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Chapter 14 Chemical Equilibrium. “Old Chemists Never Die; they just reach EQUILIBRIUM! ”. All physical and chemical changes TEND toward a state of equilibrium. A (l) A (g). A (s) A (l). Dynamic Equilibrium. The net result of a dynamic equilibrium is that no

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Chapter 14 chemical equilibrium
Chapter 14Chemical Equilibrium

“Old Chemists Never Die; they just reach

EQUILIBRIUM!”

All physical and chemical changes TEND toward a

state of equilibrium.

A (l) A (g)

A (s) A (l)


Dynamic Equilibrium

The net result of a dynamic equilibrium is that no

change in the system is evident.

Le Chatelier’s Principle - If a change is made in a

system at equilibrium, the equilibrium will shift in such

a way so as to reduce the effect of the change.

Apply

Pressure

Pressure applied to the system at equilibrium caused it to

shift until a new equilibrium was established.


Dynamic Equilibrium

Open System

(No Equilibrium)

Evaporation

Evaporation

Liquid Gas

Liquid Gas

Liquid Gas

(No Equilibrium)

(Equilibrium)

(No Equilibrium)


Dynamic Equilibrium

Ag + + Cl - AgCl (s)

Ag + Cl -

Chemical

Equilibrium

Cl - Ag +

AgCl (s)

Rate of Precipitation = Rate of Dissolving

HC2H3O2 (aq) H + + C2H3O2-

Rate of dissociation (ionization) = Rate of Association

HC2H3O2

H +

H + C2H3O2-

C2H3O2-

HC2H3O2


CHEM 1108 Lab Experiment

Orange

Red

HC2H3O2 H+ + C2H3O2-

Pink

Blue

[CoCl4]2- + 6 H2O (l) Co(H2O)62+ + 4 Cl-

Colorless Solution

White

NH4Cl (s) NH4+ + Cl-

You can actually “see” the equilibrium shift!


Reversible Reactions

R1

N2O4 (g) 2 NO2 (g)

R2

2 NO2 (g) N2O4 (g)

R1

N2O4 (g) 2 NO2 (g)

[R1 = R2]

R2

Homogeneous Equilibrium


Reversible Reactions

[N2O4]i [NO2]i [N2O4]eq [NO2]eq

Exp. 1 0.0250 M 0.0 M 0.0202 M 0.009 66 M

Exp. 2 0.0150 M 0.0125 M 0.0146 M 0.008 23 M

Exp. 3 0.0 M 0.0250 M 0.0923 M 0.006 54 M

N2O4 (g) 2 NO2 (g)

Reaction Quotient

Equilibrium Constant

QC = [NO2]2

[N2O4]

KC = [NO2]2eq

[N2O4]eq


Equilibrium Constants

Equilibrium Constant - When the rates of the forward

and reverse reactions are equal, the system is “at equil-

ibrium” and the reaction quotient = equilibrium

constant.

Experiment 1 KC = [0.009 66]2/[0.0202] = 0.004 62 M

Experiment 2 KC = [0.008 23]2/[0.0146] = 0.004 64 M

Experiment 3 KC = [0.006 54]2/[0.009 23] = 0.004 63 M

aA + bB cC + dD

KC = [C]c[D]d

[A]a[B]b


Equilibrium Constants

H2 (g) + I2 (g) 2 HI (g)

KC = [HI]2eq [H2]eq[I2]eq

QC = [HI]2 [H2][I2]

  • Exp. [H2]eq [I2]eq [HI]eq KC

  • 1 0.00291 0.00171 0.01648 54.58

  • 2 0.00356 0.00125 0.01559 54.62

  • 3 0.00225 0.00234 0.01685 53.93

  • 4 0.00183 0.00313 0.01767 54.51

  • 5 0.00114 0.00114 0.00841 54.42

  • 0.00050 0.00050 0.00366 53.58

    • Median KC = 54.47


Equilibrium Constants

4 NH3 (g) + 3 O2 (g) 2 N2 (g) + 6 H2O (g)

QC =

[N2]2[H2O]6

[NH3]4[O2]3

[N2]eq2[H2O]6

KC =

[NH3]eq4[O2]eq3


Reaction Quotient vs.

Equilibrium Constant

Class Problem 14.1 -The concentration of N2O4 =

concentration of NO2 = 0.0125 M in a reaction vessel.

The equilibrium constant for N2O4 (g) = 2 NO2 (g) is

0.004 63. Calculate QC and state which direction the

reaction will go.

Class Problem 14.2 -If [O2] = 0.21 M and [O3] = 6.0 x 10-8 M, what is the value of the KC for the equilibrium, 2 O3 (g) = 3 O2 (g)?


Chemical Equilibrium

Heterogeneous Reaction -A reaction that takes place in more than one phase or state. These reactions occur at the interface between phases - on the surface of liquids and solids.

At a constant temperature, the concentration of a solid or liquid component remains constant in a heterogeneous equilibrium. WHY?

Since the concentration is constant, it can be considered a part of the equilibrium constant and, thus, does NOT appear in the KC expression.

C (s, graphite) + CO2 (g) = 2 CO (g)

KC = [CO]2

[C][CO2]

Keq = [CO]2 [CO2]


Chemical Equilibrium

Class Problem 14.3 -A mixture that was initially 0.005 00 M in H2 (g) and 0.012 50 M in I2 (g), and contained no HI (g), was heated at 425.4oC until equilibrium was reached. The resulting equilibrium concentration of I2 (g) was found to be 0.007 72 M. What is the value of the KC for this equilibrium at 425.4oC?

Construct an “ICE” Table:

425.4oC

Equation: H2 (g) + I2 (g)  2 HI (g)

Initial (I) conc., M 0.005 00 0.012 50 0.000 00

Change (C) in conc., M

Equil. (E) conc., M 0.007 72

- 0.004 78

- 0.004 78

+ 0.009 56

+ 0.009 56

- 0.000 22


Chemical Equilibrium

Calculate KC:

[HI]2 (0.009 56)2 [H2][I2] (0.000 22)(0.007 72)

= 54

KC =

=

Class Problem 14.4a. -When 1.000 mol each of H2O (g) and CO (g) are introduced into an empty 1.000 L vessel at 959 K and allowed to come to equilibrium, the equilibrium mixture contains 0.422 mol H2O (g). Find KC for

H2O (g) + CO (g)  H2 (g) + CO2 (g)

Construct an “ICE” Table:


Chemical Equilibrium

959oC

Equation: H2O (g) + CO (g)  H2 (g) + CO2 (g)

I 1.000 1.000 0.000 0.000

C - 0.578 - 0.578 + 0.578 + 0.578

E 0.422 0.422 0.578 0.578

[H2][CO2]

[H2O][CO]

(0.578)2

(0.422)2

Kc =

=

= 1.88

Class Problem 14.4b. -Suppose that [H2O]I = 2.00 M and [CO]I = 4.00 M?

What are the equilibrium concentrations of the four species?


Chemical Equilibrium

959oC

Equation: H2O (g) + CO (g)  H2 (g) + CO2 (g)

I 2.00 4.00 0.00 0.00

C - x - x + x + x

E 2.00 - x 4.00 - x xx

[H2][CO2]

[H2O][CO]

x2

Kc =

=

= 1.88

(2.00 – x)(4.00 – x)

x2 = 1.88(2.00 – x)(4.00 – x) = 1.88(8.00 - 6x – x2)

x2 = 15.0 – 11.3x +1.88x2 0 = 0.88x2 – 11.3x +15.0


0 = 0.88 x2 – 11.3 x + 15.0 (ax2 + bx + c)

Dust off the old Quadratic Formula:

http://www.freemathhelp.com/algebra-help.html

-(-11.3) ± [(-11.3)2 – 4(0.88)(15.0)]1/2

2(0.88)

= 11 and 1.5 ! Which is RIGHT?


Chemical Equilibrium

What is ‘x’? It is the concentration of H2 and CO2

at equilibrium! But…you can’t have more hydrogen gas than

you have of reactants to begin with!

Thus, 11 M can’t be right!

1.5 M is the only sensible answer!


Chemical Equilibrium

x = 1.5 M Therefore:

[H2O]eq = 0.5 M [CO]eq = 2.5 M

[H2]eq = [CO2]eq = 1.5 M

Check: Kc = (1.5M)2/(0.5M)(2.5M) = 1.8

There are no units in this case!

What if you don’t remember the quadratic

formula??


Use Successive

Approximation!!


Class Exercise 14.5: Consider the following reaction for the decomposition of hydrogen sulfide:

2 H2S  2 H2 (g) + S2 (g) KC = 1.67 x 10-7

800o C

A 0.500-L vessel initially contains 1.25 x 10-1 mol of H2S.

Find the equilibrium concentrations of H2 and S2.

Equation: 2 H2S (g)  2 H2 (g) + S2 (g)

Initial (M) 2.50 x 10-1 0.00 0.00

Change (M) - 2x + 2x + x

Equilibrium (M) (2.50 x 10-1 – 2x) 2x x


[H2]2[S2]

(2x)2x

(2.50 x 10-1 – 2x)2

Kc =

=

H2S

4x3

(2.50 x 10-1 – 2x)2

=

= 1.67 x 10-7

Assume x is NEGLIGIBLE compared to 2.50 x 10-1 M.

Then:

4x3

(2.50 x 10-1)2

4x3

6.25 x 10-2

=

~ 1.67 x 10-7

(6.25 x 10-2)

(1.67 x 10-7)

=

4x3

=

1.04 x 10-8

x3

=

2.61 x 10-9

x = 1.38 x 10-3 M


Is x NEGLIGIBLE compared to 1.38 x 10-3 M?

Plug it back in to check:

4x3

4x3

(2.50 x 10-1 – 2x)2

=

[(2.50 x 10-1) - 2(1.38 x 10-3)]2

4x3

=

1.67 x 10-7

4.95 x 10-4

x3 = 2.07 x 10-11>>>>>

x = 2.74 x 10-4

There appears to be a mistake in these calculations!

Please check carefully and see if you can see where it is!



Do NOT Panic!

This is NOT a typical Problem!

It is a Worst Case Scenario!!!

Any Exam Problem will

be MUCH Shorter!!


Class Exercise 14.5: In an experiment starting with

[N2O4]I = 0.020 00 M and [NO2]I = 0.000 00 M, [N2O4]eq

= 0.004 52 M. (a) What is [NO2]eq? (b) What is the

value for Kc?

Equation: N2O4 (g)  2 NO2 (g)

I (M) 2.000 x 10-2 0.000 00

C (M)

E (M) 4.52 x 10-3

-0.015 48

+0.030 96

0.030 96

[NO2]2

[N2O4]

KC =

= (0.030 96)2/(0.004 52) = 0.212


What does the value of Kc MEAN?

The larger KC is, the closer to completion the rxn is!

N2 (g) + O2 (g)  2 NO (g) KC = 1 x 10-30

2 NH3 (g)  N2 (g) + 3 H2 (g) KC = 9.5

H2 (g) + Cl2 (g)  2 HCl (g) KC = 1.33 x 1034


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