so if 0.30 mol of CH
3COOH is produced, then
0.30 mol ofvC2H
5OH is also produced
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CH3COOC2H5 H2O CH3COOH C2H5OH
Initial Amount in moles
1.00 1.00 0.00 0.00
Change in amount +0.30 +0.30
Equilibrium amount in moles
0.30 0.30
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Problem Continued
If 0.30 mole are present in the equilibrium mixture, then they must have been produced from the reaction of 0.30 moles of the ester and the water.
The amount of ester and water remaining at equilibrium must be 1.00-0.30 moles of each
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CH3COOC2H5 H2O CH3COOH C2H5OH
Initial Amount in moles
1.00 1.00 0.00 0.00
Change in amount -0.30 -0.30 +0.30 +0.30
Equilibrium amount in moles
0.70 0.70 0.30 0.30
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Problem Continued
We now have the number of moles, but not the volumes of each liquid.
You can assume 1 dm3 All molar values then become concentration
values(0.30)(0.30)
0.18(0.70)(0.70)cK
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Example Problem #2 (ex. on page 194)
When a mixture initially containing 0.0200 mol dm-3 SO
2 and an equal concentration of
O2 is allowed to reach equilibrium in a closed
container of fixed volume at 1000K, it is found that 80.0% of the SO
2 is converted to
SO3. Calculate the value of the equilibrium
constant at that temperature.
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Problem #2 Continued
Write the balanced equation first 2SO
2(g) + O
2(g) 2SO
3(g)
Make your table of ICE
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SO2 O2 SO3
Initial Amount in moles
0.0200 0.0200 0.00
Change
Equilibrium amount in moles
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Continued
Equilibrium [SO3] = (0.0200) (0.800) since
80% of the the SO2 turns into SO
3
[SO3] = 0.0160 mol dm-3
The stoichimetry says a 1:1 ratio between SO
3 and SO
2, so the [SO
2] is equal to 0.0200
minus 0.0160 which is 0.0040 mol dm-3
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Continued
According to the stoichiometry, each sulfur trioxide molecule requires only ½ an oxygen molecule so:
[O2] = 0.0200 – (1/2)(0.0160) = 0.0120 mol
dm-3
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SO2 O2 SO3
Initial Amount in moles
0.0200 0.0200 0.00
Change-0.0160 -0.0080 +0.0160
Equilibrium amount in moles
0.0040 0.0120 0.0160
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Calculate Kc
=1333 mol-1dm3
1330 mol-1dm3 (to 3 sig figs)
2
2
(0.0160)
(0.0040) (0.0120)cK
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Another Fun Problem #3
SO3(g)
+ NO(g)
NO2(g)
+ SO2(g)
Kc for this reaction is 6.78 at a specified
temperature If the initial concentrations of NO and SO
3
were both 0.03 mol dm-3, what would be the equilibrium concentration of each component?
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Continued
We will have to use variables for the exponents to keep track of the concentrations and to solve
Let the change in concentration of the reactants both equal -x
Let the change in concentration of the products both equal +x
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SO3 NO NO2 SO2
Initial Amount in moles
0.03 0.03 0.00 0.00
Change in amount -x -x +x +x
Equilibrium amount in moles
0.03-x 0.03-x x x
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Continued
2 2
3
[ ][ ]
[ ][ ]c
NO SOK
SO NO
2
26.78
(0.03 )
x
x
6.780.03
x
x
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Continued
2.60 (0.03 – x) = x
0.078 – 2.60x = x
0.078 = 3.60x
0.0217 = x (realize you are not finished when you calculate “x”)
2 .600 .03
x
x
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SO3 NO NO2 SO2
Initial Amount in moles
0.03 0.03 0.00 0.00
Change in amount -x -x +x +x
Equilibrium amount in moles
0.03-x 0.03-x x x
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SO3 NO NO2 SO2
Initial Amount in moles
0.03 0.03 0.00 0.00
Change in amount -0.0217 -0.0217 +0.0217 +0.0217
Equilibrium amount in moles
0.0083 0.0083 0.022 0.022
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One more type of problem
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N2 (g) 3H2 (g) 2NH3
Initial Amount in moles 1.00 3.00 0.00
Change in amount -0.062÷2= 0.031
-(3*(0.062÷2))=0.093 +0.062
Equilibrium amount in moles
0.969 2.91 0.062
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Finish
Kc = 1.61 x 10-4
2
3
(0 .0 62 )
(0 .969 )(2 .91)cK