Transcript Equilibrium Examples

Equilibrium Examples
Ch. 15
1
Variations in Concentration
problems
1. If all components are present, find Q
before proceeding so that you know
which direction the reaction proceeds.
This is not necessary if one side of
reaction has 0 initial concentration since
you then know which way reaction must
go.
2. If K is very small, then it is possible to
approximate the term (yy – x) as yy if yy
is the initial concentration and x is the
change.
2
Calculating Equilibrium
Concentrations
The reaction of hydrogen and fluorine to
form HF gas has an equilibrium constant
of 1.15 x 102 at a given temp. 3.000 mol
of each component was added to 1.500L flask. Calculate the equilibrium
concentrations of each.
1. Equation: H2(g) + F2(g)   2HF (g)
2. K = 1.15 x 102 = [HF]2
(product is favored)
[H2][F2]
3
HF Problem, continued
(K = 1.15 x 102)
3.
[HF]0 = [H2]0 = [F2]0 = 3.000 mol = 2.000M
1.500 L
4. Q = [HF]02 ___ = 2.0002
= 1.000
[H2]0[F2]0 (2.000)(2.000)
Q < K so system shifts to products to reach equilibrium (to
the right).
R
H2
F2
2HF
I
2.000
2.000
2.000
C
-x
-x
+2x
E
2.000 - x
2.000 - x
2.000 + 2x
4
(K = 1.15 x
2
10 )
Kc = 1.15 x 102 = (2.000 + 2x)2
(2.000-x)2
How convenient! 2 squared terms, so take square root
of each side.
√1.15 x 102 = 2.000 + 2x
2.000 – x
10.7238(2-x)=2+2x
19.4476 = 12.7238x
x = 1.528
[H2] = [F2] = 2.000M – 1.528 = 0.472 M
[HF] = 2.000M + 2(1.528) = 5.056 M
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Calculating Equilibrium
Concentrations
The reaction of hydrogen and fluorine to
form HF gas has an equilibrium constant
of 1.15 x 102 at a given temp. 3.000 mol
H2 and 6.000 mol F2 are added to a
3.000-L flask. Calculate the equilibrium
concentrations of each.
1. Equation: H2(g) + F2(g)   2HF (g)
2. K = 1.15 x 102 = [HF]2
(product is favored)
[H2][F2]
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2nd HF Problem, continued (K = 1.15 x 102)
3.
[H2]0 = 3.000 mol = 1.000M [F2]0 = 6.000 mol = 2.000 M
3.000 L
3.000 L
4. System shifts to products to reach equilibrium (to
the right) since there is no HF at start.
R
H2
F2
2HF
I
1.000
2.000
0
C
-x
-x
+2x
E
1.000 - x
2.000 - x
2x
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2nd HF Problem, cont.
(K = 1.15 x 102)
Kc = 1.15 x 102 = (2x)2 _____________
(1.000 – x)(2.000-x)
115(2-3x+x2) = 4x2
111x2 – 345x + 230 = 0
X = -b ± √(b2 – 4ac)= 345 ± √ (3452-4(111)(230)
2a
2(111)
X = 2.14 and x = .968
[H2] = 1.000M-2.14 = not possible, so x = .968
[H2] = 1.000M-.968 = 0.032 M
[F2] = 2.000M – .968 = 1.032 M
[HF] = 2(0.968) = 1.936 M
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Small Equilibrium Constants
Gaseous NOCl decomposes to form the
gases NO and Cl2. At 35º C, the
equilibrium constant is 1.6 x 10-5. In an
experiment in which 1.0 mol of NOCl is
placed in a 2.0-L flask, what are the
equilibrium concentrations?
1. 2NOCl (g)   2NO (g) + Cl2 (g)
2. K = [NO]2[Cl2] = 1.6 x 10-5
[NOCl]2
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Small Equilibrium Constant,
cont.
3. [NOCl]0 = 1.0 mol/2.0 L = 0.50 M
4. System moves to right since no products were
present in the beginning.
R
2NOCl
2NO
Cl2
I
0.50 M
0
0
C
-2x
+2x
+x
E
0.50 – 2x
2x
x
10
Small Kc, continued
Kc = 1.6 x 10-5 = (2x)2 (x)___
(0.50 – 2x)2
Yikes!
But--- the system does not have to proceed far to
reach equilibrium since K is so small. X must be
very small compared to 0.50, so ignore x in the
subtraction term (but not in the cubed term.)
0.50 – 2x ~ 0.50
(0.50)2(1.6 x 10-5) = 4x3
x3 = 1x 10-6
x = 0.01
[NOCl] = 0.50 -2x ~ 0.50 M (recall we approximated)
[NO] = 2x = 2(0.01) = 0.02M
[Cl2] = x = 0.01M
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More Equilibrium Practice
(Zumdahl 13.6)
The reaction between nitrogen and oxygen
to form nitric oxide has an equilibrium
constant value of 4.1 x 10-4 at 2000 K. If 0.50
moles of nitrogen and 0.86 moles of oxygen
are put into a 2.0 L container at 2000 K, what
are the equilibrium concentrations of each
species?
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Prob. 13-6
K= 4.1 x 10-4
• What is the extent of this reaction? K is small,
so reaction stays far to left.
• [N2]0 = 0.50 mole/2.0 L = 0.25 M
• [O2]0 = 0.86 mole/2.0 L = 0.43 M
R
I
C
E
N2
0.25 M
-x
0.25 -x
O2
0.43 M
-x
0.43 - x
2NO
0
+2x
+2x
13
Prob. 13-6
K= 4.1 x 10-4
• K is small, so assume x is negligible compared to 0.25
and 0.43.
• K = [NO]2 =
4x2
= 4.1 x 10-4
[N2]0 [O2]0
(.25)(.43)
• X = 3.3 x 10-3 M
• [N2] = 0.25 M [O2] = 0.43 M [NO]= 6.6 x 10-3M
R
I
C
E
N2
0.25 M
-x
0.25 -x
O2
0.43 M
-x
0.43 - x
2NO
0
+2x
+2x
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Intermediate K: Practice 13.6 B
• Sulfurous acid dissociates in water:
H2SO3 (aq)   H+ (aq) + HSO3- (aq)
If [H2SO3]0 = 1.5 M and the other
concentrations are 0 M, find the
equilibrium concentrations of all species at
25º C if K = 1.20 x 10-2.
• Since K is small, try considering x
negligible in the reduction of [H2SO3]0 .
15
K= 1.20 x 10-2
Prob. 13-6 B
• K = [H+]0 [HSO3-]0 = x2
= 1.20 x 10-2
[H2SO3]
1.50
• X = 0.134 M
• This is 9% of 1.50—cannot consider x
negligible.
R
I
C
E
H2SO3
1.50 M
-x
1.50 - x
H+
0
+x
+x
HSO30
+x
+x
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Prob. 13-6 B
K= 1.20 x 10-2
• K = [H+]0 [HSO3-]0 = x2
= 1.20 x 10-2
[H2SO3]
1.50 - x
• x2 + 0.0120x – 0.0180 = 0
• X = 0.128 M
• [H+] = [HSO3-] = 0.128 M
• [H2SO3] = 1.50 – 0.128 = 1.37 M
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Le Châtelier’s Principle
ΔH = +1.81 kJ
Kp = 3.3 x 1030
What happens to the direction of equilibrium if
the following changes were made?
1. N2 is added
2. He is added
3. The container is made larger
4. The system is cooled
N2(g) + O2(g)   2NO(g)
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N2(g) + O2(g)   2NO(g)
Q = [NO]2
[N2][O2]
This makes Q smaller. When Q<K, more
product must be made. Shift to right.
2. He is added
Helium is inert and does not affect any partial
pressures of these gases. No change to
the equilibrium.
1. N2 is added
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N2(g) + O2(g)   2NO(g)
3. The container is made larger:
This favors the side with more gas molecules
because they can exist with fewer
collisions per unit time. Both sides have
same number of molecules at equilibrium,
so no change occurs in equilibrium.
4. The system is cooled: This is an
endothermic reaction which requires heat
to go to right. Cooling shifts equilibrium to
left.
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2H2 (g) + O2 (g)   2H2O(g)
ΔH= -484 kJ
Predict the effect of each of the following changes to the system on the
direction of the equilibrium.
1. H2O is removed as it is being generated.
Moves to right (Q < K if P removed).
2. H2 is added.
Moves to right (Q < K if H2 added).
3. The system is cooled. This exothermic
reaction, so lowering temperature moves
system to right because temperature gradient
increases between system and surroundings.
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