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CHAPTER FIFTEEN

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

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

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Indicated by double arrow

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Chem 15.1

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

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2NH 3(g)

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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

Equilibrium

N 2 HABER process: + 3H 2

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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

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)

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)

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)

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)

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)

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)

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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

= mol g/mol cm 3 the density is constant at any given temperature so...

Chem 116: Prof. T.L. Heise Chem 15.3

Equilibrium

CaCO 3 (s)

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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)

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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)

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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)

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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

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)

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Fe 3 O 4 (s) + 4H 2 (g) 28 Chem 116: Prof. T.L. Heise Chem 15.3

Equilibrium

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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)

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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)

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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

Equilibrium

2NO 2 Cl(g)

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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)

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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

Equilibrium

2NO2Cl(g)

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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

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

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use initial concentrations to determine original K eq

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obtain K eq at temperature you want

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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

Equilibrium

Predicting the Direction

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2.00 mol H 2 , 1.00 mol of N 2 , and 2.00 mol of NH 3 at 472°C

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Q = (2) 2 /(1)(2) 3 = 0.500

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according to S.E. 15.7, K eq = 0.105

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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

Equilibrium

Sample exercise: At 1000 K the value of K eq for the reaction 2SO 3 (g)

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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)

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Equilibrium

Sample exercise: At 1000 K the value of K eq for the reaction 2SO 3 (g)

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Equilibrium

Sample exercise: At 1000 K the value of K eq for the reaction 2SO 3 (g)

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[SO 3 ] 2 (2 x 10 -3 ) 2 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)



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

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

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

Equilibrium

Sample exercise:At 500 K the reaction PCl 5 (g)

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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

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add or remove a reactant or product

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change the pressure

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change the temperature Chem 116: Prof. T.L. Heise Chem 15.6

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

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