chem 35.5 equilibrium chapter 17 - · pdf file2006-02-10 · the van’t hoff...

AnnouncementsHour Exam IIThursday Feb 5, 2009CTC 102 and 105Time: 6:00 - 8:00 PM (1.5 hrs long)Bring Blue Book, calculator and writing instrument.

Tomorrow, the “Zos-man” will go over selected problems provided by all of the instructors. I think it will help.

Tomorrow we will review in class.

A + 2B AB2kf

kr

rateforward = kf [A][B]2

ratereverse = kr [AB2]

Equilibrium occurs when the rate of the forward reaction and the rate of the reverse reaction are equal!

kf [A][B]2 = kr [AB2]

kfkr

[AB2][A][B]2

= Kc =

ratef = rater

Assume a reaction that is also an elementary reaction

Because the reaction are elementary we can write rate laws from balanced equation

At equilibrium the reactions continue but at the same rate!

Kc is defined as the ratio of the rate constants and is tied to stoichiometry of the elementary event!

The van’t Hoff Equation is the mathematical relationship between the equilibrium constant and temperature.

We have the Clausius-Clapyeron Equation, the Arrehenius Equation and the van’t Hoff Equation

R = universal gas constant = 8.314 J/mol*K

lnk2

k1

= -Ea

R

1

T2

1

T1

- lnP2

P1

= -ΔHvap

R

1

T2

1

T1

-

lnK2

K1

= -ΔH0

rxn

R

1

T2

1

T1

-

K1 is the equilibrium constant at T1 and K2 is the equilibrium constant at T2

ln K2

K1= -

ΔH0rxn

R1T2

1T1

-

ln K2

K1= -

ΔH0rxn

R1T2

1T1

-

R = universal gas constant = 8.314 J/mol K

K1 is the equilibrium constant at T1 and K2 is the equilibrium constant at T2

Stearic acid, nature's most common fatty acid, dimerizes when dissolved in hexane:

2C17H35COOH <=> (C17H35COOH)2; ΔH°rxn = -172 kJ

The equilibrium constant for this reaction at 28°C is 2900. Estimate the equilibrium constant at 38°C.

Solving Equilibrium Problems• Two basic flavors of chemical equilibrium problems

1. Equilibrium quantities are given (concentrations or partial pressures) and we solve for Kc. This is easy plug and chug.

2. We use ICE table to calculate either Kc from initial quantities or we calculate equilibrium concentrations given Kc.

---it’s the algebra that’s tricky

In either case we can use a book-keeping techniqueInitial, Change Equilibrium Method

Let’s try it in an example problem!

PLAN:

Calculating equilibrium concentration using a simplfying assumption (no quadratic equation)

The reaction of nitrogen with oxygen giving nitrogen oxide contributes to air pollution whenever a fuel is burnt at high temperature. At 1500K the K = 1.0 X 10-5. Suppose a sample of air has [N2] = 0.80 M, and [O2] = 0.20M before a reaction occurs. Calculate the equilibrium concentrations at equilibrium.

Set up ICE table and begin filling it in.

N2(g) + O2(g) <==> 2NO(g) Kc = 1.0 X 10-5

= 0.020

1. Write down what we solve: Kc and what we know.

2. Set up the ICE table

Kc = [N2] [O2][NO]2 = 1.0 X 10-5

Molarity

initialchange

equilibrium

0.80-x +2x-x

0.20 0

0.80 - x 2x+

N2(g) + O2(g) <=> 2NO(g)

0.20 - x

=[.80 -x]

[2x]2Kc = [N2] [O2][NO]2 = 1.0 X 10-5

[.20 -x]

3. Use equilibrium expression and substitute values.

4. Either perfect square, quadratic equation or simplifying assumption.

=[.80 -x]

[2x]2Kc = [N2] [O2][NO]2 = 1.0 X 10-5

[.20 -x]

3. Use equilibrium expression and substitute values.

4. Either perfect square, quadratic equation or simplifying assumption.

RULE: If 100 x K < [A]0 then .80 - x = .80100 X 10-5 = 10-3 < [A]0 .......it’s ok to rid x.

=Kc = [N2] [O2][NO]2 = 1.0 X 10-5

[.80][2x]2

[.20] [.16][2x]2=

x = (1.0 X 10-5 (.16) 1/2 = 6.3 X 10-4 4

)([N2] = .80 - x = 0.80 - 6.3 X 10-4 = 0.80M[O2] = -.20 - x = 0.20 - 6.3 X 10-4 = 0.20M[NO] = 2x = 6.3 X 10-4 X 2 = 1.3 X 10-3M

Determining equilibrium concentrations from Kc

A chemical engineer mixes gaseous CH4 and H2O in a 0.32 L flask at 1200 K. At equilibrium, the flask contains 0.26 mol of CO, 0.091 mol of H2, and 0.041 mol of CH4. What is the [H2O] at equilibrium? A Handbook states that Kc = 0.26 for this reaction is:

CH4(g) + H2O(g) CO(g) + 3H2(g) Kc = 0.26

PLAN: Use the balanced equation to write the Kc expression, and then substitute values for each component.

[H2O]eq = [CO]eq[H2]eq3

[CH4]eq Kc= 0.53 M= (0.81)(0.28]3

(0.13)(0.26]

= 0.13 M

0.26mol0.32 L

= 0.81 M

0.091 mol0.32 L

= 0.28 M

CH4(g) + H2O(g) CO(g) + 3H2(g)concentration (M)

initialchange

equilibrium

?

? ??

? ?

0.041 mol/0.32 L ?

?

?

Kc =[CO][H2]3

[CH4][H2O]CH4(g) + H2O(g) CO(g) + 3H2(g)

Determining equilibrium concentrations from initial concentrations and KcFuel engineers use the extent of the change from CO and H2O to CO2 and H2 to regulate the proportions of synthetic fuel mixtures. If 0.250 mol of CO and 0.250 mol of H2O are placed in a 125 mL flask at 900K, what is the composition of the mixture at equilibrium? At 900K, Kc is 1.56 for this reaction.

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

PLAN: 1)Balance equation, 2)write equilibrium expression, 3)set up ICE table, 4)find the concentrations of all species at initial conditions or equilibrium in the problem 5)use algebra to determine equilibrium concentrations and then substitute into a Kc expression.

Kc = 1.56

[CO] = [H2O] = 2.00 - x = 0.89 Mx = 1.11M = [CO2] = [H2]

(the negative result is ignored)

Initial concentrations must be calculated as M, we have from the data given [CO] = [H2O] = 0.250/0.125L = 2.00M.

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

changeequilibrium

2.00 2.00 0 0-x -x +x +x

2.00 -x 2.00 -x x x

_________________________________________________

Kc =[CO2][H2][CO][H2O]

=x2

(2.00! x)(2.00! x)=

x2

(2.00! x)2= 1.56

x

2.00! x="

1.56 = ±1.25

x = 1.25(2.00! x) = 2.50! 1.25x

2.25x = 2.50

Same problem....but not a perfect square!Fuel engineers use the extent of the change from CO and H2O to CO2 and H2 to regulate the proportions of synthetic fuel mixtures. If 0.250 mol of CO and 0.125 mol of H2O are placed in a 125 mL flask at 900K, what is the composition of the mixture at equilibrium? At 900K, Kc is 1.56 for this reaction.

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

PLAN: 1)Balance equation, 2)write equilibrium expression, 3)set up ICE table, 4)find the concentrations of all species at initial conditions or equilibrium in the problem 5)use algebra to determine equilibrium concentrations and then substitute into a Kc expression.

Kc = 1.56

Initial concentrations must be calculated as M, we have from the data given [CO] = [H2O] = 0.250/0.125L = 2.00M.

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

changeequilibrium

2.00 1.00 0 0-x -x +x +x

2.00 -x 1.00 -x x x

_________________________________________________

Kc =[CO2][H2][CO][H2O]

=x2

(2.00! x)(1.00! x)=

x2

x2 ! 3.00x + 2.00= 1.56

x2 = 1.56(x2 ! 3.00x + 2.00)x2 = 1.56x2 ! 4.68x + 3.12)

0.56x2 - 4.68x + 3.12 = 0

ax2 + bx + c =0

-b ± b2 – 4ac √2ax =

Kc =[CO2][H2][CO][H2O]

=x2

(2.00! x)(1.00! x)=

x2

x2 ! 3.00x + 2.00= 1.56

x2 = 1.56(x2 ! 3.00x + 2.00)x2 = 1.56x2 ! 4.68x + 3.12)

0.56x2 - 4.68x + 3.12 = 0

ax2 + bx + c =0

-b ± b2 – 4ac √2ax =

- (-4.68 ± (-4.68)2 – 4(0.56)(3.12) √2(0.56)

x =

x = 7.6M and x = 0.73M

x = 7.6M makes 2.00 - x < 0 and makes no sense-toss it

x = 0.73 = [CO2] = [H2] and [CO] = [H2O] = 2.00 - 0.73

Br2 (g) 2Br (g)At 1280˚C the equilibrium constant (Kc) for the reaction

is 1.1 x 10-3. If the initial concentrations are [Br2] = 0.063 M and [Br] = 0.012 M, calculate the concentrations of these species at equilibrium.

Using the Quadratic Formula

Kc is 1.1 X 10-3

Let x be the change in concentration of Br2

[Br]2[Br2]

Kc =

The expression is not a perfect square: we therefore must use quadratic equation to solve it

Br2 (g) 2Br (g)1. Write Balanced Equation

2. Write Equilibrium Expression

3. Set Up ICE TableBr2 (g) 2Br (g)

Initial (M)

Change (M)

Equilibrium (M)

[0.063] [0.012]

-x +2x

0.063 - x 0.012 + 2x

Kc = (0.012 + 2x)2

0.063 - x= 1.1 x 10-3

[Br]2[Br2]

Kc = =

Kc = (0.012 + 2x)2

0.063 - x= 1.1 x 10-3

4x2 + 0.048x + 0.000144 = 0.0000693 – 0.0011x4x2 + 0.0491x + 0.0000747 = 0

-b ± b2 – 4ac √2ax =

x = -0.00178x = -0.0105

At equilibrium, [Br] = 0.012 + 2x = -0.009 M and 0.00844 M

At equilibrium, [Br2] = 0.062 – x = 0.06378 M

ax2 + bx + c =0

Br2 (g) 2Br (g)

Initial (M)

Change (M)

Equilibrium (M)

[0.063] [0.012]

-x +2x

0.063 - x 0.012 + 2x

Two solutions

Phosgene is a potent chemical warfare agent that is now outlawed by international agreement. It decomposes by the reaction,

COCl2(g) CO(g) + Cl2(g) Kc = 8.3 x 10-4 (at 360 oC)

Calculate [CO], [Cl2], and [COCl2] at equilibrium when the initial amount of phosgene gas is 0.100 mol in a 10.0 L flask.

[COCl2] = 0.100 mol/10.0 L = 0.0100 M

Kc = 8.3 x 10-4 (at 360 oC)

COCl2(g) CO(g) + Cl2(g) Kc = 8.3 x 10-4 =[CO] [Cl2][COCl2]

Initial (M)Change (M)

Equilibrium (M)

0.0 0.0-x +x

0.010 - x x

COCl2(g) CO(g) + Cl2(g) 0.010

+x

x

Kc = 8.3 x 10-4 =(x) (x)

(0.010 - x)x2

(0.010 - x)=

Kc = 8.3 x 10-4 (0.010 - x) = x2

Kc = 8.3 x 10-6 - 8.3 x 10-4 x = x2

Kc = 1x2 + 8.3 x 10-4 x - 8.3 x 10-6 = 0

ax2 + bx + c =0

x =

x = [CO] = [Cl2] = 2.5 x 10-3 M & 0.0100 - x = [COCl2] = 7.5 x 10-3 M.

Kc = 1x2 + 8.3 x 10-4 x - 8.3 x 10-6 = 0

ax2 + bx + c =0

-b ± b2 – 4ac √2ax =

a = 1b = 8.3 x 10-4 c = - 8.3 x 10-6

-8.3 x 10-4 ± (8.3 x 10-4) 2 – 4(1) (-8.3 x 10-6)√2(1)

x1 = 0.002495M = 0.0025Mx2 = -0.003325 which makes no sense and rejected.

The research and development unit of a chemical company is studying the reaction of CH4 and H2S, two components of natural gas.

CH4(g) + 2H2S(g) CS2(g) + 4H2(g)

In one experiment, 1.00 mol of CH4, 1.00 mol of CS2, 2.00 mol of H2S and 2.00 mol of H2 are mixed in a 250 mL vessel at 960 oC. At this temperature, Kc = 0.036.

(a) In which direction will the reaction proceed to reach equilibrium?

(b) If [CH4] = 5.56 M at equilibrium, what are the equilibrium concentrations of the other three substances?

PLAN: Find the initial molar concentrations of all components and use these to calculate Qc. Compare Qc to Kc, determine in which direction the reaction will progress, and draw up expressions for the equilibrium concentrations.

SOLUTION:

Qc = [CS2][H2]4

[CH4][H2S]2= [4.0][8.0]4

[4.0][8.0]2= 64.0

Qc of 64 is >> than Kc = 0.036

The reaction will progress to the left.

[CH4]initial = 1.00 mol/0.25 L = 4.00 M

[H2S]initial = 2.00 mol/0.25 L = 8.00 M

[H2]initial = 2.00 mol/0.25 L = 8.00 M

[CS2]initial = 1.00 mol/0.25 L = 4.00 M

CH4(g) + 2H2S(g) CS2(g) + 4H2(g)

SOLUTION:

8.00

At equilibrium [CH4] = 5.56 M, so 5.56 = 4.00 + x; thus,

x = 1.56 M

Therefore:

[H2S] = 8.00 + 2x = 11.12 M

[CS2] = 4.00 - x = 2.44 M

[H2] = 8.00 - 4x = 1.76 M

CH4(g) + 2H2S(g) CS2(g) + 4H2(g)concentrations

change

equilibrium

initial 4.00 8.00 4.00+ x + 2x - 4x

4.00 + x 8.00 + 2x

- x

4.00 - x 8.00 - 4x

______________________________________________________

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