Types of Gaseous Equilibrium Problems

Types of Gaseous Equilibrium Problems

Solution Set

1. Given the molarity (or info to find molarity) of reactants & products, find Kc.

2. Given the partial pressure (or info to find partial pressure) of reactants & products, find Kp.

3. Given Kc and all concentration except for one, solve for remaining concentration

4. Given initial concentrations, overall equation, and one equilibrium conc., solve for remaining equilibrium conc. and then Kc

5. Given equilibrium concentration, solve for percent ionization.

6. Given initial conc. and percent ionization, solve for Kc

7. Small % ionization, IRE problem, ignoring x

8. Perfect square problem

9. Given equilibrium concentrations, make a conc. change, solve for new equilibrium conc.

10. Calculating Q and comparing to K.

Examples:

1A. Consider the following reaction: 2 A + 3 B <====> 4C (all gases @35.0oC and 1 atm.)

What is the value of Kc if [A] = 0.350 M, [B] = 1.50 M, and [C] = 4.00 M when the system reaches equilibrium? The reaction vessel is a 5.00 liter container.

[C]4 [4.00]4 256

KC = ------= ------= ------= 619 M-1

[A]2[B]3 [0.350]2[1.50]3 0.413

{Note #1: volume of the container does not matter since molarity is given.}

{Note #2: although the equilibrium has a unit, because it does not have any significance,

it is usually omitted.}

1B. Consider the following reaction: 2 A + 3 B <====> 4C (all gases @35.0oC and 1 atm.)

What is the value of Kc if there are 4.00 moles of C, 1.50 moles of B and 0.350 moles of A when the system reaches equilibrium? The reaction vessel is a 5.00 liter container.

[A] = 0.350 mol / 5.00 liter = 0.0700 M

[B] = 1.50 mol / 5.00 liter = 0.300 M

[C] = 4.00 mol / 5.00 liter = 0.800 M

[0.800]4 0.410

KC = ------= ------= 3.10 x 103 M-1

[0.0700]2[0.300]3 1.32 x 10-4

1C. Consider the following reaction: A + 3 B <====> 4C (all gases @35.0oC and 1 atm.)

What is the value of Kc if there are 4.00 moles of C, 1.50 moles of B and 0.350 moles of A when the system reaches equilibrium? The reaction vessel is a 5.00 liter container.

Since Dn = 0, it doesn’t matter whether we use moles or molarity in the expression.

[C]4 [4.00]4 256

KC = ------= ------= ------= 217

[A][B]3 [0.350][1.50]3 1.18

2A. Consider the following reaction: 2 A + B <====> C (all gases @35.0oC)

What is the value of KP if PA = 3.50 atm., PB = 1.50 atm., and PC = 8.50 atm. when the system reaches equilibrium? The reaction vessel is a 10.0 liter container.

PC 8.50 8.50

KP = ------= ------= ------= 0.463 atm-2

PA2 . PB (3.50)2(1.50) 18.4

2B. Consider the following reaction: 2 A + B <====> C (all gases @35.0oC)

What is the value of Kp if there are 4.00 moles of C, 1.50 moles of B and 0.350 moles of A when the system reaches equilibrium? The reaction vessel is a 6.00-liter container.

Can be solved either of two ways: find the partial pressures, then KP

ntotal = 5.85 moles

Ptotal = nRT / V = [(5.85 mol)(0.0821)(308 K)] / (6.00 liter) = 24.7 atm.

PA = (24.7)(0.350/5.85) = 1.48 atm PB = (24.7)(1.50/5.85) = 6.33 atm. PC = 16.9 atm.

PC 16.9 16.9

KP = ------= ------= ------= 1.22 atm-2

PA2 . PB (1.48)2(6.33) 13.9

alternate method: find the molarities, then KC, then KP

[A] = 0.350 mol / 6.00 liters = 0.0583 M; [B] = 0.250 M; [C] = 0.667 M

[C] [0.667] 0.667

KC = ------= ------= ------= 785 M-2

[A]2[B] [0.0583]2[0.250] 8.50 x 10-4

KP = KC(RT)Dn = 785[(0.0821)(308)]-2 = 1.22

3. Consider the following reaction: 2 A + 2 B <====> C (all gases @25.0oC)

If the value of Kc = 45.0, and [A] = 0.350 M, [B] = 1.50 M, what is [C] when the system reaches equilibrium? The reaction vessel is a 5.00 liter container.

[C] [C]

KC = ------=> 45.0 = ------= > [C] = 12.4 M

[A] 2[B]2 [0.350] 2[1.50]2

4. Consider the following reaction: A + 2 B <====> 2 C + 3D (all gases @25.0oC)

Initially [A] = 1.50 M and [B] = 1.50 M (before the reaction begins). Once equilibrium is reached the concentration of D is found to be 1.26 M.

A. What are the equilibrium concentrations of the other reagents?

B. What is the value of Kc? C. What is the value of Kp?

A + 2 B <====> 2 C + 3D

I 1.50 1.50 0.0 0.0

R (-0.42) (-0.84) (+0.84) (+1.26)

A. = Eq = 1.08 0.66 0.84 1.26

B. = KC = [(0.84)2(1.26)3]/[(1.08)(0.66)2] = 1.41 / 0.470 = 3.00 M2

C. KP = KC(RT)Dn = 3.00[(0.0821)(298)]2 = 1.80 x 103

5. Consider the following reaction: 2 A <====> 2B + C (all gases @25.0oC and 1 atm.)

Initially [A] = 5.00 M (before the reaction begins). Once the decomposition equilibrium is reached [C] = 0.200 M. What is the percent ionization of A?

2 A <====> 2B + C

I 5.00 0.0 0.0

R (-0.400) (+0.400) (+0.200)

Eq 4.60 0.400 0.200

% ionization A = (0.400 / 5.00) x 100 = 8.00 %

6. Consider the following reaction: A <====> B + 2 C (all gases @25.0oC and 1 atm.)

Initially [A] = 5.00 M (before the reaction begins). It is known that A only decomposes 12.0% Once the decomposition equilibrium is reached, what is the value of Kc?

A <====> B + 2 C

I 5.00 0.0 0.0

R -[(0.120)(5.00]

(-0.600) (+0.600) (+1.20)

Eq 4.40 0.600 1.20

[1.20]2[0.600]

KC = ------= 0.196

[4.40]

7. Consider the following reaction: A <====> B + C (all gases @25.0oC and 1 atm.)

Initially [A] = 8.00 M (before the reaction begins). What are the equilibrium concentrations of all substances once the decomposition equilibrium is reached if the value of Kc = 1.20 x 10-3?

A <====> B + C

I 8.00 0.0 0.0

R - x +x +x

Eq 8.00 - x x x

8.00

x2

KC = ------= 1.20 x 10-3 => x = 0.0980

8.00

x = [B] = [C] = 0.0980 M

[A] = 8.00 - x = 7.90 M

Discuss the importance of the magnitude of “x” vs. the magnitude of K.

8. Consider the following reaction: 2 HI <====> H2 + I2 (all gases @550oC and 1 atm.)

Initially [HI] = 8.00 M (before the reaction begins). What are the equilibrium concentrations of all substances once the decomposition equilibrium is reached if the value of KC is 28.8?

What is the value for KP for this reaction at the same conditions?

2HI <====> H2 + I2

I 8.00 0.0 0.0

R - 2 x +x +x

Eq 8.00 - 2x x x

x2 x

KC = ------= 28.8 => ------= 5.37 => x = 3.66 M = [H2] = [I2]

[8.00 - 2x]2 8.00 - 2x

[HI] = 8.00 - 2x = 0.681 M Since Dn = 0 KP = KC

9. Consider the following reaction: A + 2 B <====> 2 C + 3D (all gases @25.0oC and 1 atm.)

Initial equilibrium values: [A] = 2.50 M, [B] = 3.50 M, [C] = 5.00 M, and [D] = 6.00 M. Moles of B are removed until the new equilibrium concentration of D is 5.25 M.

A. What was the value of Kc? KC = [5.00]2[6.00]3 / [2.50][3.50]2 = 176

B. How many moles of B were removed if the reaction took place in a 250.0-ml flask?

A + 2 B <====> 2 C + 3D

I 2.50 3.50 - x 5.00 6.00

R (+0.25) (+0.50) (-0.50) (-0.75)

Eq 2.75 4.00 - x 4.50 5.25

KC = [4.50]2[5.25]3 / [2.75][4.00-x]2 = 176 x = 1.54 M

moles B = 0.250 liter x 1.54 M = 0.384 moles

10. Consider the following reaction: 2 A + B <====> 2 C + 3D (all gases @25.0oC)

The value of Kc = 50.0. A reaction mixture is analyzed and found to contain the following concentrations: [A] = 1.35 M, [B] = 1.50 M, [C] = 2.00, and [D] = 5.00.

Determine whether or not the system is at equilibrium.

If the system is not at equilibrium, which way is the system moving - to the left or to the right?

Q = [2.00]2[5.00]3 / [1.35] 2[1.50] = 183 Q > K not at equilibrium

Rxn mixture shifts left