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CHEMICAL EQUILIBRIUM. Properties of an Equilibrium. Equilibrium systems are DYNAMIC (in constant motion) REVERSIBLE can be approached from either direction. Pink to blue Co(H 2 O) 6 Cl 2 ---> Co(H 2 O) 4 Cl 2 + 2 H 2 O. Blue to pink

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properties of an equilibrium
Properties of an Equilibrium

Equilibrium systems are

  • DYNAMIC (in constant motion)
  • REVERSIBLE
  • can be approached from either direction

Pink to blue

Co(H2O)6Cl2 ---> Co(H2O)4Cl2 + 2 H2O

Blue to pink

Co(H2O)4Cl2 + 2 H2O ---> Co(H2O)6Cl2

chemical equilibrium fe 3 scn fescn 21
Chemical EquilibriumFe3+ + SCN- FeSCN2+

e

  • After a period of time, the concentrations of reactants and products are constant.
  • The forward and reverse reactions continue after equilibrium is attained.
examples of chemical equilibria
Examples of Chemical Equilibria

Phase changes such as H2O(s) H2O(liq)



examples of chemical equilibria1
Examples of Chemical Equilibria

Formation of stalactites and stalagmites

CaCO3(s) + H2O(liq) + CO2(g) Ca2+(aq) + 2 HCO3-(aq)



chemical equilibria
Chemical Equilibria

CaCO3(s) + H2O(liq) + CO2(g) Ca2+(aq) + 2 HCO3-(aq)

At a given T and P of CO2, [Ca2+] and [HCO3-] can be found from the EQUILIBRIUM CONSTANT.

reaction quotient equilibrium constant

Equilibrium achieved

Screen 16.4

& Fig. 16.2

Reaction Quotient & Equilibrium Constant

Product conc. increases and then becomes constant at equilibrium

Reactant conc. declines and then becomes constant at equilibrium

reaction quotient equilibrium constant1
Reaction Quotient & Equilibrium Constant

At any point in the reaction

H2 + I2 ↔2 HI

the reacton quotient q
The Reacton Quotient, Q

In general, all reacting chemical systems are characterized by their REACTION QUOTIENT, Q.

If Q = K, then system is at equilibrium.

the equilibrium constant
THE EQUILIBRIUM CONSTANT

Hi Mr. Rierson! For any type of chemical equilibrium of the type

a A + b Bc C + d D

the following is a CONSTANT (at a given T)

If K is known, then we can predict concs. of products or reactants.

determining k
Determining K

2 NOCl(g) ↔2 NO(g) + Cl2(g)

Place 2.00 mol of NOCl in a 1.00 L flask. At equilibrium you find 0.66 mol/L of NO. Calculate K.

Solution

Set of an “ICE” table of concentrations

[NOCl] [NO] [Cl2]

Initial 2.00 0 0

Change

Equilibrium 0.66

determining k1
Determining K

2 NOCl(g) ↔ 2 NO(g) + Cl2(g)

Place 2.00 mol of NOCl is a 1.00 L flask. At equilibrium you find 0.66 mol/L of NO. Calculate K.

Solution

Set of a table of concentrations

[NOCl] [NO] [Cl2]

Initial 2.00 0 0

Change -0.66 +0.66 +0.33

Equilibrium 1.34 0.66 0.33

determining k2
Determining K

2 NOCl(g) ↔ 2 NO(g) + Cl2(g)

[NOCl] [NO] [Cl2]

Initial 2.00 0 0

Change -0.66 +0.66 +0.33

Equilibrium 1.34 0.66 0.33

writing and manipulating k expressions
Writing and Manipulating K Expressions

Solids and liquids NEVER appear in equilibrium expressions.

S(s) + O2(g)  SO2(g)

writing and manipulating k expressions1
Writing and Manipulating K Expressions

Solids and liquids NEVER appear in equilibrium expressions.

NH3(aq) + H2O(liq) NH4+(aq) + OH-(aq)

messing with k
Messing with K
  • If you reverse the reaction . . .
    • K (forward) = K-1(reverse)
  • If you multiply the coefficients by a factor x . .
  • If you add two reactions together . . .
    • Multiply the K’s together
  • If you switch from measuring concentration to measuring pressure . . .
    • This is related through Ideal Gas Law PV = nRT or P = (n/V) RT or P = C RT… This is primarily impacted by the change in the number of moles (n) of gas from reactants to products
    • Kp = Kc(RT)change in n
the meaning of k
The Meaning of K

1. Can tell if a reaction is product-favored or reactant-favored.

For N2(g) + 3 H2(g) ↔ 2 NH3(g)

Conc. of products is much greater than that of reactants at equilibrium.

The reaction is strongly product-favored.

the meaning of k1
The Meaning of K

For AgCl(s) ↔ Ag+(aq) + Cl-(aq)

Kc = [Ag+] [Cl-] = 1.8 x 10-5

Conc. of products is much less than that of reactants at equilibrium.

The reaction is strongly reactant-favored.

Ag+(aq) + Cl-(aq)

e AgCl(s)

is product-favored.

product or reactant favored
Product- or Reactant Favored

Product-favored

Reactant-favored

the meaning of k2
The Meaning of K

K comes from thermodynamics.

See Chapter 19, page 812-813

∆G˚ < 0: reaction is product favored

∆G˚ > 0: reaction is reactant-favored

If K > 1, then ∆G˚ is negative

If K < 1, then ∆G˚ is positive

the meaning of k3
The Meaning of K

2. Can tell if a reaction is at equilibrium. If not, which way it moves to approach equilibrium.

the meaning of k4
The Meaning of K

If [iso] = 0.35 M and [n] = 0.15 M, are you at equilibrium?

If not, which way does the reaction “shift” to approach equilibrium?

the meaning of k5
The Meaning of K

All reacting chemical systems are characterized by their REACTION QUOTIENT, Q.

If Q = K, then system is at equilibrium.

Q (2.33) < K (2.5)

Reaction is NOT at equilibrium, so [iso] must become ________ and [n] must ____________.

typical calculations
Typical Calculations

PROBLEM: Place 1.00 mol each of H2 and I2 in a 1.00 L flask. Calc. equilibrium concentrations.

H2(g) + I2(g) ↔ 2 HI(g)

h 2 g i 2 g 2 hi g k c 55 3
H2(g) + I2(g) ↔ 2 HI(g)Kc = 55.3

Step 1. Set up ICE table to define EQUILIBRIUM concentrations.

[H2] [I2] [HI]

Initial 1.00 1.00 0

Change

Equilib

h 2 g i 2 g 2 hi g k c 55 31
H2(g) + I2(g) ↔ 2 HI(g)Kc = 55.3

Step 1. Set up ICE table to define EQUILIBRIUM concentrations.

[H2] [I2] [HI]

Initial 1.00 1.00 0

Change -x -x +2x

Equilib 1.00-x 1.00-x 2x

where x is defined as am’t of H2 and I2 consumed on approaching equilibrium.

h 2 g i 2 g 2 hi g k c 55 32
H2(g) + I2(g) ↔ 2 HI(g)Kc = 55.3

Step 2. Put equilibrium concentrations into Kc expression.

h 2 g i 2 g 2 hi g k c 55 33
H2(g) + I2(g) ↔ 2 HI(g)Kc = 55.3

Step 3. Solve Kc expression - take square root of both sides.

x = 0.79

Therefore, at equilibrium

[H2] = [I2] = 1.00 - x = 0.21 M

[HI] = 2x = 1.58 M

nitrogen dioxide equilibrium n 2 o 4 g 2 no 2 g1
Nitrogen Dioxide EquilibriumN2O4(g) ↔ 2 NO2(g)

If initial concentration of N2O4 is 0.50 M, what are the equilibrium concentrations?

Step 1. Set up an ICE table

[N2O4] [NO2]

Initial 0.50 0

Change

Equilib

nitrogen dioxide equilibrium n 2 o 4 g 2 no 2 g2
Nitrogen Dioxide EquilibriumN2O4(g) ↔ 2 NO2(g)

If initial concentration of N2O4 is 0.50 M, what are the equilibrium concentrations?

Step 1. Set up an ICE table

[N2O4] [NO2]

Initial 0.50 0

Change -x +2x

Equilib 0.50 - x 2x

nitrogen dioxide equilibrium n 2 o 4 g 2 no 2 g3
Nitrogen Dioxide EquilibriumN2O4(g) ↔ 2 NO2(g)

Step 2. Substitute into Kc expression and solve.

Rearrange: 0.0059 (0.50 - x) = 4x2

0.0029 - 0.0059x = 4x2

4x2 + 0.0059x - 0.0029 = 0

This is a QUADRATIC EQUATION

ax2 + bx + c = 0

a = 4 b = 0.0059c = -0.0029

nitrogen dioxide equilibrium n 2 o 4 g 2 no 2 g4
Nitrogen Dioxide EquilibriumN2O4(g) ↔ 2 NO2(g)

Solve the quadratic equation for x.

ax2 + bx + c = 0

a = 4 b = 0.0059c = -0.0029

x = -0.00074 ± 1/8(0.046)1/2 = -0.00074 ± 0.027

nitrogen dioxide equilibrium n 2 o 4 g 2 no 2 g5
Nitrogen Dioxide EquilibriumN2O4(g) ↔ 2 NO2(g)

x = 0.026 or -0.028

But a negative value is not reasonable.

Conclusion: x = 0.026 M

[N2O4] = 0.50 - x = 0.47 M

[NO2] = 2x = 0.052 M

x = -0.00074 ± 1/8(0.046)1/2 = -0.00074 ± 0.027

solving quadratic equations
Solving Quadratic Equations
  • Recommend you solve the equation exactly on a calculator (do you have a quadratic program? Get one…) or use the “method of successive approximations”
  • See Appendix A.
your turn
Your turn:
  • Now: some group-work problems for you to work in class and if time, put solutions on board
  • Webassign tonight as HW there will be two assignments.
    • First one is practice (due Thursday)
    • Second one is graded (due Thursday)
  • Read 16.4, 16.5
  • Prepare for 1 question quiz tomorrow