Chemical Equilibrium Chemical equilibrium occurs when a reaction and - - PDF document

equilibrium­presentation­2012­02­27.notebook 1 November 09, 2012

Chemical Equilibrium

The Concept of Equilibrium

Chemical equilibrium occurs when a reaction and its reverse reaction proceed at the same rate.

Chemical Equilibrium

Chemical equilibrium occurs when a reaction and its reverse reaction proceed at the same rate. N2O4 is a colorless gas. It decomposes into NO2, a brown­colored gas. Eventually, the forward and backward reactions reach a state of equilibrium. This means that the colorless gas is decomposing at the same rate at which the brown gas is combining. N2O4(g) 2NO2(g) When both the forward and reverse reactions are being carried out, we write the equation with a double arrow.

Chemical Equilibrium

N2O4(g) 2NO2(g)

The Concept of Equilibrium

• As a system approaches

equilibrium, both the forward and reverse reactions are occurring.

• When a reaction begins,

the forward reaction

• ccurs quickly, at first then

begins to slow down.

• The reverse reaction is

non­existent at the start, but then picks up speed as more product is created.

kf[N2O4] kr[NO2]2

Time Rate

Equilibrium achieved (rates are equal)

• At equilibrium, the forward

and reverse reactions are proceeding at the same rate.

• Once equilibrium is

achieved, the amount of each reactant and product remains constant.

Chemical Equilibrium

Kf[N2O4] Kf[NO2]2 Time Rate Equilibrium achieved (rates are equal)

This is a dynamic equilibrium, meaning that the reactions do not stop, but there is no net change in any concentrations.

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Once equilibrium is achieved, two important conditions apply: (1) the forward and reverse reactions are proceeding at the same rate.

Chemical Equilibrium

(2) the amount of each reactant and product remains constant.

Kf[N2O4] Kf[NO2]2 Time Rate Equilibrium achieved (rates are equal)

(This does not imply that the amount of reactants = amount of products.)

Chemical Equilibrium

It is important to note that equilibrium can be achieved regardless of whether you start with reactants or products, as long as there is sufficient material to get both processes going. N2 (g) + 3H2 (g) 2NH3 (g)

Chemical Equilibrium

N2 (g) + 3H2 (g) 2NH3 (g) In this graph, observe that the

• nly substances initially

present are nitrogen and hydrogen; there is no ammonia at the start. In this graph, observe that the

• nly substance initially

present is ammonia; there is no nitrogen and hydrogen at the start.

Chemical Equilibrium

Regardless of the starting material, however, the same equilibrium concentrations of all three substances will eventually be reached. N2 (g) + 3H2 (g) 2NH3 (g)

1 At equilibrium, _______________________.

A all chemical reactions have ceased B the rates of forward and reverse reactions are equal C the rate constants of the forward and reverse reactions are equal D the value of the equilibrium constant is 1 E the limiting reagent has been consumed

2 Which of the following statements is NOT true about a reversible reaction at equilibrium?

A reactant and product concentrations are constant B reactants and products remain in dynamic equilibrium C reactant concentrations are equal to product concentrations D the forward reaction occurs at the same rate as the reverse reaction

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3 To achieve equilibrium, you must have only reactants to start with.

True False

4 Consider the equilibrium A + 3B <­­> 2C. Equilibrium can be reached if you start with:

A

• nly A

B

• nly B

C

• nly C

5 Consider the equilibrium A + 3B <­­> 2C. Equilibrium cannot be reached if you start with:

A A & B B

• nly B

C

• nly C

6 Consider the equilibrium 2 SO2 (g) + O2 (g) 2 SO3 (g). Equilibrium cannot be established when _________is/are in the container.

A 0.5 mol SO2 and 0.5 mol O2

B 0.5 mol SO3

C 0.5 mol O2 and 0.5 mol SO3 D

0.5 mol SO2

E 0.5 mol SO2 and 0.5 mol SO3

The Equilibrium Constant

• Recall that the definition of chemical equilibrium is when the

rate of the forward reaction equals the rate of the reverse reaction. N2O4(g) <­­>2NO2(g)

• Therefore, at equilibrium, we can set the two rate laws equal:

Rate of forward reaction = Rate of reverse reaction kf [N2O4] = kr [NO2]2

• Rewriting this, it becomes

kf [NO2]2 kr [N2O4] = The ratio of the rate constants is known as the equilibrium constant, K. The ratio simplifies to a ratio of products to reactants.

The Equilibrium Constant

kf [NO2]2 kr [N2O4] = Keq =

The value of Keq is constant for a specific reversible reaction at a certain temperature. Like rate constants, an equilibrium constant, Keq will only change when there is a change in temperature.

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The Equilibrium Constant

The equilibrium constant expression can be written from a balanced chemical equation aA + bB cC + dD according to the general format:

Kc = K = [C]c [D]d [A]a [B]b

The Equilibrium Constant

Pure solids and liquids are not included in the calculation of the equilibrium constant. You can take their concentration to be 1, and not include them. That's because their concentrations don't change during a reaction.

Molarity = density / molar mass = (g/L) / (g/mol) = mol/L Since neither density nor molar mass varies for a pure liquid or pure solid, we take their molarities to remain constant.

The equilibrium constant for the reaction is CH4 (g) + 2O2(g) CO2 (g) + 2H2O(l)

The Equilibrium Constant

K = [CO2] [CH4] [O2]2

The Equilibrium Constant

The equilibrium constant for the reaction is 2Cl2 (g) + 2H2O(g) 4HCl(g) + O2(g)

K = [HCl]4 [O2] [Cl2]2 [H2O]2

The Equilibrium Constant

The equilibrium constant for the reaction is (CH3)4Sn(s) (CH3)4Sn(g) K = [(CH3)4Sn]

7

The equilibrium­constant expression depends on the __________ of the reaction. A stoichiometry B mechanism C

stoichiometry and mechanism D the quantities of reactants and products initially present

E temperature

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8

Which one of the following will change the value of an equilibrium constant? A

changing temperature B adding other substances that do not react with any of the species involved in the equilibrium C varying the initial concentrations of reactants D varying the initial concentrations of products E changing the volume of the reaction vessel

• If K >1, the reaction is product­

favored; product predominates at equilibrium.

• If K<1, the reaction is

reactant­favored; reactant predominates at equilibrium.

What Does the Value of K Mean?

Reactants Reactants Products Products

K > 1 K < 1

The direction from which equilibrium is achieved (starting with all products or all reactants) doesn't matter. It is only a convention that we calculate K by dividing the concentrations of products over reactants, since the reaction proceeds both ways. Unless specified, read the equation left to right Left side ­ reactants; right side ­ products. But that convention makes it easy to understand the meaning of Keq.

Equilibrium

9 For a reaction at equilibrium, a value of K= 1 x 108 means that

A reactants are favored B products are favored C the reaction is spontaneous D the reaction is endothermic

10 For a reaction at equilibrium, a value of K = 4 x 10­12 means that

A reactants and products are present in equal amounts B the forward reaction is occurring faster than the reverse reaction

C

the reverse reaction is occurring faster than the forward reaction

D reactants are favored

E

products are favored

11 X <­­> Y + Z What is the equilibrium constant expression for the reaction above?

A

K = [X] / [Y] +[Z]

B

K = [X] / [Y] ­ [Z]

C

K = [X] / [Y] [Z]

D

K = [Y] + [Z] / [X]

E

K = [Y] [Z] / [X]

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12 X <­­> Y + Z Calculate the value of the equilibrium constant, K, for the reaction above if the equilibrium concentrations are: [X] = 0.3 M, [Y] = 0.5 M, [Z] = 1.2 M 13 A + 2B <­­> C Calculate the value of the equilibrium constant, K, for the reaction above if the equilibrium concentrations are: [A] = 0.1 M, [B] = 0.2 M, [C] = 0.4 M 14 2A + B <­­> C The value of the equilibrium constant, K, for the reaction above at 400 K is 100. Calculate the equilibrium concentration

• f C, given these equilibrium

concentrations at 400 K: [A] = 0.1 M, [B] = 0.4 M [C] = ____ M 15 X <­­> 2Y + Z The value of the equilibrium constant, K, for the reaction above at 350 K is 0.5. Calculate the equilibrium concentration

• f X, given these equilibrium

concentrations at 400 K: [Y] = 2 M, [Z] = 1 M [X] = ____ M

The Equilibrium Constant

Since pressure is proportional to concentration for gases in a closed system, the equilibrium expression can also be written in terms of pressure

where Pc = partial pressure of gaseous substance C PA = partial pressure of gaseous substance A and so forth for B and D Recall that PA = XA(Ptot) where XA is the mole fraction of the gas and Ptot is total pressure. Consider the generalized reaction cC + dD aA + bB

Kp = (PA)a(PB)b (PC)c (PD)d 16 What is the equilibrium expression for Kp for the reaction given. 2 O3 (g) 3 O2 (g)

A

3 PO2 / 2 PO3

B

2PO3 / 3 PO2

C

3 PO3 / 2 PO2

D

(PO3)2 / (PO2)3

E

(PO2)3 / (PO3)2

Kp = (PA)a(PB)b (PC)c (PD)d

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From the Ideal Gas Law we know that Rearranging it, we get PV = nRT

P = RT n V = [M] (the concentration) n V So, P = [M] RT

Relationship Between Kc and Kp

Kp = ([C]cRT) ([D]dRT) ([A]aRT) ([B]bRT) [C]c [D]d RTc+d [A]a[B]b RTa+b = Kc x (RT) [(c+d) ­ (a+b)]

Kp = Kc (RT) Δn

=

Relationship Between Kc and Kp

Since P = (n/V)RT = [M] RT, we can substitute as follows: Kp = (PA)a(PB)b (PC)c (PD)d Therefore, the relationship between Kc and Kp becomes

Kp = Kc (RT)Δn

Δn = (moles of gaseous product) ­ (moles of gaseous reactant) Δn is the change in the number of moles of gas, which may be either positive or negative R is the gas constant (0.0821 L­atm/mol­K) T is the absoluted temperature in Kelvin

Relationship Between Kc and Kp

where

17 Given the following reaction at equilibrium at 450oC, if PCO2 = 0.0160 atm, then what is the value of the equilibrium constant, Kp? CaCO3 (s) CaO (s) + CO2 (g)

A 2.70 x 10­4 B 0.0160

C 0.0821

D 7.23

E 723

Kp = (PA)a(PB)b (PC)c (PD)d

18 Given the following reaction at equilibrium, if Kc = 6.44 x 105 at 230

• C, what is Kp?

2 NO (g) + O2 (g) 2 NO2 (g)

A

3.67 x 10­ 2

B 6.44 x 10 ­5 C 1.56 x 10 4 D

2.66 x 106

E

2.67 x 107

Kp = Kc (RT)Δn

19 Given the following reaction at equilibrium, if Kp = 1.05 at 250

• C,

then Kc = ___? PCl5 (g) PCl3 (g) + Cl2 (g)

A

3.90 x 10­6

B

2.45 x 10­2 C 1.05 D 42.9 E 45.0

Kp = Kc (RT)Δn

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Manipulating Equilibrium Constants

We now consider how the value of Keq changes when:

• a reversible reaction is written in the reverse direction
• the coefficients of a balanced equation are increased or

decreased

• two small reactions are added to yield an overall reaction

Manipulating Equilibrium Constants

Consider the reaction A <­­> 2B Write the K expression for this reaction: Now, consider the reaction 2B <­­> A Write the K expression for this reaction:

Manipulating Equilibrium Constants

When a reaction is reversed, the new K expression is simply the reciprocal of the original. [B]2 [A]

• riginal K =

[A] [B]2 new K = A <­­> 2B 2B <­­> A 20 At 250K, the equilibrium constant for the reaction X + 3Y Z is 8. What is the value of the equilibrium constant for the reaction Z X + 3Y? A 0.125 B

0.25

C 8 D 80 21 At 1000K, the equilibrium constant for the reaction 2A + B 2C is 0.02. What is the value of the equilibrium constant for the reaction 2C 2A + B?

A 0.02 B 50 C 100 D 200

Manipulating Equilibrium Constants

Now, consider the effect of multiplying all the coefficients in a balanced equation by a small whole number.

A 2B [B]2 [A]1 Kc = = 0.2 (0.2)3 K'c = = 3A 6B ([B]2)3 ([A]1)3 The equilibrium constant, K'c, of a reaction that has been multiplied by a small, whole number is the

• riginal equilibrium constant, Kc, raised to the power of

that small, whole number.

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Manipulating Equilibrium Constants

Here is another example:

N2O4(g) 2NO2 [NO2]2 [N2O4]1 Kc = = 0.212 2N2O4(g) 4NO2 ([NO2]2)2 ([N2O4]1)2 K'c = 0.0449 (0.212)2 K'c = =

The equilibrium constant, K'c, of a reaction that has been multiplied by a small, whole number is the original equilibrium constant, Kc, raised to the power of that small, whole number.

22 The value of Keq for the reaction A + B C + D is 0.25. The value of Keq at the same temperature for the reaction 2A + 2B 2C + 2D is __________. A 0.25

B

2 x (0.25) C 1/2 x (0.25) D 1/ (0.25)

E (0.25)2

23 The value of Keq for the equilibrium H2(g) + I2(g) 2HI(g) is 900 at 25°C. What is the value of Keq for the equilibrium below? ½H2(g) + ½I2(g) HI(g)

A

900

B

900/2

C

900*2

D

√900

E

9002 24

2 Cl2 (g) + 2 H2O (g) 4 HCl (g) + O2 (g) The Keq for the equilibrium above is 7.52 x 10­2 at 480oC. What is the value of Keq at this temperature for the reaction Cl2 (g) + H2O (g) 2 HCl (g) + 1/2 O2 (g)? A

5.66 x 10­3

B 0.0376

C 0.0752

D 0.150 E 0.274

25 At 550oC, Keq = 100 for the equilibrium 6A +2B 4C + 2D. What is the value of Keq at this temperature for the reaction 2C + D 3A + B?

A

0.1

B

0.02

C

50

D

200

E

√200

Manipulating Equilibrium Constants

Consider a reaction that can be written as the sum

• f two or more steps. For example, the overall

reaction A + B D + E can be written as the sum of these two steps:

Equation (1) A + B C Equation (2) C D + E Sum of Equations (1+2) A + B D + E

Write the K expression for these three equations.

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Manipulating Equilibrium Constants

Equation (1) A + B C Equation (2) C D + E Sum of Equations (1+2) A + B D + E

The equilibrium constant for a net reaction made up

• f two or more steps is the product of the equilibrium

constants for the individual steps. Write and compare the equilibrium expressions for these 3 reactions. K3 = K3 x K2 = [C] [A][B] [D][E] [C]

X

K3 = [D][E] [A][B] [C] [A][B] K1 = [D][E] [C] K2 = The equilibrium constant for a net reaction made up

• f two or more steps is the product of the equilibrium

constants for the individual steps.

Manipulating Equilibrium Constants

26 X ­­> Y K1 = 0.3 Y ­­> Z K2 = 0.1 The equilibrium constant for the reaction X <­­> Z is _____

A

0.03

B

0.1

C

0.2

D

0.3

E

0.4 27 A ­­> C + D K1 = 5 A + B + C ­­> 2D K2 = 6 The equilibrium constant for the reaction 2A + B <­­> 3D is _____

A

5

B

6

C

11

D

30

E

Not enough information is given.

Solving Equilibrium Problems

So far, all of the equilibrium problems have been for a reaction at equilibrium. The following problem gives initial concentrations (instead of equilibrium concentrations) and asks for calculation of K. We use an "ICE" chart, like the one shown below, for problems such as these. Example: A + 2B <­­> C A 2B C (I) Initial concentration (C) Change (E) Equilibrium concentration

A closed system initially containing 1.000 x 10­3 M H2 and 2.000 x 10­3 M I2 at 448 °C is allowed to reach equilibrium. Analysis of the equilibrium mixture shows that the concentration of HI is 1.87 x 10­3 M. Calculate Kc at 448 °C for the reaction taking place, which is, H2 (g) + I2 (s) 2 HI (g)

Solving Equilibrium Problems

H2 (g) I2 (s) 2HI (g) (I) Initial concentration

1.000 x 10­3 M 2.000 x 10­3 M

(C) Change (E) Equilibrium concentration

1.87 x 10­3 M

• First, put in all of the information that is given.
• Pay close attention to whether the given values are

INITIAL concentrations or EQUILIBRIUM concentrations.

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H2 (g) I2 (s) 2HI (g) (I) Initial concentration

1.000 x 10­3 M 2.000 x 10­3 M

(C) Change (E) Equilibrium concentration

1.87 x 10­3 M

Note that the problem does not indicate that any product, HI, is initially present. Therefore, the starting concentration for HI is put down as 0.

Solving Equilibrium Problems

H2 (g) I2 (s) 2HI (g) (I) Initial concentration

1.000 x 10­3 M 2.000 x 10­3 M

(C) Change + 1.87 x 10­3 M (E) Equilibrium concentration

1.87 x 10­3 M

Now, we consider how to complete the chart. If [HI] starts at 0 and ends at 1.87 x 10­3 M , then the change

must be + 1.87 x 10­3 M.

Solving Equilibrium Problems

H2 (g) I2 (s) 2HI (g) (I) Initial concentration

1.000 x 10­3 M 2.000 x 10­3 M

(C) Change

­1/2(1.87 x 10­3) M = ­ 9.35 x 10­4 M ­1/2(1.87 x 10­3) M = ­ 9.35 x 10­4 M

+ 1.87 x 10­3 M (E) Equilibrium concentration

1.87 x 10­3 M

Now, look at the stoichiometry of the reaction. Using the ratio of the coefficients, complete the "CHANGE" row. Note that reactant concentrations will decrease as these substances get consumed.

Solving Equilibrium Problems

H2 (g) I2 (s) 2HI (g) (I) Initial concentration

1.000 x 10­3 M 2.000 x 10­3 M

(C) Change

­1/2(1.87 x 10­3) M = ­ 9.35 x 10­4 M ­1/2(1.87 x 10­3) M = ­ 9.35 x 10­4 M

+ 1.87 x 10­3 M (E) Equilibrium concentration 6.50 x 10­5 M 1.07 x 10­3 M 1.87 x 10­3 M Since reactants get consumed, subtract the CHANGE amount from the INITIAL amounts to obtain theEQUILIBRIUM amounts.

Solving Equilibrium Problems

Now, we are ready to substitute the equilibrium concentrations into the equilibrium constant expression for the reaction.

Kc =

[HI]2 [H2] [I2] (1.87x10­3)2 (6.50 x 10­5) (1.07 x 10­3) = = 51

Solving Equilibrium Problems

Note that an ICE chart may be completed using either moles or molarity. If you use moles, make sure to divide by the volume to

• btain molarity before substituting into the K

expression.

Solving Equilibrium Problems

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28 Consider the reaction X <­­> Y The initial concentration of X is 4 M and the equilibrium concentration of Y is 1 M. What is the equilibrium amount of X?

A

3

B

4

C

5

X Y Initial 4 Change Equilibrium 1

29 Consider the reaction A <­­> 2 B The initial concentration of A is 10 M and the equilibrium concentration of B is 2 M. What is the equilibrium amount of A?

A

7

B

8

C

9

D

11

E

12

A 2 B Initial 10 Change Equilibrium 2

30

Consider the reaction A + B <­­> 2 C The initial concentrations of A and B are both 5 mol and the equilibrium concentration of C is 4 mol. What is the TOTAL number of moles at equilibrium? A

4

B

6

C

10

D

14

A B 2 C Initial 5 5 Change Equilibrium 4

31 Consider the reaction 2A <­­> B The initial concentration of A is 8 M and the equilibrium concentration of B is 3 M. What is the equilibrium amount of A?

A

2

B

5

C

6

D

11

E

14

2 A B Initial 8 Change Equilibrium 3

Solving Equilibrium Problems

• If initial concentrations are given, then you should

use an ICE chart.

• Many equilibrium problems ask you to calculate

the value of the equilibrium constant.

• Other equilibrium problems give you the value of

Kc or Kp and ask you to solve for an equilibrium

• concentration. This is illustrated in the next sample

problem. PCl5 PCl3 Cl2 (I) Initial pressure 1.66 atm (C) Change (E) Equilibrium pressure

For the equilibrium PCl5 (g) <­­> PCl3 (g) + Cl2 (g) Kp = 0.497 at 500 K. A gas cylinder is charged (filled) with PCl5 to an initial pressure of 1.66 atm. What are the partial pressures of all substances at equilibrium?

Solving Equilibrium Problems

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Since there are no equilibrium concentrations given, we use variables for the CHANGE row.

Solving Equilibrium Problems

PCl5 PCl3 Cl2 (I) Initial pressure 1.66 atm (C) Change ­ x + x + x (E) Equilibrium pressure Now, we calculate equilibrium concentrations.

Solving Equilibrium Problems

PCl5 PCl3 Cl2 (I) Initial pressure 1.66 atm (C) Change ­ x + x + x (E) Equilibrium pressure 1.66 ­ x x x Write out the equilibrium constant expression, Kc.

Solving Equilibrium Problems

(x)2 (1.66 ­ x) =

= 0.497

Kp = (PCl2)(PCl3) (PPCl5) PCl5 PCl3 Cl2

(I) Initial pressure 1.66 atm (C) Change ­ x + x + x (E) Equilibrium pressure 1.66 ­ x x x

Solving Equilibrium Problems

(x)2 (1.66 ­ x) =

= 0.497

Kp = (PCl2)(PCl3) (PPCl5) which simplifies to x2 + 0.497x ­ 0.825 = 0 A = 1 B = 0.497 C = ­0.825

Use the quadratic equation to solve for x. Examine the two values that you obtain; one value will be negative and therefore incorrect. PCl5 PCl3 Cl2 (I) Initial pressure 1.66 atm (C) Change

­ x + x + x (E) Equilibrium pressure 0.967 0.693 0.693

Solving Equilibrium Problems

Once you obtain the correct values, substitute them into the K expression to check your work.

(0.693)2 (0.967) =

= 0.497

Kp = (PCl2)(PCl3) (PPCl5)

The Reaction Quotient (Q)

When a system is not at equilibrium, it is useful to try to predict which reaction (the forward or the reverse reaction) must be favored in order to reach equilibrium.

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The Reaction Quotient, Q, gives the same ratio the equilibrium expression gives, but for a system that is not at equilibrium.

The Reaction Quotient (Q)

• To calculate Q, one substitutes the initial

concentrations of reactants and products into the equilibrium expression.

• Then, by comparing the values for Q and K, one

can predict in which direction (forward or reverse) the reaction must proceed in order to achieve an equilibrium state.

The Reaction Quotient (Q) If Q = K,

then we conclude that the system is already at equilibrium.

If Q > K,

then there is too much product; the reaction must proceed to the left in

• rder to reach a state of equilibrium.

The phrase "proceed to the left" basically means that the reverse reaction must speed up while the forward reaction slows down, in order for Q=K eventually.

If Q < K,

then there is too much reactant; the reaction must proceed to the right in order to reach a state of equilibrium.

The phrase "proceed to the right" basically means that the forward reaction must speed up while the reverse reaction slows down, in

• rder for Q=K eventually.

32 If K = 50 for a reaction at a temperature of 500 K, and Q = 12, then

A

the reaction will proceed to the right to reach equilibrium

B

the reaction will proceed to the left to reach equilibrium

C

the reaction is already at equilibrium

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33 If K = 0.45 for a reaction at a temperature of 100 K, and Q = 6, then

A

the reaction is already at equilibrium

B

the reaction will proceed to the left to reach equilibrium

C

the reaction will proceed to the right to reach equilibrium

Le Chatelier's Principle

Le Châtelier’s Principle

“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.” There are 3 possible stresses that can cause a system at equilibrium to no longer be at equilibrium:

• change in concentration (of a reactant or product),
• change in pressure, and
• change in temperature

We will examine the independent effect of each stress.

Le Châtelier’s Principle

How is equilibrium affected if we change the concentration?

+

Consider the following reaction: The Haber Process Synthesis of ammonia from hydrogen and nitrogen N2 (g) + 3H2 (g) 2NH3 (g)

Le Châtelier’s Principle

"take it and go" policy

• If H2 is added to the system, the added hydrogen needs to be consumed.
• How?
• By reacting with nitrogen.
• More N2 will be consumed with the added H2
• More NH3 will be produced
• If we increase the concentration of H2, equilibrium will shift to the right.

Initial equilibrium H2 NH3 N2 H2 added at this time

Equilibrium reestablished

Time P a r t i a l p r e s s u r e

+

• What happens when you remove NH3?
• The system has to replace it to bring back the equilibrium
• More NH3 will be produced.
• As a result, [H2] and [N2] will get lower.
• We say that equilibrium will shift to the right as a result of this

stress.

• We need large amount of NH3 for making fertilizers.

Le Châtelier’s Principle

Consider a decrease in the concentration of ammonia. How will equilibrium be affected?

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34 What happens to a reaction at equilibrium when a reactant concentration is lowered?

A Equilibrium shifts right; the reaction makes more products. B Equilibrium shifts left; the reaction makes more reactants. C The reaction makes more of both products and reactants, so equilibrium is unaffected.

35 What happens to a reaction at equilibrium when a product is removed from the reaction system?

A The reaction makes more products. B The reaction makes more reactants. C The reaction makes more of both products and reactants, so equilibrium is unaffected. 36 What is the effect of adding more carbon dioxide to the following equilibrium reaction? CO2 (g) + H2O (l) H2CO3 (aq)

A More H2CO3 (carbonic acid) is produced. B Equilibrium shifts right, toward the products. C Equilibrium shifts left, toward the reactants. D Both A and B.

Le Châtelier’s Principle

How will equilirbium be affected if we change volume/pressure? Recall the Ideal Gas Law PV = nRT What is the relationship between P and n? There is a direct relationship between pressure and number of moles (or molecules).

Le Châtelier’s Principle

How will equilibrium be affected if we increase pressure? N2 + 3H2 2NH3

• If the volume is decreased, it will increase the total pressure of

the system.

• So the system will try to reduce the pressure by producing fewer

moles of gas.

• The reaction will favor producing fewer moles of gas.
• The reactant side has 4 moles of gas and the product side only 2
• moles. Therefore, equilibrium shifts to the right, in this case, to

reduce pressure. Stress: Increasing pressure +/or reducing volume Effect: Equilibrium shifts to the right ­ there are fewer moles of gas

Le Châtelier’s Principle

How will equilibrium be affected if we decrease pressure? N2 + 3H2 2NH3

• If the volume is increased, it will reduce the pressure of the

system

• So the system will try to increase the pressure by producing more

molecules or moles of gas.

• In this case, equilibrium will shift to the left, where there is a

greater number of mole of gas. Stress: Decreasing pressure/ increasing volume Effect: Equilibrium shifts to the left ­ there are more moles of gas

equilibrium­presentation­2012­02­27.notebook 17 November 09, 2012

37 How does an increase in pressure affect the following reaction at equilibrium? C2H2 (g) + H2 (g) C2H4 (g)

A

Equilibrium shifts to the right; the reaction makes more products.

B

Equilibrium shifts to the left; the reaction makes more reactants. C The reaction makes more of both products and reactants, so equilibrium is unaffected.

38 Which of the changes listed below would shift the equilibrium of this reaction to the right?

4HCl (g) + O2 (g) 2Cl2 (g) + 2H2O (g) A Addition of Cl2 gas.

B

Removal of O2 gas. C Increase in pressure.

Le Châtelier’s Principle

If the temperature of the system is increased:

• The system should accept the thermal energy supplied to favor

the endothermic reaction, and favor the forward reaction.

• Equilibrium will shift to the right.
• [PCl3] and [Cl2] increase while [PCl5] decreases.
• Therefore, the value of K will increase.

How will equilibrium be affected if we change the temperature? Consider this endothermic decomposition reaction.

PCl5 PCl3 + Cl2 ΔH = 88 KJ

Le Châtelier’s Principle

If the temperature of the system is lowered:

• The system should respond to the lowering of thermal energy

by releasing heat to surroundings in an exothermic reaction. Therefore, the reverse reaction is favored because it is exothermic.

• Equilibrium will shift to the left.
• [PCl5] increases while [PCl3] and [Cl2] decrease.
• Therefore, the value of K will decrease.

PCl5 PCl3 + Cl2 ΔH = 88 KJ If we change the temperature: At low temp, the purple hexa aqua species if favored. At high temp, the blue tetra chloro species is favored.

Le Châtelier’s Principle

at 1000C at 250C

Many ions containing transition metals produce colored solutions. In the reaction above, the cation, Co(H2O)6

2+, yields a purple solution,

while the anion CoCl4

2­ produces a blue solution.

purple blue

Co(H2O)6

2+(aq) + 4Cl(aq) CoCl4 2­(aq) + 6H2O(l)

Do you think the reaction above is endothermic or exothermic?

purple blue

Co(H2O)6

2+(aq) + 4Cl(aq) CoCl4 2­(aq) + 6H2O(l) + heat

exothermic reaction

purple blue

heat + Co(H2O)6

2+(aq) + 4Cl(aq) CoCl4 2­(aq) + 6H2O(l)

Which reaction below do you think correctly represents the

• bservation of these colors?

at 1000C at 250C

Le Châtelier’s Principle

endothermic reaction

• r

equilibrium­presentation­2012­02­27.notebook 18 November 09, 2012

39 For the reaction below, what is the effect of raising the temperature?

4A (g) + B (g) 3C ΔH = ­ 405 kJ A

Equilibrium shifts to the right; the reaction makes more products.

B

Equilibrium shifts to the left; the reaction makes more reactants. C The reaction makes more of both products and reactants, so equilibrium is unaffected.

40 For the reaction below at equilibrium when the temperature is raised, the value of K will:

4A (g) + B (g) 3C ΔH = ­ 405 kJ A

increase

B

decrease C remain constant

D

Not enough information.

41 For the reaction below, what is the effect of raising the temperature?

X 2Y + Z ΔH = + 255 kJ A

Equilibrium shifts to the right and K will increase.

B

Equilibrium shifts to the left and K will decrease. C Equilibrium is unaffected and K stays constant. Significance of K value Equilibrium lies to the right (product­

favored) or lies to the left (reactant favored). Comparing Q and K The reaction must proceed to the right or proceed to the left in order to achieve equilbrium. Le Chatelier's Principle Equilibrium shifts to the left or shifts to the right, in order to re­establish equilibrium.

Comparison of Equilibrium Terminology

Different terminology is required, depending on whether a system is: a) at equilibrium b) trying to reach equilibrium c) was at equilibrium but had its equilibrium disturbed