Unit 11: Reaction Rates &

Equilibrium Chapter 18

This tutorial is designed to help students understand scientific measurements.

Objectives for this unit appear on the next slide.

◦ Each objective is linked to its description.

◦ Select the number at the front of the slide to go directly to its description.

Throughout the tutorial, key words will be defined.

◦ Select the word to see its definition.

Objectives

27 State five factors that affect reaction rates

28 Explain activation energy, including how catalysts, inhibitors, and enzymes affect the activation energy

29 Calculate the rate of consumption or production of a chemical reaction

30 Determine the rate law equation, and calculate the rate of reaction using the rate law equation

31 Define reversible, completion, and equilibrium reactions

32 State Le Chatelier’s Principle, and describe equilibrium shifts

33 Write Keq expressions and solve them, determining if reactants or products are favored

27 Affecting the Rate

At times, it is necessary to change the

rate of the reaction.

There are five factors that affect on fast

or slow a reaction will occur.

◦ Temperature

◦ Surface Area

◦ Concentration

◦ Orientation

◦ Nature of the Reaction

Temperature

In order for chemicals to react, they must run into each other.

◦ This happens randomly but can be helped by putting the chemicals in close proximity to each other.

Recall that temperature measures the average kinetic energy of particles so the higher the temperature, the faster the particles will move.

If particles move faster, they will have more collisions thus speeding up a reaction.

If particles move slower, they will have fewer collisions thus slowing down a reaction.

Surface Area

A chemical reaction will occur faster if several collisions can take place at once.

Therefore, the more open a molecule is, the faster it will react.

By increasing the surface area, more collisions will occur thus speeding up the reaction rate.

By decreasing the surface area, less collisions will occur thus slowing down the reaction rate.

Concentration

A chemical reaction will occur more quickly if more chemicals are present

Therefore, higher concentrations result in faster rates.

Lower concentrations result in slower rates.

The concentration of the reactants have a direct affect on the rate law which will be discussed later.

Orientation

Certain molecules have only a portion of

their structure that is reactive.

Therefore, the collision must take place at

that position.

If the reactants are oriented in the

correct position, a reaction will occur.

If the reactants are oriented incorrectly, a

reaction will not occur.

Nature of the reaction

Everything in nature has particular attributes.

Some reactions will be fast and some will be slow.

For each reaction, it is possible to determine the energy needed behind each collision to make a reaction proceed.

This energy is known as the activation energy.

28 Activation Energy

Activation energy is the energy required

to make a chemical reaction proceed.

The more activation energy required, the

harder it will be for a chemical reaction

to occur.

It is often measured on a diagram that will

be shown on the next page.

Activation Energy Diagrams

Reactants and products each have a certain energy. ◦ Reactants are represented by A + B ◦ Products are represented by C + D

The activation energy is the difference between the reactant energy and the top of the curve.

Image from: http://www.saskschools.ca/curr_content/chem30_05/2_kinetics/practice/q3_2.htm

http://www.saskschools.ca/curr_content/chem30_05/2_kinetics/practice/q3_2.htm

Exothermic vs. Endothermic

An exothermic reaction has more energy released with the products than the activation energy.

An endothermic reaction has less energy released with the products than the activation energy.

Image from: http://www.saskschools.ca/curr_content/chem30_05/2_kinetics/practice/q3_2.htm

http://www.saskschools.ca/curr_content/chem30_05/2_kinetics/practice/q3_2.htm

Changing the activation energy

Certain chemicals can be added to a

system to either increase or decrease the

activation energy required.

◦ Catalysts will decrease the activation energy.

◦ Inhibitors will increase the activation energy.

Recall the lower the activation energy, the

easier it will be for the reaction to

proceed.

Enzymes are biological catalysts.

29 Rates of

Consumption/Production Chemical reactions do not always

proceed at the same speed.

The speed of the reaction is measured in

units of moles/second.

For the reactants, this is called the rate of

consumption.

◦ How fast reactants are used up.

For the products, this is called the rate of

production

◦ How fast products are made.

Reactions Rates.

To calculate a reaction rate, the number of moles and time would be required.

◦ For instance, if 15 moles were consumed in 5 seconds, the reaction rate would be 3 moles/second.

◦ It is a simple matter of dividing the moles by the time.

In the chemical reaction, each chemical can have a different rate and each can be determined if one is known.

Reaction Rates

The mole ratio will allow the rates of

each component to be calculated if one

rate is known.

By multiplying the known rate by the

mole ratio, the other rates can be

determined.

Reactions Rates

Assume you had the reaction below and you measured the rate of consumption of sodium to be 2 moles/sec.

2Na + Cl2 2NaCl

If the rates for Cl2 and NaCl were needed, just use the mole ratio.

◦2 𝑚𝑜𝑙𝑒𝑠

1 𝑠𝑒𝑐𝑜𝑛𝑑𝑥

1 𝑚𝑜𝑙𝑒 𝐶𝑙2

2 𝑚𝑜𝑙𝑒 𝑁𝑎= 1 mole/sec

◦2 𝑚𝑜𝑙𝑒𝑠

1 𝑠𝑒𝑐𝑜𝑛𝑑𝑥

2 𝑚𝑜𝑙𝑒 𝑁𝑎𝐶𝑙

2 𝑚𝑜𝑙𝑒 𝑁𝑎= 2 mole/sec

Rate of

consumption of

chlorine

Rate of production

of NaCl

Reaction Rates

It is useful to know the rate of a reaction because it will allow you to perform certain calculations.

Consider the following problem:

How long would it take to produce 20 grams of hydrogen gas if HCl is consumed at a rate of 10 moles/sec using the reaction below?

Zn + 2HCl ZnCl2 + H2

Reaction Rates

To solve the problem on the previous slide, it would be necessary to know the rate of production of H2.

◦10 𝑚𝑜𝑙𝑒 𝐻𝐶𝑙

1 𝑠𝑒𝑐𝑥

1 𝑚𝑜𝑙𝑒 𝐻2

2 𝑚𝑜𝑙𝑒 𝐻𝐶𝑙= 5 mole/sec H2

Next, we would need to know the number of moles of hydrogen we were trying to produce.

◦ 20 grams x 1 𝑚𝑜𝑙𝑒

2.02 𝑔𝑟𝑎𝑚𝑠 = 9.9 moles

Finally, we would use dimensional analysis to determine the time.

◦ 9.9 moles x 1 𝑠𝑒𝑐

5 𝑚𝑜𝑙𝑒 𝐻2= 1.98 𝑠𝑒𝑐𝑜𝑛𝑑𝑠

30 Concentration and Reaction

Rates Concentration of the reactants can have a

major affect on the rat of a reaction.

The concentration can relate to the rate

either linearly or exponentially.

◦ On rare occasions the concentration will have little affect but it is very rare.

This relationship is very consistent for

each reaction and can be used to

determine a rate law for each reaction.

Rate Laws

A rate law compares the concentration of

the reactants to the reaction rate.

It cannot be determined by simply looking

at a balanced chemical reaction.

Rather, experimental data will need to be

analyzed to determine the rate law.

Rate Laws

A rate law will have the basic set up:

◦ For one reactant problems:

Rate = k [A]n

◦ For two reactant problems Rate = k [A]n[B]m

In these rate laws:

k represents a constant

[A] represents the concentration of that reactant

n and m represent the order of the reaction

Calculating a Rate Law

For the generic reaction A B + C the following data was collected above.

To calculate the rate law, the first step is to determine the order.

◦ Look at the factor with which the concentration and the rate change.

◦ Those factors can be placed in the rate law equation and allow us to solve for n.

Trial Concentration

of A

Rate of

Reaction

1 1 M 2 mole/sec

2 2 M 16 mole/sec 8 2

Calculating a Rate Law

Using the factors: 8 = 1[2]n n = 3

◦ For this step k=1

◦ The order of this reaction is 3

Using the order and one of the trials, we can solve for k.

◦ Using trial #1: 2 mole/sec = k[1M]3

◦ K = 2

◦ K will have units but for our purposes, we will just use the number.

Trial Concentration

of A

Rate of

Reaction

1 1 M 2 mole/sec

2 2 M 16 mole/sec

Calculating Rate Laws

The rate law for the previous slides

would be:

Rate = 2[A]3

With rate law, we would know that the

reaction is a third order reaction.

We would also be able to calculate the

rate if given a concentration for [A].

Calculating Rate Laws

Single reactant rate laws are simpler to

use but less common.

More common will be two reactant

reactions.

The process will be similar though.

◦ Determine the order (exponents)

◦ Select one trial

◦ Calculate K

Calculating Rate Laws

To determine the exponent on [A],

choose trials where [B] is constant.

◦ If both [A] and [B] change, it is impossible to know which changed the rate.

To determine the exponent on [B]

choose trials where [A] is constant.

Trial Concentration of A Concentration of B Reaction

Rate

1 1 M 2 M 2 mole /sec

2 2 M 2 M 4 mole /sec

3 2 M 6 M 36 mole/sec

2 2

3 9

Calculating Rate Laws

For [A]: 2 = 1[2]n[1] n= 1

◦ B is 1 because it was constant

For [B]: 9 = 1[1][3]m m= 2

Select a trial to solve for K

◦ Trial #1: 2 mol/sec = K [1M]1[2M]2

◦ K = 0.5

The rate law would be:

Rate = 0.5[A]1[B]2

Calculating Rate Laws

Rate = 0.5[A]1[B]2

The order of this reaction would be:

◦ First in respect to A

◦ Second in respect to B

◦ Third overall

This is determined by adding the exponents.

Explaining some math

You may wonder why we stated that K=1

and [B]=1 at times.

The full math problem for calculating the

order of A for the last problem is shown

below: rate = k[A]n[B]m

4 mole/sec = k[2M]n[2M]m

2 mole/sec = k[1M]n[2M]m

2 = 1[2]n [1]

Therefore n = 1

÷

31 Equilibrium Reactions

There are a select group of reactions that can proceed in both directions. ◦ The reactants form products. ◦ The products reform the reactants. ◦ Equilibrium reactions are indicated by “ “

When a reaction can do this, it is called reversible.

Both reactions occur at the same time but the reaction is only at equilibrium if:

The forward reaction rate is equal to the reverse reactions rate.

Equilibrium Reactions

Most reactions will run to completion.

This means that one of the products is removed from the system either by turning into a gas, becoming a precipitant, or being removed as it is produced.

To make an equilibrium reaction go to completion, one part of the reaction must be removed continuously.

This would force the reaction to proceed in one direction in order to return to equilibrium.

32 Le Chatelier’s Principle

The French scientist Le Chatelier studied

equilibrium reactions and stated the

following principle:

When a stress is placed on an equilibrium

system, the system will shift until it returns

to equilibrium.

Le Chatelier’s Principle

The way an equilibrium reaction will shift

is by increasing or decreasing the rate of

either the forward reaction or the

reverse reaction.

There are a few main stresses that can be

placed on a system.

◦ Concentration of chemicals

◦ Temperature

◦ Pressure

Concentration of Chemicals

Assume there was the generic reaction:

A + B C

If more of A was added, the reaction

would have to shift towards the products

to balance the reaction rates.

If A was removed, the reaction would

have to shift towards the reactants to

balance the reaction rates.

Temperature

Most chemical reactions have either heat

as a product or as a reactant.

Thus the addition or subtraction of heat

can be treated like adding or subtracting a

reactant.

Pressure

Changing the pressure of a system can be

less intuitive to determine the shift.

To bring the system back to balance, the

number of collisions would have to be

increased or decreased.

If the pressure is increased, shift to the

side with fewer moles (coefficients).

If the pressure is decreased, shift to the

side with more moles (coefficients).

Le Chatelier’s Principle

Basic Rules for determining the shift

◦ Concentration or Temperature

If added, shift to the other side.

If removed, shift to that side.

◦ Pressure

If increased, shift towards less moles.

If decreased, shift towards more moles.

Le Chatelier’s Principle Examples

A + B C + Heat

Increase A: Shift to products

Increase C: Shift to reactants

Increase Temp: Shift to reactants

Decrease Temp: Shift to products

Increase Pressure: Shift to products

Decrease Pressure: Shift to reactants

2 moles 1 mole

33 Equilibrium Constants

The relationship between the reactant

and product concentration is equal to a

constant.

This relationship can be mathematically

determined by looking at the balanced

chemical reaction.

Equilibrium Constants

1A + 2B 3C + 4D

𝐾𝑒𝑞 =[𝐶]3[𝐷]4

[𝐴]1[𝐵]2

Products are divided by reactants Coeffecients

become

exponents.

Equilibrium Constants

𝐾𝑒𝑞 =[𝐶]3[𝐷]4

[𝐴]1[𝐵]2

This relationship can be very useful in determining unknown concentrations or conceptual ideas.

◦ For instance, if Keq is below 1, then the reactants are favored in this reaction.

◦ If Keq is above 1, then the products are favored in this reaction.

◦ If Keq is 1, then neither reactants or products are favored.

This concludes the tutorial on

measurements.

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