Transcript No Slide Title
CHAPTER FIFTEEN
Copyright © Tyna L. Gaylord 2002 - 2009 All Rights Reserved 1
Equilibrium
At equilibrium, the rate at which products form is EQUAL to the rate at which products decompose.
Forward reaction: A --> B rate = k f [A] Reverse reaction: B --> A rate = k r [B] At equilibrium, k f [A] = k r [B] [B] = k f [A] k r = a constant (K eq ) Chem 15.1
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Equilibrium
Once equilibrium is established, the concentrations of A and B do not change.
The fact that the composition remains constant with time does not mean that A and B stop reacting
Compound A is still converted into compound B, but both processes occur at the same rate
Indicated by double arrow
Chem 15.1
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Chem 15.1
Equilibrium
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Equilibrium
One of the most important chemical systems is the synthesis of ammonia from nitrogen and hydrogen (HABER process) 3 Chem 15.1
Equilibrium
HABER process: N 2 + 3H 2
2NH 3(g)
put N 2 + 3H 2 in a high pressure tank at a total pressure of several hundred atmospheres, in the presence of a catalyst, and at a temperature of several hundred degrees Celsius.
Two gases react, but does not lead to complete consumption of gases Chem 15.1
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Equilibrium
N 2 HABER process: + 3H 2
2NH 3(g) at equilibrium, the relative concentrations of H 2 , N 2 and NH 3 are the same, regardless of the starting mixture, and note that the equilibrium is reached from either direction !
Chem 15.1
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Equilibrium
The equilibrium constant expression depends only on the stoichiometry of the reaction, not on its mechanism.
The only thing can alter the proportionality constant of a balanced chemical equation is TEMPERATURE, altering anything else will cause the reaction to shift in order to get back to the same proportionality...
3 Chem 15.1
Equilibrium
…which is why the rates of the forward and reverse reactions are always EQUAL and the concentration of each chemical always remains CONSTANT 3 Chem 15.1
Equilibrium
Sample exercise: Write the equilibrium constant expression for H 2(g) + I 2(g)
2HI (g) 11 Chem 116: Prof. T.L. Heise Chem 15.2
Equilibrium
Sample exercise: Write the equilibrium constant expression for H 2(g) + I 2(g)
2HI (g) 12 K eq = [HI] 2 [H 2 ][I 2 ] Chem 116: Prof. T.L. Heise Chem 15.2
Equilibrium
When the reactants and products of an equilibrium are gases, the partial pressures can be used… K p = (P P ) p (P Q ) q (P A ) a (P B ) b 13 K p = K eq (RT)
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n Chem 116: Prof. T.L. Heise Chem 15.2
Equilibrium
Sample exercise: For the equilibrium 2SO 3(g)
2SO 2(g) + O 2(g) at a temperature of 1000K, K eq has the value of 4.08 x 10 -3 . Calculate the value for K p .
14 Chem 116: Prof. T.L. Heise Chem 15.2
Equilibrium
Sample exercise: For the equilibrium 2SO 3(g)
2SO 2(g) + O 2(g) at a temperature of 1000K, K eq has the value of 4.08 x 10 -3 . Calculate the value for K p .
K p = K eq (RT)
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n 15 Chem 116: Prof. T.L. Heise Chem 15.2
Equilibrium
Sample exercise: For the equilibrium 2SO 3(g)
2SO 2(g) + O 2(g) at a temperature of 1000K, K eq has the value of 4.08 x 10 -3 . Calculate the value for K p .
K p = K eq (RT)
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n = 4.08 x 10 -3 ((0.0821)(1000)) 1 16 Chem 116: Prof. T.L. Heise Chem 15.2
Equilibrium
Sample exercise: For the equilibrium 2SO 3(g)
2SO 2(g) + O 2(g) at a temperature of 1000K, K eq has the value of 4.08 x 10 -3 . Calculate the value for K p .
K p = K eq (RT)
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n = 4.08 x 10 -3 ((0.0821)(1000)) 1 = 4.08 x 10 -3 (82.1) 17 Chem 116: Prof. T.L. Heise Chem 15.2
Equilibrium
Sample exercise: For the equilibrium 2SO 3(g)
2SO 2(g) + O 2(g) at a temperature of 1000K, K eq has the value of 4.08 x 10 -3 . Calculate the value for K p .
K p = K eq (RT)
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n = 4.08 x 10 -3 ((0.0821)(1000)) 1 = 4.08 x 10 -3 (82.1) = 0.335
18 Chem 116: Prof. T.L. Heise Chem 15.2
Equilibrium
Equilibrium Constants can be very large or very small. The magnitude of the equilibrium constant provides us with important information about the equilibrium mixture.
19 Chem 116: Prof. T.L. Heise Chem 15.2
Equilibrium
Sample Exercise: The equilibrium constant for the reaction H 2(g) + I 2(g)
2HI (g) varies with temperature in the following way: K eq K eq = 794 at 298 K = 54 at 700 K Is the formation of products favored more at the higher or lower temperature?
20 Chem 116: Prof. T.L. Heise Chem 15.2
Equilibrium
Sample Exercise: The equilibrium constant for the reaction H 2(g) + I 2(g)
2HI (g) varies with temperature in the following way: K eq K eq = 794 at 298 K = 54 at 700 K Is the formation of products favored more at the higher or lower temperature?
Lower because K eq is larger 21 Chem 116: Prof. T.L. Heise Chem 15.2
Equilibrium
The substances in the equilibrium are of different phases: the concentration of a pure solid or liquid equals its density divided by its molar mass D = g/cm 3
M
= mol g/mol cm 3 the density is constant at any given temperature so...
Chem 116: Prof. T.L. Heise Chem 15.3
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Equilibrium
CaCO 3 (s)
CaO(s) + CO 2 (g) K eq = [CaO][CO 2 ] [CaCO 3 ]
If a pure solid or liquid is used it is NOT included in the equilibrium constant
K eq = constant 1[CO 2 ] constant 2 K eq ’ = K eq constant 1 = [CO 2 ] constant 2 23 Chem 116: Prof. T.L. Heise Chem 15.3
Equilibrium
Sample Exercise: Write the equilibrium expressions for K eq and K p 3Fe(s) + 4H 2 O(g)
for the reaction Fe 3 O 4 (s) + 4H 2 (g) 24 Chem 116: Prof. T.L. Heise Chem 15.3
Equilibrium
Sample Exercise: Write the equilibrium expressions for K eq and K p 3Fe(s) + 4H 2 O(g)
for the reaction Fe 3 O 4 (s) + 4H 2 (g) K eq = products reactants 25 Chem 116: Prof. T.L. Heise Chem 15.3
Equilibrium
Sample Exercise: Write the equilibrium expressions for K eq and K p 3Fe(s) + 4H 2 O(g)
for the reaction Fe 3 O 4 (s) + 4H 2 (g) K eq = products = [H 2 ] 4 reactants [H 2 O] 4 26 Chem 116: Prof. T.L. Heise Chem 15.3
Equilibrium
Sample Exercise: Write the equilibrium expressions for K eq and K p 3Fe(s) + 4H 2 O(g)
for the reaction Fe 3 O 4 (s) + 4H 2 (g) K eq = products = [H 2 ] 4 reactants [H 2 O] 4 K p = P (H 2 ) 4 P (H 2 O) 4 Chem 116: Prof. T.L. Heise Chem 15.3
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Equilibrium
Sample Exercise: Which of the following substances - H 2 (g), H 2 O(g), O 2 (g) - when added to Fe 3 O 4 (s) in a closed container at high temperature, permits attainment of equilibrium in the rxn 3Fe(s) + 4H 2 O(g)
Fe 3 O 4 (s) + 4H 2 (g) 28 Chem 116: Prof. T.L. Heise Chem 15.3
Equilibrium
Sample Exercise: Which of the following substances - H 2 (g), H 2 O(g), O 2 (g) - when added to Fe 3 O 4 (s) in a closed container at high temperature, permits attainment of equilibrium in the rxn 3Fe(s) + 4H 2 O(g)
Fe 3 O 4 (s) + 4H 2 (g) Chem 15.3
Only hydrogen Chem 116: Prof. T.L. Heise 29
Equilibrium
Sample Exercise: Nitryl chloride, NO 2 Cl, is in equilibrium with NO 2 2NO 2 Cl(g)
2NO 2 and Cl 2 (g) + Cl : 2 (g) At equilibrium the concentrations of the substances are [NO 2 Cl] = 0.00106 M, [NO 2 ] = 0.0108 M, and [Cl 2 ] = 0.00538 M. From these data calculate the equilibrium constant, K eq .
30 Chem 116: Prof. T.L. Heise Chem 15.4
Equilibrium
2NO 2 Cl(g)
2NO 2 (g) + Cl 2 (g) At equilibrium the concentrations of the substances are [NO 2 Cl] = 0.00106 M, [NO 2 ] = 0.0108 M, and [Cl 2 ] = 0.00538 M. From these data calculate the equilibrium constant, K eq .
Keq = products reactants Chem 116: Prof. T.L. Heise Chem 15.4
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Equilibrium
2NO 2 Cl(g)
2NO 2 (g) + Cl 2 (g) equilibrium the concentrations of the substances are [NO 2 Cl] = 0.00106 M, [NO 2 ] = 0.0108 M, and [Cl 2 ] = 0.00538 M. From these data calculate the equilibrium constant, K eq .
At Keq = products = [NO 2 ] 2 [Cl 2 ] reactants [NO 2 Cl] 2 32 Chem 116: Prof. T.L. Heise Chem 15.4
Equilibrium
2NO2Cl(g)
2NO2(g) + Cl2(g) At equilibrium the concentrations of the substances are [NO2Cl] = 0.00106 M, [NO2] = 0.0108 M, and [Cl2] = 0.00538 M. From these data calculate the equilibrium constant, Keq.
Keq = [NO 2 ] 2 [Cl 2 ] = (0.0108) 2 (0.00538) [NO 2 Cl] 2 (0.00106) 2 Chem 116: Prof. T.L. Heise Chem 15.4
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Equilibrium
2NO2Cl(g)
2NO2(g) + Cl2(g) At equilibrium the concentrations of the substances are [NO2Cl] = 0.00106 M, [NO2] = 0.0108 M, and [Cl2] = 0.00538 M. From these data calculate the equilibrium constant, Keq.
Keq = [NO 2 ] 2 [Cl 2 ] = (0.0108) 2 (0.00538) [NO 2 Cl] 2 (0.00106) 2 = 0.558
Chem 116: Prof. T.L. Heise Chem 15.4
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Equilibrium
Rarely are we given all the information as in the first example problem, normally, one concentration will be given and stoichiometric calculations will give you the others.
35 Sample Exercise pg 572 on board.
Chem 116: Prof. T.L. Heise Chem 15.4
Equilibrium
If K is very large, the reaction will tend to proceed far to the right creating a large amount of products.
If K is very small, the reaction will not proceed, staying far to the left, leaving a large amount of reactants unused.
36 Chem 116: Prof. T.L. Heise Chem 15.5
Equilibrium
Predicting the Direction
use initial concentrations to determine original K eq
obtain K eq at temperature you want
compare calculated K eq to actual K eq and it will tell you how the reaction is going Chem 116: Prof. T.L. Heise Chem 15.5
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Equilibrium
Predicting the Direction
2.00 mol H 2 , 1.00 mol of N 2 , and 2.00 mol of NH 3 at 472°C
Q = (2) 2 /(1)(2) 3 = 0.500
according to S.E. 15.7, K eq = 0.105
our actual is much smaller than the original which telling me there is quite a bit less product and more reactant, so reaction is proceeding to the left Chem 116: Prof. T.L. Heise Chem 15.5
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Equilibrium
Sample exercise: At 1000 K the value of K eq for the reaction 2SO 3 (g)
2SO 2 (g) + O 2 (g) is 4.08 x 10 -3 . Calculate the value for Q, and predict the direction in which the reaction will proceed toward equilibrium if the initial concentrations are: [SO 3 ] = 2 x 10 -3 [SO 2 ] = 5 x 10 -3 M, and [O 2 ] = 3 x 10 M, -2 M. 39 Chem 116: Prof. T.L. Heise Chem 15.5
Equilibrium
Sample exercise: At 1000 K the value of K eq for the reaction 2SO 3 (g)
2SO 2 (g) + O 2 (g) is 4.08 x 10 -3 . Calculate the value for Q, and predict the direction in which the reaction will proceed toward equilibrium if the initial concentrations are: [SO 3 ] = 2 x 10 -3 [SO 2 ] = 5 x 10 -3 M, and [O 2 ] = 3 x 10 M, -2 M. Q = [SO 2 ] 2 [O 2 ] [SO 3 ] 2 Chem 116: Prof. T.L. Heise Chem 15.5
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Equilibrium
Sample exercise: At 1000 K the value of K eq for the reaction 2SO 3 (g)
2SO 2 (g) + O 2 (g) is 4.08 x 10 -3 . Calculate the value for Q, and predict the direction in which the reaction will proceed toward equilibrium if the initial concentrations are: [SO 3 ] = 2 x 10 -3 [SO 2 ] = 5 x 10 -3 M, and [O 2 ] = 3 x 10 M, -2 M. Q = [SO 2 ] 2 [O 2 ] = (5 x 10 -3 ) 2 (3 x 10 -2 ) [SO 3 ] 2 (2 x 10 -3 ) 2 Chem 116: Prof. T.L. Heise Chem 15.5
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Equilibrium
Sample exercise: At 1000 K the value of K eq for the reaction 2SO 3 (g)
2SO 2 (g) + O 2 (g) is 4.08 x 10 -3 . Calculate the value for Q, and predict the direction in which the reaction will proceed toward equilibrium if the initial concentrations are: [SO 3 ] = 2 x 10 -3 [SO 2 ] = 5 x 10 -3 M, and [O 2 ] = 3 x 10 M, -2 M. Q = [SO 2 ] 2 [O 2 ] = (5 x 10 -3 ) 2 (3 x 10 -2 ) = 0.187
[SO 3 ] 2 (2 x 10 -3 ) 2 Chem 116: Prof. T.L. Heise Chem 15.5
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Equilibrium
Sample exercise: At 1000 K the value of K eq for the reaction 2SO 3 (g)
2SO 2 (g) + O 2 (g) is 4.08 x 10 -3 . Calculate the value for Q, and predict the direction in which the reaction will proceed toward equilibrium if the initial concentrations are: [SO 3 ] = 2 x 10 -3 [SO 2 ] = 5 x 10 -3 M, and [O 2 ] = 3 x 10 M, -2 M. Q = 0.2 ; 4.08 x 10 -3 is much smaller, so I need less products to form and more reactants which indicates a shift to the left Chem 116: Prof. T.L. Heise Chem 15.5
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Equilibrium
It is possible to know the K eq and some of the concentrations and need to work backwards to find a missing concentration.
Example: For the Haber process, K p = 1.45 x 10 -5 at 500°C. Partial pressures known are H 2 0.928 atm and N 2 0.432 atm. What is the pressure of NH 3 ?
Chem 116: Prof. T.L. Heise Chem 15.5
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Equilibrium
It is possible to know the K eq and some of the concentrations and need to work backwards to find a missing concentration.
Example: For the Haber process, K p = 1.45 x 10 -5 at 500°C. Partial pressures known are H 2 0.928 atm and N 2 0.432 atm. What is the pressure of NH 3 ?
K p = 1.45 x 10 -5 Chem 15.5
= [NH 3 ] 2 [H 2 ] 3 [N 2 ] 45
Equilibrium
It is possible to know the K eq and some of the concentrations and need to work backwards to find a missing concentration.
K p Example: For the Haber process, K p = 1.45 x 10 -5 at 500°C. Partial pressures known are H 2 0.928 atm and N 2 0.432 atm. What is the pressure of NH 3 ?
= 1.45 x 10 -5 = x 2 (0.928) 3 (0.432) Chem 116: Prof. T.L. Heise Chem 15.5
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Equilibrium
It is possible to know the K eq and some of the concentrations and need to work backwards to find a missing concentration.
Example: For the Haber process, K p = 1.45 x 10 -5 at 500°C. Partial pressures known are H 2 0.928 atm and N 2 0.432 atm. What is the pressure of NH 3 ?
x = 5.01 x 10 -6 Chem 116: Prof. T.L. Heise Chem 15.5
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Equilibrium
Sample exercise:At 500 K the reaction PCl 5 (g)
PCl 3 (g) + Cl 2 (g) has K p = 0.497
. In an equilibrium mixture at 500 K, the partial pressure of PCl 5 that of PCl 3 is 0.860 atm and is 0.350 atm. What is the partial pressure of Cl mixture?
2 in the equilibrium 47 Chem 116: Prof. T.L. Heise Chem 15.5
Equilibrium
Sample exercise:At 500 K the reaction PCl 5 (g)
PCl 3 (g) + Cl 2 (g) has K p = 0.497
. In an equilibrium mixture at 500 K, the partial pressure of PCl 5 that of PCl 3 is 0.860 atm and is 0.350 atm. What is the partial pressure of Cl mixture?
2 in the equilibrium K p = [PCl 3 ][Cl 2 ] [PCl 5 ] Chem 15.5
Chem 116: Prof. T.L. Heise 48
Equilibrium
Sample exercise:At 500 K the reaction PCl 5 (g)
PCl 3 (g) + Cl 2 (g) has K p = 0.497
. In an equilibrium mixture at 500 K, the partial pressure of PCl 5 that of PCl 3 is 0.860 atm and is 0.350 atm. What is the partial pressure of Cl mixture?
2 in the equilibrium K p = [PCl 3 ][Cl 2 ] [PCl 5 ] Chem 15.5
0.497 = (.350)(x) (0.860) Chem 116: Prof. T.L. Heise 49
Equilibrium
Sample exercise:At 500 K the reaction PCl 5 (g)
PCl 3 (g) + Cl 2 (g) has K p = 0.497
. In an equilibrium mixture at 500 K, the partial pressure of PCl 5 that of PCl 3 is 0.860 atm and is 0.350 atm. What is the partial pressure of Cl mixture?
2 in the equilibrium K p = [PCl 3 ][Cl 2 ] [PCl 5 ] Chem 15.5
0.497(0.860) = (.350)(x) Chem 116: Prof. T.L. Heise 50
Equilibrium
51 Sample exercise:At 500 K the reaction PCl 5 (g)
PCl 3 (g) + Cl 2 (g) has K p = 0.497
. In an equilibrium mixture at 500 K, the partial pressure of PCl 5 that of PCl 3 is 0.860 atm and is 0.350 atm. What is the partial pressure of Cl mixture?
2 in the equilibrium K p = [PCl 3 ][Cl 2 ] [PCl 5 ] Chem 15.5
0.497(0.860) = (x) (0.350) Chem 116: Prof. T.L. Heise
Equilibrium
52 Sample exercise:At 500 K the reaction PCl 5 (g)
PCl 3 (g) + Cl 2 (g) has K p = 0.497
. In an equilibrium mixture at 500 K, the partial pressure of PCl 5 that of PCl 3 is 0.860 atm and is 0.350 atm. What is the partial pressure of Cl mixture?
2 in the equilibrium K p = [PCl 3 ][Cl 2 ] [PCl 5 ] Chem 15.5
1.22 = (x) Chem 116: Prof. T.L. Heise
Equilibrium
If a system at equilibrium is disturbed by a change in temperature, pressure, or the concentration of one of the components, the system will shift its equilibrium position so as to counteract the effect of the disturbance.
3 ways to change an equilibrium
add or remove a reactant or product
change the pressure
change the temperature Chem 116: Prof. T.L. Heise Chem 15.6
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Equilibrium
Change in reactant or product concentrations
adding a reactant or product forces a shift that will use up what has been added
removing a reactant or product forces a shift that will create more of what was taken 54 Chem 116: Prof. T.L. Heise Chem 15.6
Equilibrium
Change in volume or pressure
decreasing volume causes increasing pressure and vice versa
increasing pressure forces a shift in equilibrium towards the production of fewer moles of gas
decreasing pressure forces a shift in equilibrium towards the production of more moles of gas Chem 116: Prof. T.L. Heise Chem 15.6
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Equilibrium
Change in temperature
increasing temperature forces a shift in the endothermic direction so increased energy can be used up
decreasing temperature forces a shift in the exothermic direction so more energy can be produced to replace lost energy Chem 116: Prof. T.L. Heise 56 Chem 15.6
Chem 15.6
Equilibrium
Concentration of H 2 is altered and affects are graphed 57 Chem 116: Prof. T.L. Heise
Equilibrium
Sample Exercise: For the reaction PCl 5 (g) + energy
PCl 3 (g) + Cl 2 (g) in what direction will the equilibrium shift when (a) Cl 2 (g) is added (b) the temperature is increased (c ) the volume is decreased (d) PCl 5 (g) is added 58 Chem 116: Prof. T.L. Heise Chem 15.6
Equilibrium
Sample Exercise: For the reaction PCl 5 (g) + energy
PCl 3 (g) + Cl 2 (g) in what direction will the equilibrium shift when (a) Cl 2 (g) is added PCl 5 (g) + energy
PCl 3 (g) + Cl 2 (g) 59 Chem 116: Prof. T.L. Heise Chem 15.6
Equilibrium
Sample Exercise: For the reaction PCl 5 (g) + energy
PCl 3 (g) + Cl 2 (g) in what direction will the equilibrium shift when (a) Cl 2 (g) is added PCl 5 (g) + energy
PCl 3 (g) + Cl 2 (g) 60 Chem 116: Prof. T.L. Heise Chem 15.6
Equilibrium
Sample Exercise: For the reaction PCl 5 (g) + energy
PCl 3 (g) + Cl 2 (g) in what direction will the equilibrium shift when (b) the temperature is increased PCl 5 (g) + energy
PCl 3 (g) + Cl 2 (g) 61 Chem 116: Prof. T.L. Heise Chem 15.6
Equilibrium
Sample Exercise: For the reaction PCl 5 (g) + energy
PCl 3 (g) + Cl 2 (g) in what direction will the equilibrium shift when (b) the temperature is increased PCl 5 (g) + energy
PCl 3 (g) + Cl 2 (g) 62 Chem 116: Prof. T.L. Heise Chem 15.6
Equilibrium
Sample Exercise: For the reaction PCl 5 (g) + energy
PCl 3 (g) + Cl 2 (g) in what direction will the equilibrium shift when (c ) the volume is decreased 63 Chem 116: Prof. T.L. Heise Chem 15.6
Equilibrium
Sample Exercise: For the reaction PCl 5 (g) + energy
PCl 3 (g) + Cl 2 (g) in what direction will the equilibrium shift when (c ) the volume is decreased ...so pressure is increased, favors less moles PCl 5 (g) + energy
PCl 3 (g) + Cl 2 (g) 1 mole 1 mole + 1 mole Chem 116: Prof. T.L. Heise Chem 15.6
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Equilibrium
Sample Exercise: For the reaction PCl 5 (g) + energy
PCl 3 (g) + Cl 2 (g) in what direction will the equilibrium shift when (c ) the volume is decrease ...so pressure is increased, favors less moles PCl 5 (g) + energy
PCl 3 (g) + Cl 2 (g) 1 mole 1 mole + 1 mole Chem 116: Prof. T.L. Heise Chem 15.6
65
Equilibrium
Sample Exercise: For the reaction PCl 5 (g) + energy
PCl 3 (g) + Cl 2 (g) in what direction will the equilibrium shift when (d) PCl 5 (g) is added 66 Chem 116: Prof. T.L. Heise Chem 15.6
Equilibrium
Sample Exercise: For the reaction PCl 5 (g) + energy
PCl 3 (g) + Cl 2 (g) in what direction will the equilibrium shift when (d) PCl5(g) is added PCl 5 (g) + energy
PCl 3 (g) + Cl 2 (g) 67 Chem 116: Prof. T.L. Heise Chem 15.6
Equilibrium
Sample Exercise: For the reaction PCl 5 (g) + energy
PCl 3 (g) + Cl 2 (g) in what direction will the equilibrium shift when (d) PCl5(g) is added PCl 5 (g) + energy
PCl 3 (g) + Cl 2 (g) 68 Chem 116: Prof. T.L. Heise Chem 15.6
Equilibrium
Sample Exercise: Using the thermodynamic data in Appendix C, determine the enthalpy change for the reaction 2POCl 3 (g)
2PCl 3 (g) + O 2 (g) use this result to determine how the equilibrium constant for the reaction should change with temperature.
69 Chem 116: Prof. T.L. Heise Chem 15.6
Equilibrium
Sample Exercise: Using the thermodynamic data in Appendix C, determine the enthalpy change for the reaction 2POCl 3 (g)
2PCl 3 (g) + O 2 (g) -542.2 -288.07 + 0
D
H = products - reactants 2(-288.07) - 2(-542.2) 508.26 kJ Chem 116: Prof. T.L. Heise Chem 15.6
70
Equilibrium
Sample Exercise: Using the thermodynamic data in Appendix C, determine the enthalpy change for the reaction 2POCl 3 (g)
2PCl 3 (g) + O 2 (g) -542.2 -288.07 + 0
D
H = products - reactants 2(-288.07) - 2(-542.2) 508.26 kJ Endothermic Chem 116: Prof. T.L. Heise Chem 15.6
71
Equilibrium
Sample Exercise: Using the thermodynamic data in Appendix C, determine the enthalpy change for the reaction 2POCl 3 (g) + energy
2PCl 3 (g) + O 2 (g) use this result to determine how the equilibrium constant for the reaction should change with temperature.
72 Chem 116: Prof. T.L. Heise Chem 15.6
Equilibrium
Sample Exercise: Using the thermodynamic data in Appendix C, determine the enthalpy change for the reaction 2POCl 3 (g) + energy
2PCl 3 (g) + O 2 (g) use this result to determine how the equilibrium constant for the reaction should change with temperature.
K eq = [PCl 3 ] 2 [O 2 ] [POCl 3 ] 2 73 Chem 116: Prof. T.L. Heise Chem 15.6
Equilibrium
Sample Exercise: Using the thermodynamic data in Appendix C, determine the enthalpy change for the reaction 2POCl 3 (g) + energy
2PCl 3 (g) + O 2 (g) use this result to determine how the equilibrium constant for the reaction should change with temperature.
K eq = [PCl 3 ] 2 [O 2 ] [POCl 3 ] 2 if energy inc., equilibrium shifts right, products are bigger, K eq is larger Chem 116: Prof. T.L. Heise Chem 15.6
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Equilibrium
Effects of a Catalyst: a catalyst changes the rate of a reaction only, not the amounts of the compounds!!
Remember - Rates are Equal Concentrations are Constant Only temp. can change a constant Chem 116: Prof. T.L. Heise Chem 15.6
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