ap e unit 7 - equilibrium

AP Chemistry

EquilibriumISPS Chemistry Jan/Feb 2021 page 1

Unit 7 - Equilibrium

7.1 Introduction to Equilibrium 7.2 Direction of Reversible Reactions 7.3 Reaction Quotient and Equilibrium Constant 7.4 Calculating the Equilibrium Constant 7.5 Magnitude of the Equilibrium Constant 7.6 Properties of the Equilibrium Constant 7.7 Calculating Equilibrium Concentrations 7.8 Representations of Equilibrium 7.9 Introduction to Le Châtelier’s Principle 7.10 Reaction Quotient & Le Châtelier’s Principle 7.11 Introduction to Solubility Equilibria 7.12 Common-Ion Effect 7.13 pH and Solubility 7.14 Free Energy of Dissolution

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7.7 Calculating Equilibrium Concentrations

We have a number of tools we can use to calculate, or predict, concentrations or partial pressures at equilibrium. Most, if not all, have already been met and used.

Equilibrium Constant: K = [products] / [reactants] and

K > 1 means [products] > [reactants] and K < 1 means [reactants] > [products]

Reaction Quotient:

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Stoichiometry: Though chemicals will not necessarily be mixed in stoichiometric proportions initially, reactions will still obey the stoichiometrical relationships.

For example, 2 H2S(g) + CH4(g) ⇋ CS2(g) + 4 H2(g)

does not mean that your mixture will have ratios = 2 : 1 : 1 : 4 for H2S(g) : CH4(g) : CS2(g) : H2(g)

However, if CH4(g) decreases by x then H2S(g) decreases by 2x and CS2(g) increases by x and H2(g) increases by 4x

A common procedure for solving equilibrium concentrations is sometimes referred to as the ICE method, where the acronym stands for Initial, Change, and Equilibrium.

Example 1: The following gases were placed in a 4.00 L flask: 8.00 mol N2 and 10.00 mol H2.

After equilibrium was achieved, 1.20 M NH3 was found in the flask. Complete the ICE table below and determine the equilibrium constant, Keq.

K = [NH3]2 = ( 1.20 )2 = 1.44 = 3.00

[N2] [H2]3 (1.40)(0.70)3 0.480

010/4 = 2.50 M8/4 = 2.0 M

+ 1.20 M- 3x- x

1.20 M2.50 - 1.80 = 0.70

2.00 - 0.60 = 1.40

+ 2x

x = 0.60 M

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Example 2: A mixture of 0.500 mol H2 and 0.500 mol I2 was placed in a 1.00-L stainless-steel flask at 430 °C. The equilibrium constant Kc for the reaction

H2(g) + I2(g) ⇋ 2HI(g) is 54.3 at this temperature.

Calculate the concentrations of H2, I2, and HI at equilibrium.

H2(g) + I2(g) ⇋ 2HI(g)

Initial 0.500 0.500 0.000

Change - x - x + 2x

Equilibrium 0.500 - x 0.500 - x + 2x

K = [HI]2 = ( 2x )2 = 54.3

[H2] [I2] (0.500 - x)(0.500 - x)

taking √ of both sides 2x = 7.37

(0.500 - x)

rearranging 2x = 7.37 (0.500 - x) = 3.69 - 7.37x

9.37x = 3.69

x = 3.69/9.37 = 0.394 M

so [H2] = 0.500 - 0.394 = 0.106 M

[I2] = 0.500 - 0.394 = 0.106 M

[HI] = 2 x 0.394 = 0.788 M

This particular example 'solved easier' due to the fact that the x2 term could be easily dealt with.

On occasions, it might be necessary to rearrange into the quadratic equation:

and solve using the quadratic formula:

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Temperature & Pressure: Changes in temperature and/or pressure can also be used to indicate which direction an equilibrium is shifting before applying the ICE method.Remember, Increasing temperature will favour the endothermic reaction

Decreasing temperature will favour the exothermic reaction

Increasing pressure will favour the reaction that leads to a decrease in gas volumes

Decreasing pressure will favour the reaction that leads to an increase in gas volumes

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7.7 Practice Problems1.

A 0.10 mol sample of each of the four species in the reaction represented above is injected into a rigid, previously evacuated 1.0 L container. Which of the following species will have the highest concentration when the system reaches equilibrium?

A H2S(g) B CH4(g) C CS2(g) D H2(g)

2. 2 XY(g) ⇋ X2(g) + Y2(g) KP = 230

A certain gas, XY(g), decomposes as represented by the equation above. A sample of each of the three gases is put in a previously evacuated container. The initial partial pressures of the gases are shown in the table opposite.

The temperature of the reaction mixture is held constant. In which direction will the reaction proceed?

A The reaction will form more products.

B The reaction will form more reactant.

C The mixture is at equilibrium, so there will be no change.

D It cannot be determined unless the volume of the container is known.

3. 4 HCl(g) + O2(g) ⇋ 2 Cl2(g) + 2 H2O(g)

Equal numbers of moles of HCl and O2 in a closed system are allowed to reach equilibrium as represented by the equation above. Which of the following must be true at equilibrium?

I. [HCI] must be less than [Cl2]. II. [O2] must be greater than [HCl].

III. [Cl2] must equal [H2O].

A I only B II only C I & III only

D II & III only E I, II & III only

O

O

O

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4. CO(g) + 2 H2(g) ⇋ CH3OH(g) ΔH < 0

The synthesis of CH3OH(g) from CO(g) and H2(g) is represented by the equation above. The value of Kc for the reaction at 483 K is 14.5.

A 1.0 mol sample of CO(g) and a 1.0 mol sample of H2(g) are pumped into a rigid, previously evacuated 2.0 L reaction vessel at 483 K. Which of the following is true at equilibrium?

A [H2] = 2[CO] B [H2] < [CO]

C [CO] = [CH3OH] < [H2] D [CO] = [CH3OH] = [H2]

5. H2(g) + I2(g) ⇋ 2 HI(g)

At 450 °C, 2.0 moles each of H2(g), I2(g), and HI(g) are combined in a 1.0 L rigid container.

The value of Kc at 450 °C is 50. Which of the following will occur as the system moves toward equilibrium?

A More H2(g) and I2(g) will form.

B More HI(g) will form.

C The total pressure will decrease.

D No net reaction will occur, because the number of molecules is the same on both sides of the equation.

6. PCl3(g) + Cl2(g) ⇋ PCl5(g) Kc = 6.5 At a certain point in time, a 1.00 L rigid reaction vessel contains 1.5 mol of PCl3(g), 1.0 mol of Cl2(g), and 2.5 mol of PCl5(g). Which of the following describes how the measured pressure in the reaction vessel will change and why it will change that way as the reaction system approaches equilibrium at constant temperature?

A The pressure will increase because Q < Kc.

B The pressure will increase because Q > Kc.

C The pressure will decrease because Q < Kc.

D The pressure will decrease because Q > Kc.

O

O

O

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7. The compound butane, C4H10 , occurs in two isomeric forms, n-butane and isobutane (2-methyl propane). Both compounds exist as gases at 25°C and 1.0 atm.

a) Draw the structural formula of each of the isomers (include all atoms). Clearly label each structure.

b) On the basis of molecular structure, identify the isomer that has the higher boiling point. Justify your answer.

The isomer n-butane has the higher boiling point. London (dispersion) forces are greater among molecules of n-butane than they are among molecules of isobutane because molecules of n-butane, with its linear structure, can approach one another more closely and can form a greater number of induced temporary dipoles than molecules of isobutane, with its more compact structure, can form.

The two isomers exist in equilibrium as represented by the equation below.

n-butane(g) ⇋ isobutane(g) Kc = 2.5 at 25°C

Suppose that a 0.010 mol sample of pure n-butane is placed in an evacuated 1.0 L rigid container at 25°C.

c) Write the expression for the equilibrium constant, Kc , for the reaction.

Kc = [isobutane] / [n-butane]

d) Calculate the initial pressure in the container when the n-butane is first introduced (before the reaction starts).

PV = nRT so P = nRT / V

P = (0.010 mol)(0.0821 L atm mol-1 K-1)(273 + 25 K) / (1.0 L) = 0.24 atm

e) The n-butane reacts until equilibrium has been established at 25°C .

i) Calculate the total pressure in the container at equilibrium. Justify your answer.

The total pressure in the container remains the same, 0.24 atm. As the reaction proceeds, the number of molecules in the container remains constant; one molecule of isobutane is produced for each molecule of n-butane consumed.

H H H H | | | |H — C — C — C — C — H | | | | H H H H

n-butane

H CH3 H | | | H — C — C — C — H | | | H H H

2-methylpropane

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Q7 e) contd

ii) Calculate the molar concentration of each species at equilibrium.

1 point is earned for the correct setup. 1 point is earned for both correct numerical answers.

n-butane(g) isobutane(g)

I 0.010 M 0 MC -x +xE 0.010 - x x

K = [isobutane] / [n-butane] = 2.5 x / (0.010 -x ) = 2.5

so x = 0.025 - 2.5 x 3.5 x = 0.025 x = 0.025 / 3.5 = 0.007

isobutane = x = 0.007 M n-butane = 0.010 - x = 0.003 M

iii) If the volume of the system is reduced to half of its original volume, what will be the new concentration of n-butane after equilibrium has been reestablished at 25°C ? Justify your answer.

1 point is earned for the correct answer with justification.

Halving the volume of the container at equilibrium doubles the pressure of both isobutane and n-butane, which has no effect on the equilibrium because the stoichiometry of the reaction is one mole of product produced for each mole of reactant consumed.

Since the number of moles of each isomer is unchanged but the volume is reduced by half, concentrations of both isomers are doubled and the concentration of n-butane will be 2 x 0.003 M = 0.006 M.

Suppose that in another experiment a 0.010 mol sample of pure isobutane is placed in an evacuated 1.0 L rigid container and allowed to come to equilibrium at 25°C .

f) Calculate the molar concentration of each species after equilibrium has been established.

1 point is earned for correct numerical answers or a correct statement regarding their equivalence to values obtained in part e) ii).

The concentrations of isobutane and n-butane would be the same as they were calculated in part e) ii), 0.007 M and 0.003 M, respectively.

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8. C(s) + CO2(g) ⇄ 2 CO(g)

Solid carbon and carbon dioxide gas at 1,160 K were placed in a rigid 2.00 L container, and the reaction represented above occurred.

As the reaction proceeded, the total pressure in the container was monitored. When equilibrium was reached, there was still some C(s) remaining in the container.

Results are recorded in the table opposite.

a) Write the expression for the equilibrium constant, Kp , for the reaction.

KP = (PCO)2 / (PCO2

)

b) Calculate the number of moles of CO2(g) initially placed in the container. (Assume that the volume of the solid carbon is negligible.)

PV = nRT so n = PV / VRT

n = (5.00 atm)(2.00 L) / (0.0821 L atm mol-1 K-1) (1160 K) = 0.105 mol

c) For the reaction mixture at equilibrium at 1,160 K, the partial pressure of the CO2(g) is 1.63 atm. Calculate

i) the partial pressure of CO(g) , and

Ptotal = PCO + PCO2 so PCO = Ptotal - PCO2 = 8.37 - 1.63 = 6.74 atm

ii) the value of the equilibrium constant, Kp .

KP = (PCO)2 / (PCO2

) so KP = (6.74)2 / (1.63

) = 27.9

d) If a suitable solid catalyst were placed in the reaction vessel, would the final total pressure of the gases at equilibrium be greater than, less than, or equal to the final total pressure of the gases at equilibrium without the catalyst?

Justify your answer. (Assume that the volume of the solid catalyst is negligible.)


The total pressure of the gases at equilibrium with a catalyst present would be equal to the total pressure of the gases without a catalyst.

Although a catalyst would cause the system to reach the same equilibrium state more quickly, it would not affect the extent of the reaction, which is determined by the value of the equilibrium constant, Kp.

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Q8 contd.

In another experiment involving the same reaction, a rigid 2.00 L container initially contains 10.0 g of C(s), plus CO(g) and CO2(g), each at a partial pressure of 2.00 atm at 1,160 K.

e) Predict whether the partial pressure of CO2(g) will increase, decrease, or remain the same as this system approaches equilibrium.

Justify your prediction with a calculation.

1 point is earned for a correct calculation involving Q or ICE calculation. 1 point is earned for a correct conclusion based on the calculation.

Q = (PCO)2 / (PCO2

) = (2.00)2 / (2.00) = 2.00 (but KP = 27.9)

Q < KP so reactants → products as reaction approaches equilibrium

so PCO2 will decrease

9. CaSO4 . 2 H2O(s) ⇄ CaSO4(s) + 2 H2O(g)

The hydrate CaSO4 . 2 H2O(s) can be heated to form the anhydrous salt, CaSO4(s), as shown by the reaction represented above.

a) Write the expression for the equilibrium constant, Kp, for the reaction.

KP = (PH2O )2

b) Given that the equilibrium constant, Kp , is 6.4 x 10–4 at 298 K, determine the partial pressure, in atm, of water vapor in the cylinder at equilibrium at 298 K.

KP = 6.4 x 10-4 = (PH2O )2 so PH2O

= √(6.4 x 10-4) = 0.025 atm

c) If the volume of the system is reduced to one-half of its original volume and the system is allowed to reestablish equilibrium at 298 K, what will be the pressure, in atm, of the water vapor at the new volume? Justify your answer.


The PH2O at equilibrium at the new volume will be 0.025 atm. Equilibrium vapor pressure is dependent on Kp, which in turn is a function of temperature, not volume.

Because the temperature is still 298 K, the vapor pressure of H2O remains 0.025 atm in the new volume.

ap e unit 7 - equilibrium

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