Unit 5Properties of Solutions and Chemical

Kinetics

A solution forms when one substance evenly disperses in another.

The ability for two substances to form solutions depends on two general factors:◦ The types of intermolecular forces involved.◦ The natural tendency of substances to disperse

throughout one another.

The Solution Proces

Any of the intermolecular forces we discussed earlier in the year can play a part in the solution process.

For example Ion-Dipole forces dominate in solutions made from ionic compounds dissolved in water.

London Dispersion forces dominate when a nonpolar substance dissolves in a nonpolar solvent.

The forces we must consider when trying to determine if a solution will form are:

Solute-solvent, solute-solute, and solvent-solvent

The Effect Of Intermolecular Forces

These interactions can be analyzed by examining the energy changes associated with the solvation process.

We have seen that NaCl dissolves in water because the ion-dipole interactions between the water and the separate sodium and chlorine ions is stronger than the ion-ion interactions.

The water molecules don’t just have to form ion-dipole interactions with the Na+ and Cl- ions.

They must also break the hydrogen bonds they have between them.

Energy Changes and Solution Formation

Three Stages of Solvation

As a solid solute begins to dissolve in a solvent the concentration of solute particles in solution increases.

Sometimes previously dissolved solute particles come into contact with the surface of the still undissolved solid and can undergo a process called crystallization.

A solution that has reached this equilibrium is said to be saturated.

A super saturated solution is one that is under very specific conditions that allows it to contain more solute than should be possible.

Saturated Solutions

Solubility is defined as the maximum amount of solute that will dissolve in a given amount of solvent at a specific temperature (solubility changes with temperature)

Solubility

We already know how to use molarity (mol/L)

Mass Percent:

Mole Fraction:

Molality:

Ways of Expressing Concentration

Chemical KineticsChapter 14

Chemical kinetics is the branch of chemistry concerned with the speed, or rate, of reactions.

Chemical Kinetics

1. The physical state of the reactants.2. The concentration of the reactants.3. The temperature at which the reaction

occurs.4. The presence of a catalyst.

Factors That Affect Reaction Rates

The speed of any even is defined as the change that occurs in a given interval of time.

The speed, or rate, of a chemical reaction is the change in the concentration of reactants per unit time.

The units for rate are usually molarity per second (M/s)

Reaction Rates

Example

Consider the reaction between butyl chloride and water:C4H9Cl(aq) + H2O(l) C4H9OH(aq) + HCl(aq)

Suppose we prepare a 0.100 M aqueous solution of butyl chloride, mix with water and measure the concentration of butyl chloride at various times after t = 0 sec.

It is typical for rates to decrease as a reaction proceeds.

Change of Rate with Time

In the butyl chloride example the rate of disappearance of reactant was equal in magnitude but opposite in sign to the rate of appearance of the product.

This is because the stoichiometry was all one to one .

Consider the reaction:2 HI(g) H2(g) + I2(g)

Reaction Rates and Stoichiometry

(a) How is the rate at which ozone disappears related to the rate at which oxygen appears in the reaction:

2 O3(g) 3 O2(g)

(b) If the rate at which O2(g) appears is 6.0 x 10-5 M/s at a particular instant what is the rate of disappearance of O3(g)?

Sample Exercise 14.3

One way of studying the effect of concentration on reaction rate is to determine how the initial concentrations of reactants affect the initial rate.

The general form of a rate law is:

The Rate Law

Experiment Number

[A](M) [B](M) Initial rate (M/s)

1 0.100 0.100 4.0 x 10-5

2 0.100 0.200 4.0 x 10-5

3 0.200 0.100 16.0 x 10-5

Using Experimental Data To Determine Rate Law

Experiment Number

Initial [NH4+]

(M)Initial [NO2

-] (M)

Observed Initial Rate

(M/s)

1 0.0100 0.200 5.4 x 10-7

2 0.0200 0.200 10.8 x 10-7

3 0.0400 0.200 21.5 x 10-7

4 0.200 0.0202 10.8 x 10-7

5 0.200 0.0404 21.6 x 10-7

6 0.200 0.0808 43.3 x 10-7

Initial Reaction Rate Data

Reaction Order

A first-order reaction is one whose rate depends on the concentration of a single reactant:

A Products

The integrated rate law relates the concentration of A at the start of the reaction to its concentration at any other time (t).

The Change Of Concentration With Time

The decomposition of a certain insecticide in water follows first-order kinetics with a rate constant of 1.45 yr-1 at 12oC. A quantity of this insecticide was washed into a lake on June 1st, leading to a concentration of 5.0 x 10-7 g/cm3. Assume that the lake holds a constant temperature of 12oC. (a) What is the concentration of the insecticide on June 1st of the following year. (b) How long will it take for the concentration of the insecticide to decrease to 3.0 x 10-7 g/cm3?

Sample Exercise 14.7

A second order reaction is one whose rate depends on the concentration of two different reactants.

Second Order Reactions

The half life of a reaction, t1/2, is the time required for the concentration of a reactant to reach one-half of its initial value.

Half-life is especially useful for first order reactions.

Half-Life

The rates of most chemical reactions increase as the temperature rises.

This is due to the “Collision Model” The collision model states that: In order for particles to react, they must

contact one another in the correct orientation and with sufficient energy.

Increasing the temperature increases the kinetic energy of the particles as well as the probability that they will hit each other with the correct orientation.

Temperature And Rate

Activation energy is defined as the difference in energy between the starting molecule and the highest energy point along the reaction pathway.

This is the energy required to create the activated complex.

The activation energy of any reaction is calculated using the Arrhenius equation:

k = A-Ea/RT

Where k is the rate constant, A is the frequency factor for the reaction in question, Ea is activation energy, R is the gas constant (8.314 J/mol-K) and T is the temperature in Kelvin.

Activation Energy

A more useful form of the equation is as follows:

Using The Arrhenius Equation

Temperature (oC) k (s-1)

189. 7 2.52 x 10-5

198.9 5.25 x 10-5

230.3 6.30 x 10-4

251.2 3.16 x 10-3

Determining the Energy of Activation. Sample Exercise 14.11

The process by which a reaction occurs is called the reaction mechanism.

At the most sophisticated level a reaction mechanism will describe in great detail the order in which bonds are broken and formed during a chemical reaction.

Reaction Mechanisms

We have seen the reaction that occurs when two molecules of CH3NC collide with one another:

Along the same lines the reaction between NO and O3 appears to occur as the result of a single collision between an NO and an O3 molecule with sufficient energy and in the right orientation.

Elementary Reactions

Reactions that occur in a single event or step are called elementary reactions.

The number of molecules that participate as reactants in an elementary reaction defines its molecularity.

If a single molecule is involved the reaction is said to be unimolecular.

If two molecules are involved the reaction is said to be bimolecular.

If three molecules are involved it is said to be termolecular.

Elementary Reactions

Many reactions often occur as the result of multistep mechanisms.

A Multistep mechanism is simply a series of elementary reactions.

For example:

Multistep Mechanisms

It has been proposed that the conversion of ozone in to O2 proceeds by a two-step mechanism:

O3(g) O2(g) + O(g)

O3(g) + O(g) O2(g)

(a) describe the molecularity of each step (b) write the over all reaction. (c) Identify any intermediates.

Sample Exercise 14.12

For any elementary reaction its rate law is based on its molecularity

Example: A Products rate = k[A] This is a first order, unimolecular reaction. For a bimolecular elementary reaction: A+A Products, Rate = k[A]2

A+B Products, Rate = k[A][B] For a termolecular elementary reaction: A+A+A products, Rate = k[A]3

A+A+B products, Rate = k[A]2[B] B+B+B Products, Rate = k[B]3

Rate Laws For Elementary Reactions

If the following reaction occurs in a single elementary reaction, predict its rate law:

H2(g) + Br2(g) 2 HBr

Sample Exercise 14.13

Most chemical reactions occur by multistep mechanisms.

Each step of a multistep mechanism has its own rate constant (k) and its own activation energy.

Often times one of the steps moves much slower than the others.

The over all rate of any reaction cannot exceed the rate of the slowest elementary step.

The slow step of any multistep reaction mechanism is referred to as the rate limiting step.

The rate-determining step for a multistep mechanism

The decomposition of nitrous oxide, N2O, is believed to occur by a two-step mechanism:

N2O(g) N2(g) + O(g) Slow

N2O(g) + O(g) N2(g) + O2(g) Fast

(a) Write the equation for the over all reaction. (b) Write the rate law for the over all reaction.

Mechanisms With a Slow Initial

For a reaction with a fast initial step it is more difficult to figure out the rate law.

This is because one of the reactants in the rate determining step (slow step) is actually an intermediate.

Mechanisms with a fast initial step

Consider: 2 NO(g) + Br2(g) 2 NOBr(g) When the rate law is determined

experimentally we get: rate = k[NO]2[Br] This rate law could indicate that the

reaction occurs as a termolecular elementary reaction.

However these are very rare. This reaction really occurs as two steps

Example

Step 1: NO(g) + Br2(g) NOBr2(g) (Fast)

Step 2: NOBr2(g) + NO(g) 2 NOBr(g) (Slow)

Because step 2 is the slow step the rate law for the over all reaction should be the rate law of step 2:

But because NOBr2 is an intermediate it is never present in any measurable quantity.

This means that as soon as it is formed it is used up. So how can step 2 be the slow step?

When ever the first step of a reaction is the fast step we can solve for [intermediate] by assuming that equilibrium occurs during that first step.