15.2 solving equilibrium problems

September 201 Solving Equilibrium Problems

15.2 Solving Equilibrium ProblemsThe iDe Process

Dr. Fred Omega GarcesChemistry 201Miramar College


Solving Equilibrium Problems

There are many type of equilibrium problems which arise in the real world and in chemistry exams. These problems can be broken down to two basic types.

1. The first type is one in which the equilibrium concentrations are given, and the equilibrium constant must be solved.

Given: [Conc] at equilb Solve: Keq

2. The second type is one in which both the equilibrium constant and the initial concentration are given and the concentrations at equilibrium are solved for.

Given: [Conc]o and Keq Solve: [Conc] at equilb


Type 1: Equilibrium

The first type of problem is the type in which the equilibrium concentrations

are given and the equilibrium constant, Keq, must be solved.

Given: [Conc] at equilibrium Solve: Keq

Steps involved in the equilibrium calculations.

1. Write the balanced equation for the reaction.

2. Write the equilibrium expression using the Mass Action Expression.

3. Plug the equilibrium concentration into the Mass Action and solve for the

equilibrium constant, Keq .


1. Calculating Keq: Kc - Kp relationship (Type1)

Carbon dioxide is placed in a 5.00-L high-pressure reaction vessel at 1400.°C. Given the following reaction, CO (g) + 1/2 O2 (g) D CO2 (g) , calculate the equilibrium constant Kc and Kp after equilibrium is achieved with 4.95 mol of CO2 , 0.0500 mol of CO, and 0.0250 mol of O2 found in the container.

€

T = 1400°C CO(g) + 12 O2 (g) ! CO2 (g)

V = 5.0 L i Δ mol e 0.050 mol 0.025mol 4.95molVol = 5.0 L e[ ] 1.00 ⋅10-2M 5.00 ⋅10-3M 9.90 ⋅10-1M

Supplemental: Calculate the partial pressure of each gas at equilibrium


1. Calculating Keq: Kc - Kp relationship (Type1)

Carbon dioxide is placed in a 5.00-L high-pressure reaction vessel at 1400.°C. Given the following reaction, CO (g) + 1/2 O2 (g) D CO2 (g) , calculate the equilibrium constant Kc and Kp after equilibrium is achieved with 4.95 mol of CO2 , 0.0500 mol of CO, and 0.0250 mol of O2 found in the container.

€

T = 1400°C CO(g) + 12 O2 (g) ! CO2 (g)

V = 5.0 L i Δ mol e 0.050 mol 0.025mol 4.95molVol = 5.0 L e[ ] 1.00 ⋅10-2M 5.00 ⋅10-3M 9.90 ⋅10-1M

€

kc =CO2[ ]

CO[ ]∗ O2[ ]1/2 =9.90 ⋅10-1M[ ]

1.00 ⋅10-2 M[ ]∗ 5.00 ⋅10-3 M[ ]1/2

kc =1.40 •103

€

kp = kcRT+Δn Δn = 1 -1.5 = -0.5

kp = 1.40 • 103 • RT+ −.5( ) = 1.40 • 103 • 0.08206 ⋅1673( )−.5

kp = 1.20 ⋅102


CO2 (g) + H

2 (g) ! CO

(g) + H

2O

(g)

i 0.1000M 0.0500 0 0.1000V = 1.0 L Δ -X -X +X +X e!"

#$ 0.0954 ? ? ?

2. Calculating Keq iCe - method (Type1 & 2)

€

0.1000 - x = .0954x =.0046

€

Kc = 0.1046M[ ]∗ 0.0046M[ ]0.0454M[ ]∗ 0.0954M[ ]

= 0.11

€

Kp =Kc RT( )0= 0.11

€

e[ ] 0.0954 0.0454 0.0046 0.1046

Temp isn’t known, but Dn = 0What is the total pressure in the vessel at equilibrium?

A mixture of 0.1000 mol of CO2 , 0.05000 mol of H2, and 0.1000 mol of H2O is placed in a 1.000 L vessel. The following equilibrium reaction is established:

CO2 (g) + H2 (g) D CO(g)+ H2O (g)

At equilibrium [CO2] = 0.0954 M. (a) Calculate the equilibrium concentrations of H2, CO, and H2O. (b) Calculate the Kc for the reaction (c) Calculate Kp for the reaction?


Type 2: EquilibriumThe second type of problems provides information on both the equilibrium constant and the

initial concentration. The concentrations at equilibrium must be solved for.

Given: [Conc]o and Keq Solve: [Conc] at equilb

Steps involve in the equilibrium calculations.

1. Write the balanced equation for the reaction.2. Write the equilibrium expression using the Mass Action Expression.

3. List the initial conditions.

4. Set up the iDe table and solve for the equilibrium concentration in terms of a variable (x).

5. Plug the equations expressing the equilibrium concentration into the mass action expression and solve for x.

6. Assign the value of x to the equilibrium concentration equation and determine the numerical value

for the equilibrium concentration for each specie in the reaction.


3a. Calculation: Product formed @ Equilibrium (Type2)How much SO3 is formed when 0.50 mol of SO2 and 0.50 mol NO2 are in 1.00 L at 821 °C. Kc = 6.85 at this temperature.

€

SO2(g) + NO2(g) ! SO3 (g) + NO(g)

i 0.50 M 0.50M 0.00 0.00Δ −X −X + X +Xe[ ] 0.50 - X 0.50 - X +X +X

€

Kc = SO3[ ]∗ NO[ ]SO2[ ]∗ NO2[ ]

= 6.85 = X[ ]∗ X[ ]0.50 - X[ ]∗ 0.50 - X[ ]

€

X[ ]2

0.50 - X[ ]2 = 6.85 X[ ]0.50 - X[ ]

= 2.617Perfect square type problem

X = 1.3 - 2.617X, 3.617X = 1.3, X = 0.36M = [SO3] = [NO]


3b. Calculation: Product formed @ Equilibrium (Type2)

How much SO3 is formed when 0.50 mol of SO2 and 1.00 mol NO2 are in 1.00 L at 821 °C. Kc = 6.85 at this temperature. (Non-perfect square)

€

SO2(g) + NO2(g) ! SO3 (g) + NO(g)

i 0.50 M 1.00M 0.00 0.00Δ −X −X + X +Xe[ ] 0.50 - X 1.00 - X +X +X

€

Kc = SO3[ ]∗ NO[ ]SO2[ ]∗ NO2[ ]

= 6.85 = X[ ]∗ X[ ]0.50 - X[ ]∗ 1.00 - X[ ]

€

X[ ]2

0.50 - X[ ] • 1.00 - X[ ]= 6.85

€

X[ ]2

0.50 -1.50X +X2[ ]= 6.85

€

5.85X2 − 10.28X + 3.43 = 0a = 5.85 b = −10.28 c = 3.43a =1 b = -1.757 c = 0.586

€

X = 10.28 ± 10.282 - 4(3.43)(5.85)2(5.85)

€

X = 10.28 ± 5.045.85

=1.309M or 0.447M

Non-perfect square problem. Solve quadratic

€

X = 0.447MOnly answer which makes sense chemically

€

Quadratic Equation :

x = -b ± b2 −4ac2a


4. Calculating Keq: Method of Approximation (Type2)

NOCl decomposes to form the gases NO and Cl2. At 35°C the equilibrium constant is 1.6•10-5. If

1.0 mol of NOCl is placed in a 2.0 L flask what are the concentration of all specie at equilibrium?

€

2NOCl ! 2NO + Cl2 Kc = 1.6 ⋅10-5

i 0.50 M 0 0V = 2.0 L Δ −2X + 2X + X e[ ] 0.5 -2X +2X +X

€

Kc = 1.6 ⋅10-5 = (2X)2 ∗ X(0.5 -2X)2

€

Assume 0.5 -2X ~ 0.5since Kc =1.6 ⋅10-5

€

1.6 ⋅10-5 = 4X2 ∗ X(0.5)2 ,

€

4X3 = 4 ⋅10-6, X =1 ⋅10-2

€

Check assumption : 0.5 -2X = 0.5 -2(1 ⋅10-2) = 0.48 ~ 0.50


Assumption CheckWhen making assumptions, if a reaction has a relatively small keq and a relatively large initial reactant concentration, then the concentration change (x) can often be neglected without introducing significant error. This does not mean x = 0, because then this would mean there is no reaction. It means that if a reaction proceeds very little (small k) and if you start with a high reactant concentration, very little will be used up, so the following holds.

[react]o -x ≈ [react]eq ≈ [react]o

When making the assumption that x is negligible, you must check that the error introduced is not significant. If the assumption results in a change (error) in concentration of less than 5%, the error is not significant and the assumption is justified.

To test the assumption, use the following formula:

D conc / initial concentration) • 100 < 5%

In the previous problem, the assumption is check by the following calculations:

[2x / (0.5) ] • 100 = [ 2(1•10-2) / 0.5 ]•100 = 4% < 5% The assumption is valid in this example.


Assumption Check [React]initial / keq

In general, simplifying assumptions works when the initial concentration of the reactant is high but not when it is low. To summarize, we assume that x (D [React]) can be neglected if Keq is small relative to [React]initial. or [React]initial / keq = ?.

The benchmark in justifying the assumption is-

What is the threshold for

[React]initial / keq so that the assumption is justified.

If [React]initial / keq > 400, the assumption is justified; neglecting x introduces an error of < 5%

If [React]initial / keq < 400, the assumption is not justified; neglecting x introduces an error of > 5%


5a. Calculating Keq: Quadratic or iteration?Hydrogen a potential fuel is found in abundance in water. It must first be separated from oxygen, however. One possibility is thermal decomposition, but this requires very high temperature. Even at 1000°C, Kc = 7.3•10-18 for the reaction below. What is the hydrogen concentration for 0.100M water?

2H2O

(g) ! 2H

2(g) + O

2(g)

i 0.100M 0 0Δ -2x +2x +x[e] 0.100-2x 2x x

approximate

Kc = 7.3•10-18

= (2x )2 ⋅x(0.100 −2x )2

≈(2x )2 ⋅x(0.100)2

≈4x 3

0.0100 7.3•10-18(0.0100) = 4x 3 , x = 2.6•10-7

Assumption Check:

Δ

[Rct]o

⋅100 = 2x[0.100]

⋅100= 2(2.6•10-7)[0.100]

⋅100 =2(2.6•10-7)

[0.100]⋅100 = 0.00052%

[H2] = 2x = 5.2•10-7 M


5. Calculating Keq: Quadratic (Type2)

At 250 ° C, the reaction PCl5 (g) D PCl3 (g) + Cl2 (g) has an equilibrium constant Kc = 1.80. If 0.100 mol PCl5 is added to a 5.00-L vessel, what are the concentrations of all specie at equilibrium ? What % of the reactant remains?

€

PCl5 (g) ! PCl3 (g) + Cl2(g)

i 0.02 M 0 0Δ −X + X + Xe[ ] 0.02- X +X +X

€

X = -1.80 ± 1.802 + 4(0.036)2(1)

X = - 1.80 ± 1.839562(1)

= .0198M

€

1.80 (0.02- X) = X2

0.036 - 1.80 X = X2

X2 +1.80 X - 0.036 = 0a X2 + bX + c = 0

€

Kc =1.80 =PCl3[ ]∗ Cl2[ ]

PCl5[ ]= X[ ]2

0.2- X[ ]

€

PCl3[ ] = Cl2[ ] = 0.0198 M

[PCl5]eq = 0.0200-0.0198 = 2•10-4M

% [PCl5] = (2•10-4/0.02) * 100 = 1.00%for this initial condition, the reaction essentially goes to 99% completion.


6. Calculation: Product formed @ Equilibrium (Type2)Exactly 4 mol of SO3 is sealed in a 5.0 L container @1500K. Kc is9.34•103 for the rxn, what are the conc. of all specie at equilb?


•103

6. Calculation: Product formed @ Equilibrium (Type2)Exactly 4 mol of SO3 is sealed in a 5.0 L container @1500K. Kc is 9.34•103 for the rxn, what are the conc. of all specie at equilb? (use as many significant figures as calculator allow then round off your final answer.)

€

2SO3(g) ! 2 SO2(g) + O2(g)

i 0.80 M 0.00 0.00Δ −2x +2x +xe[ ] 0.80 -2x +2x +x

€

4x3" # $

% & '

0.80 -2x[ ]2≈

4 × 0.43" # $

% & '

0.80 -2x[ ]2

= 9.34

€

5.235810-3 = 0.80 −2x2x = 0.80 −5.235810-3M = 0.794765 x = 0.3974 = 0.40

€

.256[ ]9.34•103

= 2.74⋅ 10−5 = 0.80 - 2x[ ]2

The reaction is practically stoichiometric as indicated by Keq

[SO3] = 5.2e-3 M, [SO2]=0.794M, [O2]= 0.397 M

€

Kc =SO2[ ]

2∗ O2[ ]

SO3[ ]2

= 9.34 •103 =2x[ ]

2∗ x[ ]

0.80 - 2x[ ]2

>> 1

Note: Kc>>1, Rxn favors forwardAssume reaction goes to completion or 80M-2x ~ 0, therefore x ~ 0.4M. Using the iteration method, x is solved for in the equation .80M-2x.

Assumption Check:(Dx / .80) < 5% = (5.2e-3/.80)•100 = 0.65 %

.


6. Calculation: Product formed @ Equilibrium (Type2)Exactly 4 mol of SO3 is sealed in a 5.0 L container @1500K. Kc is 9.34•103 for the rxn, what are the conc. of all specie at equilb? (use as many significant figures as calculator allow then round off your final answer.)

€

2SO3(g) ! 2 SO2(g) + O2(g)

i 0.80 M 0.00 0.00Δ −2x +2x +xe[ ] 0.80 -2x +2x +x

€

5.235810-3 = 0.80 −2x2x = 0.80 −5.235810-3M = 0.794765 x = 0.3974 = 0.40

€

.256[ ]9.34•103

= 2.74⋅ 10−5 = 0.80 - 2x[ ]2

The reaction is practically stoichiometric as indicated by Keq

[SO3] = 5.2e-3 M, [SO2]=0.794M, [O2]= 0.397 M

€

Kc =SO2[ ]

2∗ O2[ ]

SO3[ ]2

= 9.34 •103 =2x[ ]

2∗ x[ ]

0.80 - 2x[ ]2

>> 1

Note: Kc>>1, Rxn favors forwardAssume reaction goes to completion or 80M-2x ~ 0, therefore x ~ 0.4M. Using the iteration method, x is solved for in the equation .80M-2x.

Assumption Check:(Dx / .80) < 5% = (5.2e-3/.80)•100 = 0.65 %

€

4x3" # $

% & '

0.80 -2x[ ]2≈

4 × 0.43" # $

% & '

0.80 -2x[ ]2

= 9.34 •103


7. ... Another equilibrium problem (difficult)

An engineer is studying the oxidation of SO2 to produce the precursor, SO3, in the manufacture of sulfuric acid. The Kp = 1.7 • 108 at 600. K for the reaction.

2SO2 (g) + O2(g) D 2SO3 (g)

i) At equilibrium pSO3 = 300. atm and pO2 = 100. atm. Calculate pSO2

ii) To this system, the engineer next add 0.0040 mol of SO2(g) and 0.0028 mol of O2(g) to the 1.0-L container and raises the temperature from 600 K to 1000K. The reaction reaches equilibrium and 0.0020 mol of SO3 (g) is present. What would the engineer calculate the Kc and PSO2 for the reaction to be 1000K. Does the engineer expect the reaction to be exothermic or endothermic?

7i pSO3 = 300 atm, pO2 = 100 atm, pSO2 = 2.3"10-3 atm

7ii 2SO2 (g) + O2(g) -> 2SO3 (g)

pi 2.3•10-3 atm 100 atm 300 atm Convert to mol via PV = nRT

P -> mol 4.67•10-5 mol 2.03 mol 6.0931 mol add 4.0•10-3 2.8•10-3 [i] 4.047•10-3 M 2.0328 M 6.0931 M C -2x -x +2x [e] 4.047•10-3 -2x 2.0328 x 0.0020 M x = -3.0456 M 6.092 5.0768 0.0020

Kc = = 2.12"10-8

pSO2 = 6.092 (.08206) 1000 / 1 = 500 atm


Procedure for Solving Equilibrium Problems

1. Write the balanced equation for the reaction.2. Write the equilibrium expression using the Mass Action Expression.3. List the initial conditions.4. Determine the type of equilibrium problem you have.

Type 1 - Given the equilibrium concentration, solve for Keq . Type 2 - Given Keq and initial concentration, solve for conc. @ equilb.

Type1Plug the equilibrium concentration into the Mass Action and solve for the equilibrium constant, Keq .

Type2a) Set up the iDe table and solve for the equilibrium concentration in terms of a variable (x).b) Plug the equations expressing the equilibrium concentration into the mass action expression and solve for xc) Assign the value of x to the equilibrium concentration equation and determine the numerical value for the equilibrium concentration for each specie in the reaction.

15.2 solving equilibrium problems

Documents