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CHAPTER 14

Chemical Equilibrium

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How Fast? Kinetics applies to the speed of a reaction, the concentration of product that appears (or of reactant that disappears) per unit time.

What happens? Stoichiometry

To What Extent? Equilibrium applies to the extent of a reaction, the concentrations of reactant product present after an unlimited time, or once no further change occurs.

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Equilibrium and Reversible Rxns.

double arrows = "reversible reaction"

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Equilibrium

A system at equilibrium is dynamic on the molecular level; no further net change is observed because changes in one direction are balanced by changes in the other.

At equilibrium: rateforward = ratereverse

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= = K the equilibrium constant kforward kreverse

[products]n

[reactants]m

This is also known as the LAW OF MASS ACTION.

aA + bB cC + dD

rate of forward reaction = kf[A]a[B]b rate of reverse reaction = kr[C]c[D]d ([..] = concentration (M) at “equilibrium”) kf[A]a[B]b = kr[C]c[D]d

kf

kr

3

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Approach to equilibrium…..

Equilibrium achieved

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-Pure solids and pure liquids are materials whose concentration doesn’t change during the course of a reaction  its amount can change, but the amount of it in solution doesn’t

• because it isn’t in solution

-Because their concentration doesn’t change, solids and liquids are not included in the equilibrium constant expression

-For the reaction aA(s) + bB(aq) ⇔ cC(l) + dD(aq) the equilibrium constant expression is

Equilibrium Constants for Heterogeneous Systems

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Figure 17.4

solids do not change their

concentrations

The Equilibrium Constant for a heterogeneous system.

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Write the equilibrium equation for each of the following reactions:

(a) CO2(g) + C(s) 2 CO(g)

(b) Hg(l) + Hg2+(aq) ! !Hg22+(aq)

(c) 2 Fe(s) + 3 H2O(g) ! Fe2O3(s) + 3 H2(g)

(d) 2 H2O(l) 2 H2(g) + O2(g)

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Figure 17.2

small K < 10-3

large K >10+3

intermediate 10-3 < K < 10+3

The range of equilibrium constants.

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Product- or Reactant-Favored.

Product-favored K > 1

Reactant-favored K < 1

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aA + bB cC + dD Kc = [C]c[D]d

[A]a[B]b

cC + dD aA + bB K' = 1/Kc

aA + bB cC + dD n Kc' = (Kc)n

For a sequence of equilibria, Koverall = K1 x K2 x K3 x …

Calculating Variations on K

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K1 = [SO2][O2]

K2 = [SO3][SO2][O2]1/2

Knet = [SO3][O2]3/2 Knet = K1 • K2

Net

Multiply their respective K's

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Writing and Manipulating K Expressions

K = [SO3][O2]3/2

Knew = [SO3]2

[O2]3

Knew = [SO3]2

[O2]3 = (Kold)2

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Writing and Manipulating K Expressions

K = [SO2][O2]

Knew = [O2][SO2]

= 1Kold

Knew = [O2][SO2]

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At any time, t, the system can be sampled to determine the amounts of reactants and products present. A ratio of products to reactants, calculated in the same manner as K tells us whether the system has come to equilibrium (Q = K) or whether the reaction has to proceed further from reactants to products (Q < K) or in the reverse direction from products to reactants (Q > K).

For a general reaction aA + bB cC + dD

where a, b, c, and d are the numerical coefficients in the balanced equation, Q (and K) can be calculated as

Q = (C)c(D)d

(A)a(B)b

e.g., initial concentrations

Q - The Reaction Quotient

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Figure 17.5

Equilibrium: no net change

Reactants Products

Reaction Progress

reactants products

Reaction Progress

Reaction direction and the relative sizes of Q and K.

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Figure 17.6

PRELIMINARY SETTING UP

SOLVING FOR X AND EQUILIBRIUM QUANTITIES

1. Write the balanced equation. 2. Write the Eq. Constant, K. 3. Convert all of the amounts into

the correct units (M or atm).

6.  Substitute the quantities into K. 7.  To simplify the math, assume that x is

negligible. 8.  [A]ini - x = [A]eq = [A]init 9.  Solve for x. 10. Find the equilibrium quantities.

 Check to see that calculated values give the known K.

 Check that assumption is justified (<5% error). If not, solve quadratic equation for x.

Steps in solving equilibrium problems.

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Determining K

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Another Example of Calculating K

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

Kc = [HI]2

[H2][I2 ] = 55.3

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Nitrogen Dioxide Equilibrium N2O4(g) 2 NO2(g)

Kc = [NO2]2

[N2O4] = 0.0059 at 298 K

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When a chemical system at equilibrium is subjected to a stress, the system will return to equilibrium by shifting to reduce the stress.

If the concentration increases, the system reacts to consume some of it.

If the concentration decreases, the system reacts to produce some of it.

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The effect of a change in concentration.

– no change in K – only the equilibrium composition changes.

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Concentration (I) PCl3(g) + Cl2(g) PCl5(g)

Disturbance

New initial

Change

New equilibrium

*Experimentally determined value.

Original equilibrium 0.200 0.125 0.600

+0.075

0.200 0.200 0.600

-x -x +x

0.200 - x 0.200 - x 0.600 + x

(0.637)*

The Effect of Added Cl2 on the PCl3-Cl2-PCl5 System

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Figure 17.8

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

The effect of added Cl2 on the PCl3-Cl2-PCl5 system.

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Figure 17.9

+

lower P (higher V)

more moles of gas

higher P (lower V)

fewer moles of gas

The effect of pressure (volume) on a system at equilibrium.

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Increase P in the system by reducing the volume (at constant T).

Kc = [NO2]2

[N2O4] = 0.0059 at 298 K

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Only temperature changes can alter K.

Consider heat as a product or a reactant.

In an exothermic reaction, heat is a product. In an endothermic reaction, heat is a reactant.

• A temperature rise will increase K for a system with a positive ΔHorxn.

• A temperature rise will decrease K for a system with a negative ΔHorxn.

What will decreases in temperature do to K?

The Effect of a Change in Temperature on an Equilibrium

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Kc = [NO2]2

[N2O4]

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K = [isobutane][butane]

= 2.5

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•  At equilibrium with [iso] = 1.25 M and [butane] = 0.50 M. K = 2.5.

•  Add 1.50 M butane. •  When the system comes to equilibrium again,

what are [iso] and [butane]?

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