General organic and biological chemistry structures off life 5th CH10 rates of reactions GOB structures 5th ed

General, Organic, and Biological Chemistry: Structures of Life, 5/eKaren C.. General, Organic, and Biological Chemistry: Structures of Life, 5/eKaren C.. General, Organic, and Biological

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General, Organic, and Biological Chemistry: Structures of Life, 5/e

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Karen C Timberlake

A neonatal nurse works with newborns that are

premature or have birth defects, cardiac

malformations, and surgical problems

Most neonatal nurses care for infants from the time

of birth until the time they are discharged from the

hospital

Chapter 4 Reaction Rates and Chemical Equilibrium

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• Writing Numbers in Scientific Notation (1.4F)

Core Chemistry Skills

• Using Significant Figures in Calculations (2.3)

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Karen C Timberlake

10.1 Rates of Reactions

Reaction rates vary greatly for everyday

processes A banana ripens in a few days,

silver tarnishes in a few months, while the

aging process of humans takes many years.

Learning Goal Describe how temperature, concentration, and catalysts affect the rate of a

reaction.

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Karen C Timberlake

Activation Energy

• Even when a collision has the proper orientation, there still must be sufficient energy to break the

bonds between the atoms of the reactants

Three Conditions Required for a Reaction to Occur

1 Collision The reactants must collide

2 Orientation The reactants must align properly to break and form bonds

3 Energy The collision must provide the energy of activation

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Rate of Reaction

The rate (or speed) of a reaction is determined by measuring the amount of

• reactant used up in a certain period of time.

• product formed in a certain period of time.

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Factors That Affect the Rate of a Reaction

Reactions with low activation energies go faster than reactions with high activation

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Karen C Timberlake

Rate of Reaction: Temperature

At higher temperatures the increase in kinetic energy of the reactant molecules

For every 10 °C increase in temperature, most reaction rates approximately double.

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Karen C Timberlake

Rate of Reaction: Reactant Concentration

When there are more reacting

molecules, more collisions that

form products can occur, and the

reaction goes faster.

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Rate of Reaction: Catalysts

• Adding a catalyst speeds up the rate of the

reaction by providing an alternative pathway that

has a lower activation energy

• When activation energy is lowered, more

collisions provide sufficient energy for reactants to

form product

• During a reaction, a catalyst is not changed or

consumed

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Karen C Timberlake

Factors That Affect Reaction Rates

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2CO(g) + O2(g)  2CO2 (g)

A raising the temperature

B removing O2

C adding a catalyst

D lowering the temperature

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2CO(g) + O2(g)  2CO2 (g)

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A decreasing the temperature

B removing one of the reactants

C adding a catalyst

D placing the reaction flask in ice

E increasing the concentration of a reactant

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Karen C Timberlake

Study Check

State the effect of each on the rate of reaction as (I) increases, (D) decreases, or (N) no change

A decreasing the temperature (D) decreases

B removing one of the reactants (D) decreases

C adding a catalyst (I) increases

D placing the reaction flask in ice (D) decreases

E increasing the concentration of a

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Karen C Timberlake

Catalytic converters are used in automobile engines to reduce pollutants such as carbon monoxide (CO), hydrocarbons such as octane (C8H18), and nitrogen oxide (NO)

Chemistry Link to the Environment:

Catalytic Converters

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Karen C Timberlake

When pollutants pass through the surface, they react with the catalysts and are converted to CO2, N2, O2, and H2O

Chemistry Link to the Environment:

Catalytic Converters

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Karen C Timberlake

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Karen C Timberlake

10.2 Chemical Equilibrium

In most chemical reactions, the reactants are not completely converted to products

because a reverse reaction takes place in which products collide to form the reactants

Learning Goal Use the concept of reversible reactions to explain chemical equilibrium.

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Reversible Reactions

When a reaction proceeds in both a forward and a reverse direction, it is said to be a reversible reaction.

As the reactants, H2 and I2, collide, the forward reaction begins

HI molecules begin to form and collide with each other to form reactants in the reverse reaction This

reversible reaction is written with a double arrow

H2(g) + I2(g) 2HI(g)

forward

reverse

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Rate of Reversible Reactions

As the reaction progresses, the rate of the forward reaction decreases and that of the reverse reaction

increases At equilibrium, the rates of the forward and reverse reactions are equal

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Equilibrium and Reversible Reactions

Equilibrium is reached when there are no further changes in the concentrations of reactants and products

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Rates of Forward and Reverse Reactions

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• the forward reaction is H2(g) + I2(g) 2HI(g).

• the reverse reaction is 2HI(g) H2(g) + I2(g).

As HI product builds up, the rate of the reverse reaction increases, while the rate of the

forward reaction decreases

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Karen C Timberlake

Equilibrium

A reaction reaches equilibrium when no further changes take place in the

concentration of the reactants and products.

At equilibrium,

• the rate of the forward reaction is equal to the rate of the reverse reaction.

• the forward and reverse reactions continue at the same rate.

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Forward and Reverse Reactions

• If we start with reactants SO2 and O2, the reaction to form SO3 takes place until

equilibrium is reached

• If we start with only the product SO3, the reaction to form SO2 and O2 takes place until

equilibrium is reached.

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Karen C Timberlake

Study Check

Complete each sentence with

1) equal 2) not equal 3) forward

4) reverse 5) changes 6) does not change

A Reactants form products in the _ reaction

B At equilibrium, the reactant concentration _

C Products form reactants in the _ reaction

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Karen C Timberlake

Solution

Complete each sentence with

1) equal 2) not equal 3) forward

4) reverse 5) changes 6) does not change

A Reactants form products in the forward reaction

B At equilibrium, the reactant concentration does not change

C Products form reactants in the reverse reaction

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Karen C Timberlake

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Karen C Timberlake

10.3 Equilibrium Constants

At equilibrium, the number of

people riding up the lift and the

number of people skiing down the slope are constant.

Learning Goal Calculate the equilibrium constant for a reversible reaction given the

concentrations of reactants and products at equilibrium.

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Equilibrium Constants

An equilibrium constant for a reversible chemical reaction

• multiplies the concentrations of the products together and divides by the concentrations of the

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Guide to Writing an Equilibrium Expression

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Study Check

Write an equilibrium expression for the following reaction:

2CO(g) + O2(g) 2CO2(g)

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Solution

2CO(g) + O2(g) 2CO2(g)

STEP 1 Write the balanced chemical equation

2CO(g) + O2(g) 2CO2(g)

STEP 2 Write the concentrations of the products as the numerator and the reactants as the

denominator

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Solution

2CO(g) + O2(g) 2CO2(g)

STEP 3 Write any coefficient in the equation as an exponent.

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Calculating Equilibrium Constants

The equilibrium constant, Kc, is the numerical value obtained by substituting experimentally measured

molar concentrations at equilibrium into the expression

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For example,

• the concentrations of each species in Experiment 1 are

[H2] = 0.10 M, [I2] = 0.20 M, and [HI] = 1.04 M, and the Kc is

• in additional Experiments 2 and 3, the mixtures have different equilibrium concentrations for the

system at equilibrium at the same temperature, but they have the same value of Kc.

Thus, a reaction at a specific temperature can have only one value for the equilibrium constant.

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The units of Kc depend on the specific equation In this example, the units of [M2]/[M2] cancel out to give a

value of 54 In this text, the numerical value will be given without any units

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Guide to Calculating the Kc

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0.31 M NO2?

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0.31 M NO2?

STEP 1 State the given and needed quantities

ANALYZE Given Need

THE PROBLEM 0.45 M N2O4 Kc

0.31 M NO2 Equation

N2O4(g) 2NO2(g)

THE PROBLEM 0.45 M N2O4 Kc

0.31 M NO2 Equation

N2O4(g) 2NO2(g)

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mixture at equilibrium contains 0.45 M N2O4 and 0.31 M NO2?

STEP 2 Write the Kc expression for the equilibrium.

STEP 3 Substitute equilibrium (molar)

concentrations and calculate Kc.

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[H2] = 1.2 M [I2] = 1.2 M [HI] = 0.35 M

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[H2] = 1.2 M [I2] = 1.2 M [HI] = 0.35 M

STEP 1 State the given and needed quantities.

THE PROBLEM 1.2 M H2, 1.2 M I2, Kc

0.35 M HI

Equation

H2(g) + I2(g) 2HI(g)

THE PROBLEM 1.2 M H2, 1.2 M I2, Kc

0.35 M HI Equation

H2(g) + I2(g) 2HI(g)

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[H2] = 1.2 M [I2] = 1.2 M [HI] = 0.35 M

STEP 2 Write the Kc expression for equilibrium

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[H2] = 1.2 M [I2] = 1.2 M [HI] = 0.35 M

STEP 3 Substitute equilibrium (molar) concentrations

and calculate Kc.

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Karen C Timberlake

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Karen C Timberlake

In the reaction of SO2(g) and O2(g), the equilibrium mixture contains

mostly product SO3(g), which results in a large Kc.

Learning Goal Use an equilibrium constant to predict the extent of the reaction and to calculate

equilibrium concentrations

10.4 Using Equilibrium Constants

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Karen C Timberlake

The values of Kc can be large or small, depending on whether equilibrium is reached with

• more products than reactants.

• more reactants than products.

However, the size of the equilibrium constant does not affect how fast equilibrium is

reached.

Using Equilibrium Constants

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Karen C Timberlake

Reactions with a large Kc have large amounts of products produced from

the forward reaction at equilibrium.

The equilibrium constant for the reaction of SO2 and O2 has a large Kc At

equilibrium, the reaction mixture contains mostly product and few

reactants.

2SO2(g) + O2(g) 2SO3(g)

Equilibrium with a Large Kc

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