Chem 109 - General Chemistry - Spring 2013

Lecture Notes 34: 26 April

Solids, cont.

Types of Solids, cont.

Molecular solids: made up of a single type of covalent molecules, the crystal is held together by weak bonds which hold the molecules together. These solids tend to have low melting points.

Ionic solids: made up of two or more types of ions (e.g. Na⁺ and Cl^-) held together by electrostatic attractions. (Figure 10.35-37, p 456-7 of Zumdahl). Note that cations (which tend to be small) often fit into the "holes of the unit cell made up the (larger) anions. (Figure 10.35, 10.37 p 456-7 of Zumdahl)

Phase Diagrams

Phase diagrams enable us to predict the behavior of a substance under different conditions of temperature and pressure. We will base our discussions on the behavior of two classic cases: water and carbon dioxide. (Note that phase diagrams strictly describe behavior only for pure substances, so they only hold exactly in closed laboratory systems. The main features do describe much of the contaminated world.)

Water (see Figure 10.49, Zumdahl p 479).

public domain image via Wikipedia Creative Commons

First note that water is a bit unusual in that there is a negative slope for the transition from solid to liquid. This is a reflection of the fact that solid water is less dense than liquid water (a rare situation). (example: Ice skating)
Note the unique points:

Critical point. Defined by two values:

the Critical Temperature = the temperature above which liquid cannot exist,
the Critical Pressure = the pressure required to give a liquid at the critical temperature. Note that above the critical point liquids no longer have an equilibrium vapor pressure, nor do they boil, rather they undergo a fluid transition to a gas as the temperature is raised.

Triple Point. This is a unique set of values at which the gas, liquid, and solid phases can co-exist in equilibrium. The triple point of water is used as a standard and calibration point for temperature scales because it is precisely defined and reproducible anywhere.

Carbon Dioxide (see Figure 10.52, Zumdahl p 468).
- Note that the solid-liquid transition has a positive slope in CO₂, which is the common situation for most substances.
- Note that the triple point for CO₂ is above 1 atm, thus liquid CO₂ is not seen in the open on Earth's surface. Rather the solid evaporates directly to a gas (sublimes) at -78 °C at 1 atm.

Chemical Equilibrium

As a general statement can say that all chemical reactions are equilibrium reactions and go toward a state of equilibrium or approach equilibrium. Some reactions reach equilibrium rapidly, some slowly, and some favor products to such an extent that the reactions "go to completion." Regardless of initial concentrations, systems will reach equilibrium given time.

What is an equilibrium - very dynamic situation in chemistry .

Let's look at a classic example of reactions approaching equilibrium, ammonia synthesis. This takes place in the gas phase at high temperature and pressure (Fig 13.5, Zumdahl, p 597; drawn on board):

N_{2 (g)} + 3 H_{2 (g)}

2 NH_{2 (g)}

Vast numbers of chemical reactions operate at equilibrium in the natural world, and it is frequently essential to be able to understand and to predict their behavior.

Quantitatively we can look at the relationship in a simple reaction at equilibrium represented by

At equilibrium this system consists of two reactions proceeding in opposite directions at the identical rate:

C characterized by a constant, k₁ to give a rate, r₁ = k₁[A]

C characterized by a constant, k₂ to give a rate, r₂ = k₂[C]

But if r₁ = r₂, then

k₁[A] = k₂[C], and gathering constants

k₁/k₂ = [C]/[A], the ratio of constants is given a new name, the equilibrium constant, K_eq.

Equilibrium Expression

The Equilibrium Expression is then:

K_eq = [C]/ [A].

Generalizing for the equation (Note that the simple derivation above does not generalize, since rates and stoichiometry are not generally correlated - the relationships between rates at equilibrium and stoichiometry are beyond our study.):

aA + bB + ... cC + dD + ...

K_eq = [C]^c[D]^d/ [A]^a[B]^b.

Mass Action Expression

A similar expression is the Mass Action Expression:

Q = [C]^c[D]^d/ [A]^a[B]^b.

The mass action expression is algebraically identical to the equilibrium expression, but it applies to a more general case. That is, the equilibrium expression requires that the values in the expression give the equilibrium constant, whereas the mass action expression allows any set of values. Thus the mass action expression is used to describe a system which has not yet reached equilibrium, while the equilibrium expression is a special case of the mass action expression for a system at equilibrium.

Qualitatively we can get an idea of how an equilibrium system will behave by using Le Châtelier's Principle.

Le Châtelier's Principle

If stress is applied to a system at equilibrium, the equilibrium will shift in such a way as to relieve the stress. e.g. if the pressure of carbon dioxide is increased over a solution of carbon dioxide in water, more carbon dioxide will dissolve, reducing the pressure increase. Of course there are other consequences, the system can't just abosorb the carbon dioxide. Note the reaction of carbon dioxide and water,

CO₂ + H₂O HCO₃^- + H⁺

So we should also see a drop in pH, that is it becomes more acidic.

CO₂ + H₂O HCO₃^- + H⁺

So we should also see a drop in pH, that is it becomes more acidic.

Note that the system we looked at last time (and in your text), the response of the ammonia synthesis reaction to the introduction of nitrogen, is readily understood using Le Châtelier's Principle. The system reponds to the addition of nitrogen by using some, but not all, up. Note that other aspects of the system also reponded.

Let's look at an equilibrium system, and try to predict its response using Le Châtelier's Principle and/or the equilibrium expression.

Example: Consider the reaction

CO + NO₂ NO + CO₂ + heat (226 kJ)

Note that heat appears on the product side - the system is giving up heat, H is negative, H = - 226 kJ

Writing the equilibrium expression then gives:

K = [NO][CO₂]/[CO][NO₂]

So, what will happen to [CO₂] if we manipulate the reaction in the ways described below? Two ways of approaching: physically or mathematically, that is, we can look at the equation and ask which way it will shift due to concentration change, collisions etc., OR we can look at teh equiilibrium expression and ask how the concentrration terms will respond in order to keep K constant.

CO is added?

[CO₂] will increase. The increased [CO] will react more rapidly with NO₂, driving the reaction to the right and reducing the amount of CO added. (Looking at K, [CO₂] must increase to keep K constant.)

NO₂ is added?

[CO₂] will increase. The increased [NO₂] will react more rapidly with CO, driving the reaction to the right and reducing the amount of NO₂ added. (Looking at K, [CO₂] must increase to keep K constant.)

NO is added?

[CO₂] will decrease. The increased [NO] will react with CO₂, driving the reaction to the left. (Looking at K, [CO₂] must decrease to keep K constant.)

T is increased?

[CO₂] will decrease. If T increases heat must have been added, therefore the reaction will use up some of the extra heat, driving the reaction to the left and lowering the temperature back toward its original value.

V is increased?

[CO₂] will decrease in proportion. Although a larger volume will result in a lower pressure (fewer collisions), we have equal numbers of reactants on both sides, so neither side will be favored. (Looking at K, all concentrations reduced by the same factor, so K remains constant without changing proportions of players.)

Ar is added?

No change. Ar is unreactive and will have no effect.