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Unit A: Reaction Kinetics Chemistry 12

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Page 1: Unit A: Reaction Kinetics - GETTIN' SMART with Mrs. Rrenooy.weebly.com/.../13790098/unit_a-_reaction_kinetics.pdf · 2018. 10. 13. · reaction to occur. •Very few molecules have

Unit A: Reaction KineticsChemistry 12

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Introduction to Reaction Kinetics

• Chemical Kinetics deals with the empirical research of reaction times and changes in measurable properties

• We can use this empirical knowledge to form theories to explain and predict rates of reaction

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Measuring Reaction Rates

• Reaction Rates are usually obtained by measuring how a property changes per unit of time.

• Any property may be used as long as it is OBSERVABLE and MEASURABLE.

• Consider the reaction of Copper metal with Nitric acid to illustrate the various methods that can be employed to measure reaction rates:

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• To measure the rate, the change needs to be measured during a certain time period.

• Any unit of time may be used (minutes, seconds, hours, days, etc.)

• The general way to express a rate is as follows:

• Recall:

• Δ “delta” means: final – initial

• Therefore: Δ time = tfinal - tinitial

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• For Example:

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• If we were to plot the graph of property vs. time for any reaction, the rate would be represented by the slope of the graph.

• For Example:

• Consider two tangents

drawn at different points

on the graph.

• For most reactions, the

concentration changes are

faster at the beginning of the

reaction and tend to decrease

as time elapses.

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Factors Affecting Reaction Rates

When making chocolate milk, does the powder dissolve faster in hot milk or cold milk?

When Magnesium metal is placed in equal volumes of Hydrochloric acid, will the reaction

be faster in 3.0mol/L or 6.0 mol/L HCl(aq)?

• Several factors (5) can affect the rate at which a reaction takes place.

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1. Temperature:

• Reactant molecules posses more kinetic energy at higher temperature.

• More energy means a greater number of successful collisions will occur per unit of time.

• Since the timer required for the reaction to reach completion decreases, the rate of the reaction increases.

2. Pressure / Concentration: (same effect)

• Increasing the pressure/concentration increases the number of reactant species in the solution.

• Having more of the reactant species present means that more will react per unit of time.

• More reactions per unit of time means a higher rate.

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3. Nature of the Reactants:

• Not all reactants are the same. Therefore, not all reactants will react the same way.

• 2 things to consider:

• Chemical Properties:

• Some reactions are naturally fast (combustion of gasoline), and some are naturally slow (rusting of Iron).

• We can only change the rate of these reactions by a small degree by manipulating temperature, pressure, and concentration.

• Number and Types of Bonds Broken and Formed:

• The number (one, two, three, etc.) and type of bond (single, double, triple) can dictate the speed of the reaction.

• The stronger the bond, and the more bonds involved, the slower the reaction.

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4. Surface Area and Phase:

• Surface area• Increasing the surface area have the same effect as increasing

the concentration.

• The ability for the reactants to meet is greater because there are more sites for the reaction to take place.

• More available sites = faster reaction rates.

• Phase of the Reactants • HOMOGENEOUS REACTION: a reaction in which all the

reactants are in the same phase.

• For example:

• Two gasses

• Two substances dissolved in water

• Two liquids that completely dissolve in each other (miscible)

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• HETEROGENEOUS REACTION: a reaction in which the reactants are present in different phases.

• For example:

• A solid and a liquid

• A liquid and a gas

• A solid and a gas

• Two liquids that do not dissolve in each other (immiscible)

• To Summarize Phase:

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5. Catalysts and Inhibitors:

• A CATALYST is a chemical that can be added in order to increasethe rate of the reaction (more to come later).

• After the reaction is complete, the same amount of catalyst will be present as was added in the beginning, it is not consumed during the reaction.

• An INHIBITOR is a chemical that can be added in order to slow a reaction down

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Assignment

• Read: 18-1 to 18-4 on pages 497-501

• Do:

• Workbook #1-8

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Reaction Rate Theory

• Molecules cannot react with one another at a distance.

• In order for there to be a reaction, molecules must collide with each other.

• The KINETIC MOLECULAR THEORY states this and several other important points.

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• A chemical sample consists of particles (molecules, atoms, or ions). They act as small, hard spheres that collide and bounce off each other transferring energy among themselves during collisions.

• An effective collision requires sufficient energy and correct positioning of the particles so that bonds can be broken, and new ones can be formed.

• Ineffective collisions result in the particles rebounding off each other unaltered in nature.

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Enthalpy Changes in Chemical Reactions• Enthalpy

• The total kinetic and potential energy of a system under constant pressure.

• Potential Energy (PE)

• A stored form of energy.

• Exists as a result of the objects position in space and as a sum of all the repulsive and attractive forces amongst the particles that make up the object.

• Directly related to the energy of the electrons in the chemical bonds and the number and type of atoms in the molecule.

• Kinetic Energy (KE)

• A form of energy related to the motion of a particle.

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• When chemical reactions occur:

• Bonds are broken and formed.

• Phases may change.

• Work may be done on or is done by system.

• Heat may be transferred into or out of the system.

• To make all these changes in energy easier to keep track of, we can find that CHANGE IN ENTHALPY (ΔH) of the open system at constant atmospheric pressure.

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The Sign of ΔH

• The sign of ΔH can either be + or – depending whether the reaction is ENDOTHERMIC or EXOTHERMIC

• Endothermic:

• Energy is absorbed by a system

from the surroundings.

• Results in an increase in the

potential energy of the system.

• The products have more energy

than the reactants.

• The loss of energy from the

surroundings to the system

makes the surroundings feel cooler.

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• Therefore:

• The two ways to express this in a chemical equation are:

• In endothermic reactions, energy in the form of heat enters (+) the system.

• The energy term is on the reactants side of the equation.

• You are putting in the energy to make the reaction happen.

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• Exothermic:

• Change in which energy is released

from a system into the surroundings,

resulting in a decrease in potential

energy of the system.

• The products have less energy than the reactants.

• Therefore:

• The two ways to express this chemical equation are:

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• In exothermic reactions, energy in the form of heat exits (-) the system.

• In this case, energy is on the products side of the equation. Energy in the form of heat is one of the products.

• Energy is being lost to the surroundings, thus making the surroundings feel warmer.

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Assignment

• Read:

• 19-4 pages 523-524

• Do:

• #1-2 on page 524

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Kinetic Energy Distributions

• The average kinetic energy system is perceived as the temperature.

• Consider the reaction:

• At room temperature (20°C), the reaction occurs at an undetectable rate.

• At 200°C, the reaction proceeds at a very slow rate.

• At 400°C, the reaction is fairly rapid

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• Consider the kinetic energy distribution of molecules in a system:

• Some molecules have a lot of kinetic energy, some have little.

• When we increase the temperature, we increase the average kinetic energy of the entire system

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• If we change the temperature, as stated before, the KE distribution of the molecules changes:

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• Using the pervious graph:

• @25°C:

• The average kinetic energy is far below the minimum needed for the reaction to occur.

• Very few molecules have sufficient kinetic energy to react; therefore we have a very low reaction rate.

• @200°C:

• The increase in temperature has increased the average kinetic energy of the molecules.

• A higher percentage of molecules now have the minimal amount of kinetic energy to react so the reaction rate increases.

• @400°C:

• At this temperature, the most number of molecules have sufficient kinetic energy to react.

• Here we get the greatest reaction rate.

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• In conclusion…

The increased reaction rate

due to an increase in

temperature is not due

solely to increased collisions

between molecules, but also

to an increase in the

number of molecules that

have sufficient kinetic

energy to react.

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The “Rule of Thumb” for

temperature/rate”:

For a slow reaction, a 10°C temperature increase doubles the

rate.

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Activation Energy

• Every reaction has some sort of barrier that molecules must pass before they can react.

• Imagine rolling a ball down a smooth track shaped as below…

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• Molecules behave much in the same way:

• Point A:

• As molecules begin to approach each other, the like charges of the electrons create repulsive forces which push the molecules apart

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• Point B:

• As the molecule slows down due to those repulsive forces, the KE it possesses begins to convert to PE.

• If the molecule can obtain enough PE by converting KE gained into PE, and adding it to the PE it already possesses, an ACTIVATED COMPLEX is formed.

• Point C:

• Activated Complex:• The structure arrangement of particles representing the

highest potential energy (changeover) point in a chemical reaction.

• Also called an “intermediate” particle.

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• Activation Energy (Ea):• The minimum amount of energy required to change the

reactants into the activated complex.

• At this point, bonds are broken and new ones are formed.

• Point D

• The electrons in the outer shell of the new products begin to repel each other and the products push themselves apart.

• They lose their excess PE by changing back into KE.

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• The imaginative barrier that keeps reactions from occurring is due to the repulsive forces between molecules, as well as the energy required to break the bonds between the atoms.

• The higher the “hill”, the more KE is needed for the reaction to occur, and the slower the reaction

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• There are three possible cases that can occur when two molecules react with one another:

• PE < Ea:

• Molecules do not posses sufficient KE to change to PE to reach the top of the PE hump.

• Molecules will come to a stop before they collide, and then repel away from each other.

• This is call an INNEFECTIVE COLLISION.

• PE = Ea:

• Molecules posses just enough KE to change to PE to reach the top of the PE hump.

• Molecules will approach each other and come to a stop. Since PE = Ea

a reaction may occur, but they may also just move away without reacting.

• PE > Ea:

• Molecules have enough KE to change to PE, plus some KE left over.

• This scenario results in an EFFECTIVE COLLISION.

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• Considering case number three from before, there are two requirements for a successful reaction:

1. Sufficient KE as outline in case 3 • Insufficient KE = insufficient PE

2. Correct Alignment • If the reactants are not correctly aligned, more energy is

required for the reaction to take place

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• PE diagrams give a snapshot of what energy is needed for a reaction to take place.

• They can be read from left to right (forward), or right to left (reverse).

• Reactants can form products and products can re-form reactants

REACTANTS PRODUCTS

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• Endothermic Reactions:

• Energy is absorbed from the surroundings.

• From the direction of the arrow, we can see ΔH < 0

• Exothermic Reactions:

• Energy is released to the surroundings.

• From the direction of the arrow, we can see ΔH < 0

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• Memorizing equations sucks! By sketching a potential energy diagram and labeling the important parts, you can just figure out what information is needed for the question.

1. If ΔH = -32kJ and Ea(f) = 72kJ, what is the value of Ea(r)?

2. Draw and label a PE diagram for the reaction:

CO(g) + NO2(g) CO2(g) + NO(g) + 227kJ

in which Ea(f) = 133kJ. Indicate on your diagram, the point at which the activated complex exists.

3. Draw and label a PE diagram showing the enthalpy change and activation energies for a reaction in which Ea(f) = 135kJ and Ea(r) = 82kJ.

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Reaction Mechanisms

• Reaction Mechanism:

• The individual steps during the progress of a reaction.

• A collision between two particles lasts for a very short time period.

• The chances of having more than two particles colliding in space with enough energy to react and with the correct orientation is very low.

• Consider this:

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• The chances of the twenty three particles coming together, such as the reaction above, are ZERO

• This leads us to conclude that complex reactions cannot go in a single step; most chemical reactions must occur as a sequence of individual steps.

• Consider the following example:

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• The preceding reaction is broken down into a series of three steps:

Step 1: HBr + O2 HOOBr (slow)

Step 2: HOOBr + HBr 2HOBr (fast)

Step 3: 2HOBr + 2HBr H2O + 2Br2 (fast)

Overall: 4HBr + O2 2H2O + 2Br2

The overall reaction is comprised of these three steps

put together

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• There are a couple of important things to remember when talking about reaction mechanisms:

• The slowest step in the reaction is called the RATE DETERMINING STEP.

• The reaction can become instantly “frozen”. This happens because HOOBr is produced very slowly and HOBr is consumed very fast. Thus the concentration of these reagents cannot build up.

• To determine the overall equation for the reaction, add up all the steps in the reaction and then cross out any species that appears on both sides of the final equation.

• For Example:

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• Substances such as HOOBr and HOBr which are formed during the reaction, but react immediately and are not present at the end of the reaction are called REACTION INTERMIDIATES.

• To find the activated complex in a reaction, add up all the atoms involved in that step:

• For Example:

Step 2: HOOBr + HBr 2HOBr

Activated Complex: H2O2Br2

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• What if we have to find a missing step?

• For Example:

The suggested equation:

Overall: CFCl3 + O3 + O CFCl2 + Cl + 2O2

Has a three step mechanism.

If the proposed 1st and 3rd steps are:

Step 1: CFCl3 CFCl2 + Cl

Step 2: ???

Step 3: ClO + O Cl + O2

What is the 2nd step in the proposed reaction?

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Step 1: CFCl3 CFCl2 + Cl

Step 2:

Step 3: ClO + O Cl + O2

Overall: CFCl3 + O3 + O CFCl2 + Cl + 2O2

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• The PE diagram for this type of reaction get a little more complicated.

• Each step has its own activated complex and activation energy.

• Since there are three steps, there will be three “humps” in the PE diagram

Overall: 4HBr + O2 2H2O + 2Br2

• The rate determining step is always the highest peak

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Example 1

Step1: Pd + 2 NO2 PdNO3 + NO

Step 2: CO + PdNO3 NO2 + CO2 + Pd

Overall Rn: CO + NO2 CO2 + NO

a. Find the first elementary process.

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Example 1

Step 1: Pd + 2 NO2 PdNO3 + NO

Step 2: CO + PdNO3 NO2 + CO2 + Pd

Overall Rn: CO + NO2 CO2 + NO

b. When [CO] is increased, the reaction rate doesn’t change. How is this possible?

We conclude that CO is not a reactant in the rate-determining step.

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Example 1

Step 1: Pd + 2 NO2 PdNO3 + NO

Step 2: CO + PdNO3 NO2 + CO2 + Pd

Overall Rn: CO + NO2 CO2 + NO

c. What will happen if the [NO2] is increased, and

why?

Reaction rate will increase because NO2 is a reactant in

a step which must be the rate-determining step

(since step 2 is not)

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56

Example 1

Step 1: Pd + 2 NO2 PdNO3 + NO

Step 2: CO + PdNO3 NO2 + CO2 + Pd

Overall Rn: CO + NO2 CO2 + NO

d. Identify any intermediates.

PdNO3

Why not NO2?

It’s a reactant!

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57

Example 1

Step 1: Pd + 2 NO2 PdNO3 + NO

Step 2: CO + PdNO3 NO2 + CO2 + Pd

Overall Rn: CO + NO2 CO2 + NO

e. Identify any catalysts.

Pd

It was used before it was produced, so it couldn’t be an intermediate.

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Example 2

Given the following P.E. diagram, determine

a. How many elementary processes involved?

Four

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Example 2

Given the following P.E. diagram, determine

b. The rate determining step.

E

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Example 2

Given the following P.E. diagram, determine

c. The activation energy for the overall reaction.

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Example 2

Given the following P.E. diagram, determine

d. Is there more than one activated complex?

Yes, A, C, E and G.

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Given the following P.E. diagram, determine

e. Which locations are associated with intermediates?

B, D, F

Example 2

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Example 2

Given the following P.E. diagram, determine

f. Is the reaction endo- or exo-thermic?

Endothermic. H is positive.

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The Effects of Catalysts

• Catalyst:

A substance that speeds up a reaction rate by providing

an overall reaction with an alternative mechanism having

lower activation energy.

Catalysts provide an alternate pathway by inserting different

intermediate steps and loweringthe activation energy for the reaction to occur. ΔH is not

changes, but the “energy hump” is lowered, therefore, the reaction

rate INCREASES.

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• By lowering the “energy hump” for the forward reaction, we have also lowered it for the reverse reaction

• Since the greater fraction of the product molecules will have sufficient KE to form the activated complex, the reverse reaction rate increases also.

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• For Example:

• Consider the decomposition of Formic acid which occurs at a very slow rate at room temperature:

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• As soon as we acidify the solution with Sulfuric acid, it begins to bubble.

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• Some important things to remember about catalysts:

• The catalyst is an active participant in a reaction which is regenerated in a later step of the reaction mechanism.

• ΔH for the overall reaction is the same for both the catalyzed and un-catalyzed reaction; only the intermediate reactions differ.

• All intermediates and catalysts cancel out when the individual steps are added up to get the overall reaction