The molecular nature of matter and change bài giảng môn học general chemistry ii

Kinetics: Rates and Mechanisms of Chemical Reactions16.1 Focusing on Reaction Rate 16.2 Expressing the Reaction Rate 16.3 The Rate Law and Its Components 16.4 Integrated Rate Laws: Conce

Kinetics: Rates and Mechanisms of Chemical Reactions

16.1 Focusing on Reaction Rate

16.2 Expressing the Reaction Rate

16.3 The Rate Law and Its Components

16.4 Integrated Rate Laws: Concentration Changes over Time

16.6 Catalysis: Speeding Up a Reaction

16.5 Theories of Chemical Kinetics

Factors That Influence Reaction Rate

• Particles must collide in order to react

• The higher the concentration of reactants, the greater

the reaction rate

– A higher concentration of reactant particles allows a greater number of collisions.

• The physical state of the reactants influences reaction

rate

– Substances must mix in order for particles to collide.

• The higher the temperature, the greater the reaction

rate

– At higher temperatures particles have more energy and therefore collide more often and more effectively.

Expressing the Reaction Rate

Reaction rate is measured in terms of the changes in

concentrations of reactants or products per unit time

For the general reaction A → B, we measure the

concentration of A at t1and at t2:

The negative sign is used because the concentration of A

is decreasing This gives the rate a positive value.

10.0

Concentration of O 3

(mol/L)

3.20x10 -5 2.42x10 -5 1.95x10 -5 1.63x10 -5 1.40x10 -5 1.23x10 -5 1.10x10 -5

where a, b, c, and d are the coefficients for the balanced

equation, the rate is expressed as:

16-15

Concentration with Time PROBLEM: Hydrogen gas has a nonpolluting combustion product

(water vapor) It is used as a fuel abord the space shuttle

and in earthbound cars with prototype engines:

PLAN: We choose O2 as the reference because its coefficient is 1 For

every molecule of O 2 that disappears, two molecules of H 2

disappear, so the rate of [O 2 ] decrease is ½ the rate of [H 2 ]

decrease Similarly, the rate at which [O 2 ] decreases is ½ the rate

at which [H 2 O] increases.

16-16

The Rate Law

For any general reaction occurring at a fixed temperature

aA + bB → cC + dD

Rate = k[A] m[B]n The term k is the rate constant, which is specific for a

given reaction at a given temperature

The exponents m and n are reaction orders and are determined by experiment.

The values of m and n are not necessarily related in any way to

the coefficients a and b.

16-17

Reaction Orders

A reaction has an individual order “with respect to” or “in”

each reactant

For the simple reaction A → products:

If the rate doubles when [A] doubles, the rate depends on

[A]1and the reaction is first order with respect to A.

If the rate quadruples when [A] doubles, the rate depends

on [A]2and the reaction is second order with respect to [A].

If the rate does not change when [A] doubles, the rate does

not depend on [A], and the reaction is zero order with

respect to A

16-18

Figure 16.7 Plots of reactant concentration, [A], vs time for

first-, second-, and zero-order reactions.

first-, second-, and zero-order reactions.

16-20

Individual and Overall Reaction Orders

For the reaction 2NO(g) + 2H2(g) → N2(g) + 2H2O(g):

The rate law is rate = k[NO]2[H2]

The reaction is second order with respect to NO, first

order with respect to H2and third order overall.

Note that the reaction is first order with respect to H2

even though the coefficient for H2in the balanced equation is 2

Reaction orders must be determined from experimental

data and cannot be deduced from the balanced

PLAN: We inspect the exponents in the rate law, not the coefficients

of the balanced equation, to find the individual orders We add

the individual orders to get the overall reaction order.

PROBLEM: For each of the following reactions, use the give rate law

to determine the reaction order with respect to each

reactant and the overall order.

(a) 2NO(g) + O2(g) → 2NO2(g); rate = k[NO]2 [O2]

(b) CH3CHO(g) → CH4(g) + CO(g); rate = k[CH3 CHO] 3/2

(c) H2 O 2(aq) + 3I-(aq) + 2H+(aq) →I3-(aq) + 2H2O(l); rate = k[H2 O 2 ][I - ]

16-22

Determining Reaction Orders

For the general reaction A + 2B → C + D, the rate law will have the form

Rate = k[A] m[B]n

To determine the values of m and n, we run a series of

experiments in which one reactant concentration changes while the other is kept constant, and we measure the effect on the initial rate in each case

[B] is kept constant for experiments 1 and 2, while [A] is doubled

Then [A] is kept constant while [B] is doubled.

16-24

Rate 2 Rate 1 =

Finding m, the order with respect to A:

We compare experiments 1 and 2, where [B] is kept constant but [A] doubles:

= [A]2[A]1

m

3.50x10 -3 mol/L·s 1.75x10 -3 mol/L·s =

5.00x10 -2 mol/L 2.50x10 -2 mol/L

m

Dividing, we get 2.00 = (2.00)m so m = 1

n

Finding n, the order with respect to B:

We compare experiments 3 and 1, where [A] is kept

constant but [B] doubles:

= [B]

[B]

n

n = [B]3[B] 1

n

3.50x10 -3 mol/L·s

1.75x10 -3 mol/L·s =

6.00x10 -2 mol/L 3.00x10 -2 mol/L

PROBLEM: Many gaseous reactions occur in a car engine and exhaust

system One of these is

NO 2(g) + CO(g) → NO(g) + CO2(g) rate = k[NO2 ]m[CO]n

Use the following data to determine the individual and

overall reaction orders:

Experiment

Initial Rate (mol/L·s)

Initial [NO 2 ] (mol/L) Initial [CO]

PLAN: We need to solve the general rate law for m and for n and

then add those orders to get the overall order We proceed by taking the ratio of the rate laws for two experiments in which only the reactant in question changes concentration.

SOLUTION:

16-29

Scenes

PROBLEM: At a particular temperature and volume, two gases, A (red)

and B (blue), react The following molecular scenes

represent starting mixtures for four experiments:

(a) What is the reaction order with respect to A? With respect to B?

The overall order?

(b) Write the rate law for the reaction.

(c) Predict the initial rate of experiment 4.

PLAN: We find the individual reaction orders by seeing how a change

in each reactant changes the rate Instead of using

concentrations we count the number of particles.

3 2.0x10 -4

Reaction Order Units of k (t in seconds)

General formula:

L mol

unit of t

order-1

Units of k =

The value of k is easily determined from experimental rate

data The units of k depend on the overall reaction order.

and actual rate law

Determine slope of tangent at t 0 for each plot.

Compare initial rates when [A] changes and

[B] is held constant (and vice versa).

Substitute initial rates, orders, and concentrations into rate = k[A] m[B]n , and

solve for k.

16-32

Integrated Rate Laws

An integrated rate law includes time as a variable.

First-order rate equation:

rate = - [A]

t

= k [A]

ln[A]0[A]t = - kt

Second-order rate equation:

after a Given Time PROBLEM: At 1000o C, cyclobutane (C4H8) decomposes in a first-order

reaction, with the very high rate constant of 87 s -1 , to two

molecules of ethylene (C2H4).

(a) If the initial C4H8concentration is 2.00 M, what is the

concentration after 0.010 s?

(b) What fraction of C4 H 8 has decomposed in this time?

PLAN: We must find the concentration of cyclobutane at time t, [C4 H 8 ]t

The problem tells us the reaction is first-order, so we use the

integrated first-order rate law:

ln [C4H8 ]0[C4H8 ]t = - kt

16-34

Figure 16.10A Graphical method for finding the reaction order

from the integrated rate law.

First-order reaction

ln [A]0 [A]t

= - kt

integrated rate law

ln[A] t= -kt + ln[A]0 straight-line form

A plot of ln [A] vs time gives a straight line for a first-order reaction.

16-35

Figure 16.10B Graphical method for finding the reaction order

from the integrated rate law.

Second-order reaction

integrated rate law 1

[A]t1 [A]0

straight-line form 1

[A]t

1 [A]0

= kt +

A plot of vs time gives a straight line for a second-order reaction.1

[A]

16-36

Figure 16.10C Graphical method for finding the reaction order

from the integrated rate law.

Zero-order reaction

A plot of [A] vs time gives a straight line for a first-order reaction.

integrated rate law [A]t- [A]0= - kt

straight-line form [A]t = - kt + [A]0

decomposition of N 2 O 5 .

The concentration data is used to construct three different plots Since the plot of ln [N 2 O 5 ] vs time gives a straight line, the reaction is first order.

16-38

Reaction Half-life

The half-life (t1/2) for a reaction is the time taken for the

concentration of a reactant to drop to half its initial value.

For a first-order reaction, t1/2does not depend on the starting concentration

t1/2 = ln 2

0.693

k

The half-life for a first-order reaction is a constant.

Radioactive decay is a first-order process The half-life for a radioactive nucleus is a useful indicator of its stability.

PROBLEM: Substance A (green) decomposes to two other

substances, B (blue) and C (yellow), in a first-order

gaseous reaction The molecular scenes below show a portion of the reaction mixture at two different times:

(a) Draw a similar molecular scene of the reaction mixture at t = 60.0 s.

(b) Find the rate constant of the reaction.

(c) If the total pressure (Ptotal ) of the mixture is 5.00 atm at 90.0 s, what

is the partial pressure of substance B (PB)?

16-41

Reaction PROBLEM: Cyclopropane is the smallest cyclic hydrocarbon

Because its 60° bond angles allow poor orbital overlap, its

bonds are weak As a result, it is thermally unstable and

rearranges to propene at 1000°C via the following

first-order reaction:

The rate constant is 9.2 s -1, (a) What is the half-life of the reaction?

(b) How long does it take for the concentration of cyclopropane to

reach one-quarter of the initial value?

PLAN: The reaction is first order, so we find t1/2using the half-life

equation for a first order reaction Once we know t1/2we can

calculate the time taken for the concentration to drop to 0.25 of

its initial value.

16-42

Half-life Equations

For a first-order reaction, t1/2does not depend on the initial concentration

For a second-order reaction, t1/2is inversely proportional

to the initial concentration:

1

k[A]0

t1/2= (second order process; rate = k[A]2 )

For a zero-order reaction, t1/2is directly proportional to

the initial concentration:

[A]0

2k0

t1/2 = (zero order process; rate = k)

Simple Second-Order Reactions

Zero Order First Order Second Order

Half-life

Integrated rate law

in straight-line form

[A]t= -kt + [A]0 ln[A]t= -kt + ln[A]0

Plot for straight line [A]tvs t ln[A]tvs t vs t

Slope, y intercept -k, [A]0 -k, ln[A]0 k,

Collision Theory and Concentration

The basic principle of collision theory is that particles

must collide in order to react

An increase in the concentration of a reactant leads to

a larger number of collisions, hence increasing reaction rate

The number of collisions depends on the product of

the numbers of reactant particles, not their sum

Concentrations are multiplied in the rate law, not added.

16-45

Figure 16.13 The number of possible collisions is the product,

not the sum, of reactant concentrations.

Temperature and the Rate Constant

Temperature has a dramatic effect on reaction rate

For many reactions, an increase of 10°C will double or triple the rate.

Experimental data shows that k increases exponentially

as T increases This is expressed in the Arrhenius equation:

Figure 16.14 Increase of the rate constant with temperature for

the hydrolysis of an ester.

Expt [Ester] [H 2O] T (K) Rate

The lower the activation energy, the faster the reaction.

When particles collide effectively, they reach an

activated state The energy difference between the

reactants and the activated state is the activation

energy (E a) for the reaction

Smaller Ea larger f larger k increased rate

Collisions must occur with

sufficient energy to reach an

activated state.

This particular reaction is reversible and is exothermic in the forward direction.

16-50

Temperature and Collision Energy

An increase in temperature causes an increase in the

kinetic energy of the particles This leads to more

frequent collisions and reaction rate increases.

At a higher temperature, the fraction of collisions with

sufficient energy equal to or greater than Eaincreases

Reaction rate therefore increases

16-51

Figure 16.16 The effect of temperature on the distribution of

collision energies.

16-52

Molecular Structure and Reaction Rate

For a collision between particles to be effective, it must

have both sufficient energy and the appropriate relative

orientation between the reacting particles.

k = Ae -Ea/RT

The term A in the Arrhenius equation is the frequency

factor for the reaction.

A = pZ p = orientation probability factor

Z = collision frequency

The term p is specific for each reaction and is related

to the structural complexity of the reactants

16-53

Catalysis: Speeding up a Reaction

A catalyst is a substance that increases the reaction rate

without itself being consumed in the reaction

In general, a catalyst provides an alternative reaction

pathway that has a lower total activation energy than the

uncatalyzed (red) process.

2 2

A small amount of NaBr is

added to a solution of H 2 O 2

Oxygen gas forms quickly as Br -(aq)

catalyzes the H 2 O 2 decomposition; the intermediate Br 2 turns the solution orange.

A homogeneous catalyst is in the same phase as the reaction mixture.

16-56

Figure B16.1 Satellite images of the increasing size of the

Antarctic ozone hole (purple).

16-57

by Cl atoms (Not drawn to scale.)

Chemical Connections

Lecture PowerPoint

Chemistry

The Molecular Nature of

Matter and Change

Equilibrium: The Extent of Chemical Reactions

17.1 The Equilibrium State and the Equilibrium Constant

17.2 The Reaction Quotient and the Equilibrium Constant

17.3 Expressing Equilibria with Pressure Terms:

Relation between Kcand Kp

17.4 Comparing Q and K to Determine Reaction Direction

17.5 How to Solve Equilibrium Problems

17.6 Reaction Conditions and Equilibrium:

Le Châtelier’s Principle

17-4

The Equilibrium State

All reactions are reversible and under suitable conditions will reach a state of equilibrium.

At equilibrium, the concentrations of products and reactants

no longer change because the rates of the forward and

reverse reactions are equal

At equilibrium: rate forward= ratereverse

Chemical equilibrium is a dynamic state because reactions

continue to occur, but because they occur at the same rate,

no net change is observed on the macroscopic level

17-5

molecular levels.

17-6

The Equilibrium Constant

At equilibrium ratefwd= raterev

so k[N2O4]eq = k[NO2] 2eq

kfwd

krev

[NO2] [N2O4]eq

2

eq

=then

The ratio of constants gives a new constant, the equilibrium constant:

kfwd

krev

[NO 2 ] [N 2 O 4 ]eq

K and the extent of reaction

K reflects a particular ratio of product concentrations to

reactant concentrations for a reaction

A small value for K indicates that the reaction yields little

product before reaching equilibrium The reaction favors

the reactants.

K therefore indicates the extent of a reaction, i.e., how

far a reaction proceeds towards the products at a

given temperature

A large value for K indicates that the reaction reaches

equilibrium with very little reactant remaining The

reaction favors the products.

17-8

small K

The reaction mixture contains mostly reactants.

large K

The reaction mixture contains mostly products.

intermediate K

17-9

The Reaction Quotient Q

For the general reaction

the reaction quotient Q =[C]c[D]d

[A]a[B]b

Q gives the ratio of product concentrations to reactant

concentrations at any point in a reaction.

At equilibrium: Q = K

For a particular system and temperature, the same

equilibrium state is attained regardless of starting

concentrations The value of Q indicates how close the

reaction is to equilibrium, and in which direction it must

proceed to reach equilibrium

17-10

Figure 17.3 The change in Q during the N2 O 4 -NO 2 reaction.

17-11

Table 17.1 Initial and Equilibrium Concentration Ratios for

the N 2 O 4 -NO 2 System at 200°C (473 K)

The form of Q and K depend on the direction in which the

balanced equation is written:

17-15

K for an Overall Reaction

PROBLEM: Nitrogen dioxide is a toxic pollutant that contributes to

photochemical smog One way it forms is through the

following sequence:

(a) Show that the overall Qcfor this reaction sequence is the same as

the product of the Qc 's of the individual reactions.

(b) Given that both reactions occur at the same temperature, find Kc

for the overall reaction.

17-16

Equation Multiplied by a Common Factor

Kcis 2.4x10 -3at 1000 K What are the values of K cfor the following balanced equations:

equation is

17-17

K and Q for hetereogeneous equilibrium

A hetereogeneous equilibrium involves reactants

and/or products in different phases

A pure solid or liquid always has the same “concentration”,

i.e., the same number of moles per liter of solid or liquid

The expressions for Q and K include only species whose

concentrations change as the reaction approaches

equilbrium

Pure solids and liquids are omitted from the expression for Q or K..

For the above reaction, Q = [CO2]

17-18

depends only on concentrations that change.

solids do not change their concentrations

K for a reaction may be expressed using partial pressures

of gaseous reactants instead of molarity

The partial pressure of each gas is directly proportional to its molarity.

Qc = [NO2 ] 2 [NO] 2 [O2]

PROBLEM: A chemical engineer injects limestone (CaCO3) into the hot

flue gas of a coal-burning power plant for form lime (CaO),

which scrubs SO 2 from the gas and forms gypsum

(CaSO 4 ·2H 2O) Find Kc for the following reaction, if CO 2

pressure is in atmospheres.

17-22

Determining the Direction of Reaction

The value of Q indicates the direction in which a reaction

must proceed to reach equilibrium

If Q < K, the reactants must increase and the products

decrease; reactants → products until equilibrium is reached

If Q > K, the reactants must decrease and the products

increase; products → reactants until equilibrium is reached

If Q = K, the system is at equilibrium and no further net

change takes place

PLAN: We must compare Qcwith Kcto determine the reaction direction,

so we use the given equilibrium concentrations to find Kc Then we

count spheres and calculate Q for each mixture

PROBLEM: For the reaction A(g) B(g), the equilibrium mixture at

175°C is [A] = 2.8x10 -4M and [B] = 1.2x10-4M The

molecular scenes below represent mixtures at various times

during runs 1-4 of this reaction (A is red; B is blue) Does the

reaction progress to the right or left or not at all for each mixture to reach equilibrium?

Direction PROBLEM: For the reaction N2 O 4(g) 2NO2(g), Kc = 0.21 at

100°C At a point during the reaction, [N 2 O 4] = 0.12 M

and [NO 2] = 0.55 M Is the reaction at equilibrium? If not,

in which direction is it progressing?

17-26

Solving Equilibrium Problems

If equilibrium quantities are given, we simply substitute

these into the expression for Kcto calculate its value

If only some equilibrium quantities are given, we use a

reaction table to calculate them and find Kc

A reaction table shows

• the balanced equation,

• the initial quantities of reactants and products,

• the changes in these quantities during the reaction, and

• the equilibrium quantities.

17-27

CO 2(g) + C(graphite) 2CO(g)

Example: In a study of carbon oxidation, an evacuated vessel

containing a small amount of powdered graphite is heated to 1080 K

Gaseous CO2is added to a pressure of 0.458 atm and CO forms At

equilibrium, the total pressure is 0.757 atm Calculate Kp.

The amount of CO 2will decrease If we let the decrease in CO2be x,

then the increase in CO will be +2x.

Equilibrium amounts are calculated by adding the change to the

The total pressure at equilibrium is 0.757 atm = P + PCO 2 (eq) CO(eq)

0.757 atm = 0.458 – x + 2x (from reaction table)

= 2.25

17-29

At equilibrium, [HI] = 0.078 M Calculate Kc

PLAN: To calculate Kcwe need equilibrium concentrations We can

find the initial [HI] from the amount and the flask volume, and

we are given [HI] at equilibrium From the balanced equation,

when 2x mol of HI reacts, x mol of H2and x mol of I2form We

use this information to set up a reaction table.

PROBLEM: In order to study hydrogen halide decomposition, a

researcher fills an evacuated 2.00-L flask with 0.200 mol

of HI gas and allows the reaction to proceed at 453°C.

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

17-30

from Kc

PROBLEM: In a study of the conversion of methane to other fuels, a

chemical engineer mixes gaseous CH 4 and H 2 O in a 0.32-L flask at 1200 K At equilibrium the flask contains 0.26 mol

of CO, 0.091 mol of H 2 , and 0.041 mol of CH 4 What is the [H2O] at equilibrium? Kc= 0.26 for this process at 1200 K.

Initial Concentrations and Kc

PROBLEM: Fuel engineers use the extent of the change from CO and

H 2 O to CO 2 and H 2 to regulate the proportions of synthetic

fuel mixtures If 0.250 mol of CO and 0.250 mol of H 2 O are

placed in a 125-mL flask at 900 K, what is the composition of

the equilibrium mixture? At this temperature, Kc is 1.56.

17-33

The Simplifying Assumption

We assume that x([A]reactingcan be neglected if

• Kcis relatively small and/or

• [A]initis relatively large

Calculating Equilibrium Concentrations PROBLEM: The research and development unit of a chemical company is

studying the reaction of CH4and H2S, two components of natural gas: CH4(g) + 2H2S(g) CS2(g) + 4H2(g)

In one experiment, 1.00 mol of CH 4 , 1.00 mol of CS 2 , 2.00 mol

of H 2 S, and 2.00 mol of H 2 are mixed in a 250-mL vessel at

960°C At this temperature, Kc = 0.036.

(a) In which direction will the reaction proceed to reach equilibrium?

(b) If [CH4] = 5.56 M at equilibrium, what are the equilibrium

concentrations of the other substances?

PLAN: (a) To find the direction of reaction we determine the initial

concentrations from the given amounts and volume, calculate Qc

and compare it with Kc

(b) Based on this information, we determine the sign of each

concentration change for the reaction table and hence calculate equilibrium concentrations.

17-35

Figure 17.6 Steps in solving equilibrium problems.

PRELIMINARY SETTING UP

1 Write the balanced equation.

2 Write the reaction quotient, Q.

3 Convert all amounts into the

correct units (M or atm).

WORKING ON THE REACTION

TABLE

4 When reaction direction is not

known, compare Q with K.

5 Construct a reaction table.

 Check the sign of x, the

change in the concentration

6 Substitute the quantities into Q.

7 To simplify the math, assume that x is

negligible:

([A] init– x = [A]eq ≈ [A] init )

8 Solve for x.

9 Find the equilibrium quantities.

 Check to see that calculated

values give the known K.

 Check that assumption is justified (<5% error) If not, solve

quadratic equation for x.

Le Châtelier’s Principle

When a chemical system at equilibrium is disturbed, it

reattains equilibrium by undergoing a net reaction that

reduces the effect of the disturbance.

A system is disturbed when a change in conditions forces

it temporarily out of equilibrium

A shift to the left is a net reaction from product to reactant.

The system responds to a disturbance by a shift in the

equilibrium position.

A shift to the right is a net reaction from reactant to product.

17-38

The Effect of a Change in Concentration

If the concentration of A increases, the system reacts to consume some of it

• If a reactant is added, the equilibrium position shifts to the right.

• If a product is added, the equilibrium position shifts to the left.

If the concentration of B decreases, the system reacts to consume some of it

• If a reactant is removed, the equilibrium position shifts to the left.

• If a product is removed, the equilibrium position shifts to the right.

Only substances that appear in the expression for Q can

part of yellow curve) and then falls

gradually as it reacts with PCl 3 to form more PCl 5 Equilibrium is re-established at

new concentrations but with the same

value of K.

17-42

Concentration on the Equilibrium Position

(d) [H2 S] if sulfur is added?

PLAN: We write the reaction quotient to see how Qc is affected by

each disturbance, relative to Kc This effect tells us the direction in which the reaction proceeds for the system to reattain equilibrium and how each concentration changes.

PROBLEM: To improve air quality and obtain a useful product, chemists

often remove sulfur from coal and natural gas by treating the contaminant hydrogen sulfide with O 2 :

What happens to

(a) [H2 O] if O 2 is added? (b) [H2 S] if O 2 is added?

(c) [O2 ] if H 2 S is removed?

2H 2S(g) + O2(g) 2S(s) + 2H2O(g)

The Effect of a Change in Pressure (Volume)

Changes in pressure affect equilibrium systems containing

gaseous components

• Changing the concentration of a gaseous component

causes the equilibrium to shift accordingly

• Adding an inert gas has no effect on the equilibrium

position, as long as the volume does not change

– This is because all concentrations and partial pressures remain

unchanged.

• Changing the volume of the reaction vessel will cause

equilibrium to shift if Dngas≠ 0

• Changes in pressure (volume) have no effect on the

value of K.

17-44

system at equilibrium.

17-45

(Pressure) on the Equilibrium Position PROBLEM: How would you change the volume of each of the

following reactions to increase the yield of the products?

PLAN: Whenever gases are present, a change in volume causes a

change in concentration For reactions in which the number of

moles of gas changes, a decrease in volume (pressure increase)

causes an equilibrium shift to lower the pressure by producing

fewer moles of gas A volume increase (pressure decrease) has

the opposite effect.

(a) CaCO3(s) CaO(s) + CO2(g)

(b) S(s) + 3F2(g) SF6(g)

(c) Cl2(g) + I2(g) 2ICl(g)

17-46

The Effect of a Change in Temperature

To determine the effect of a change in temperature on equilibrium, heat is considered a component of the system

Heat is a product in an exothermic reaction (DH°rxn< 0)

Heat is a reactant in an endothermic reaction (DH°rxn> 0)

An increase in temperature adds heat, which favors the

The only factor that affects the value of K for a given

equilibrium system is temperature.

For a reaction with DH°rxn> 0, an increase in temperature

will cause K to increase.

For a reaction with DH°rxn< 0, an increase in temperature

will cause K to decrease.

Temperature on the Equilibrium Position

PROBLEM: How does an increase in temperature affect the

equilibrium concentration of the underlined substance

and K for each of the following reactions?

PLAN: We write each equation to show heat as a reactant or product

The temperature increases when we add heat, so the system shifts to absorb the heat; that is, the endothermic reaction is

favored Thus, K will increase if the forward reaction is

endothermic and decrease if it is exothermic.

(a) CaO(s) + H2O(l) Ca(OH)2(aq) DH° = -82 kJ

(b) CaCO3(s) CaO(s) + CO2(g) DH° = 178 kJ

(c) SO2(g) S(s) + O2(g) DH° = 297 kJ

Catalysts and Equilibrium

A catalyst speeds up a reaction by lowering its activation

energy A catalyst therefore speeds up the forward and

reverse reactions to the same extent

A catalyst causes a reaction to reach equilibrium more

quickly, but has no effect on the equilibrium position.

(a) If K = 2 at the temperature of the reaction, which scene represents

the mixture at equilibrium?

(b) Will the reaction mixtures in the other two scenes proceed toward

reactant or toward products to reach equilibrium?

(c) For the mixture at equilibrium, how will a rise in temperature affect

[Y 2 ]?

PROBLEM: For the reaction X(g) + Y2(g) XY(g) + Y(g) DH > 0

the following molecular scenes depict different reaction

mixtures (X = green, Y = purple):

17-52

The Synthesis of Ammonia

Ammonia is synthesized industrially via the Haber process:

N2(g) + 3H2(g) 2NH3(g) DH°rxn= -91.8 kJThere are three ways to maximize the yield of NH3:

• Decrease [NH3] by removing NH3as it forms

• Decrease the volume (increase the pressure)

• Decrease the temperature

An increase in temperature causes the equilibrium to shift towards

the reactants, since the reaction is exothermic.

Figure 17.12 Key stages in the Haber process for

synthesizing ammonia.

17-56

Figure B17.1 The biosynthesis of isoleucine from threonine.

Isoleucine is synthesized from threonine in a sequence of five enzyme-catalyzed reactions Once enough isoleucine is present, its concentration builds up and inhibits threonine dehydratase, the first

enzyme in the pathway This process is called end-product feedback inhibition.

17-57

Figure B17.2 The effect of inhibitor binding on the shape of an

active site.

Chemical Connections

18-1

Lecture PowerPoint

Chemistry

The Molecular Nature of

Matter and Change

18.1 Acids and Bases in Water

18.2 Autoionization of Water and the pH Scale

18.3 Proton Transfer and the Brønsted-Lowry Acid-Base Definition

18.4 Solving Problems Involving Weak-Acid Equilibria

18.5 Weak Bases and Their Relation to Weak Acids

18.6 Molecular Properties and Acid Strength

18.7 Acid-Base Properties of Salt Solutions

18.8 Generalizing the Brønsted-Lowry Concept: The Leveling

Arrhenius Acid-Base Definition

An acid is a substance with H in its formula that dissociates

to yield H3O+

This is the earliest acid-base definition, which classifies

these substances in terms of their behavior in water.

A base is a substance with OH in its formula that

Strong and Weak Acids

A strong acid dissociates completely into ions in water:

HA(g or l) + H2O(l) → H3O+(aq) + A-(aq)

A dilute solution of a strong acid contains no HA molecules.

A weak acid dissociates slightly to form ions in water:

HA(aq) + H2O(l) H3O+(aq) + A-(aq)

In a dilute solution of a weak acid, most HA molecules are

undissociated.

[H3O+][A-][HA][H2O]

18-7

Strong acid: HA(g or l) + H2O(l) → H3O+(aq) + A-(aq)

Figure 18.1A The extent of dissociation for strong acids.

There are no HA molecules in solution.

18-8

Figure 18.1B The extent of dissociation for weak acids.

Weak acid: HA(aq) + H2O(l) H3 O +(aq) + A-(aq)

Most HA molecules are undissociated.

18-9

Figure 18.2 Reaction of zinc with a strong acid (left) and a

weak acid (right).

Zinc reacts rapidly with the strong acid, since [H 3 O + ] is much higher.

18-10

The Acid Dissociation Constant, Ka

[H3O+][A-][HA][H2O]

+ ][A - ] [HA]

The value of Kais an indication of acid strength

Stronger acid higher [H 3 O + ] larger Ka

Weaker acid lower % dissociation of HA smaller Ka

HA(aq) + H2O(l) H3O+(aq) + A-(aq)

18-11

18-12

Classifying the Relative Strengths of Acids

• Strong acids include

– the hydrohalic acids (HCl, HBr, and HI) and – oxoacids in which the number of O atoms exceeds the number

of ionizable protons by two or more (eg., HNO3, H2SO4, HClO4.)

• Weak acids include

– the hydrohalic acid HF, – acids in which H is not bonded to O or to a halogen (eg., HCN), – oxoacids in which the number of O atoms equals or exceeds the number of ionizable protons by one (eg., HClO, HNO2), and

– carboxylic acids, which have the general formula RCOOH (eg.,

CH3COOH and C6H5COOH.)

18-13

Classifying the Relative Strengths of Bases

• Strong bases include

– water-soluble compounds containing O 2- or OH - ions.

– The cations are usually those of the most active metals:

• M 2 O or MOH, where M = Group 1A(1) metal (Li, Na, K, Rb, Cs)

• MO or M(OH)2where M = group 2A(2) metal (Ca, Sr, Ba).

• Weak bases include

– ammonia (NH3),

– amines, which have the general formula

– The common structural feature is an N atom with a lone

electron pair.

18-14

the Chemical Formula PROBLEM: Classify each of the following compounds as a strong

acid, weak acid, strong base, or weak base.

(c) H2 SeO 4 (d) (CH3 ) 2 CHNH 2

18-15

Autoionization of Water

Water dissociates very slightly into ions in an equilibrium

process known as autoionization or self-ionization.

K c =

K c[H2O]2= Kw = [H 3 O + ][OH - ] = 1.0x10 -14 (at 25°C)

= 1.0x10-7(at 25°C)

In pure water,[H3O+] = [OH-] =

Both ions are present in all aqueous systems.

18-17

Higher [H 3 O + ] lower [OH - ]

Higher [OH - ] lower [H3O + ]

A change in [H3O+] causes an inverse change in [OH-],

and vice versa

We can define the terms “acidic” and “basic” in terms of

the relative concentrations of H3O+and OH-ions:

In an acidic solution, [H3O+] > [OH-]

In a neutral solution, [H3O+] = [OH-]

In basic solution, [H3O+] < [OH-]

18-18

Figure 18.3 The relationship between [H 3 O + ] and [OH - ] and the

relative acidity of solutions.

18-19

Sample Problem 18.2 Calculating [H 3 O + ] or [OH - ] in an Aqueous

Solution PROBLEM: A research chemist adds a measured amount of HCl gas

to pure water at 25°C and obtains a solution with [H 3 O + ] =

3.0x10 -4 M Calculate [OH- ] Is the solution neutral,

The higher the pH, the lower the [H3O+] and the less

acidic the solution.

Table 18.3 The Relationship between Kaand pKa

Hydrogen sulfate ion (HSO 4-) 1.0x10 -2 1.99

Since Kwis a constant, the values of pH, pOH, [H3O+],

and [OH-] are interrelated:

• If [H3O+] increases, [OH-] decreases (and vice versa)

• If pH increases, pOH decreases (and vice versa)

18-24

Figure 18.5 The relations among [H 3 O + ], pH, [OH - ], and pOH.

18-25

Sample Problem 18.3 Calculating [H 3 O + ], pH, [OH - ], and pOH

PROBLEM: In an art restoration project, a conservator prepares

copper-plate etching solutions by diluting concentrated

HNO 3to 2.0 M, 0.30 M, and 0.0063 M HNO3 Calculate

[H3O + ], pH, [OH - ], and pOH of the three solutions at 25°C.

Brønsted-Lowry Acid-Base Definition

An acid is a proton donor, any species that donates an

H + ion.

• An acid must contain H in its formula

A base is a proton acceptor, any species that accepts

Figure 18.7 Dissolving of an acid or base in water as a

Brønsted-Lowry acid-base reaction.

(acid, H + donor) (base, H+ acceptor)

Lone pair binds H +

(base, H + acceptor) (acid, H + donor)

Lone pair binds H +

18-29

Conjugate Acid-Base Pairs

NH3accepts a H+to form NH4

In the forward reaction:

In the reverse reaction:

H2S and HS-are a conjugate acid-base pair:

HS-is the conjugate base of the acid H2S

A Brønsted-Lowry acid-base reaction occurs when an

acid and a base react to form their conjugate base and conjugate acid, respectively.

NH3and NH4 are a conjugate acid-base pair:

NH4 is the conjugate acid of the base NH3

PROBLEM: The following reactions are important environmental

processes Identify the conjugate acid-base pairs.

(b) H2O(l) + SO32-(aq) OH-(aq) + HSO3-(aq)

(a) H2 PO 4-(aq) + CO32-(aq) HPO42-(aq) + HCO3-(aq)

18-33

Net Direction of Reaction

The net direction of an acid-base reaction depends on

the relative strength of the acids and bases involved.

A reaction will favor the formation of the weaker acid

and base

H 2 S + NH 3 HS - + NH 4

stronger base

weaker basestronger acid

weaker acid

This reaction favors the formation of the products.

18-34

The stronger the acid is, the weaker its conjugate base

When an acid reacts with a base that is farther down the list, the reaction proceeds to

the right (Kc > 1).

18-35

Acid-Base Reaction

PROBLEM: Predict the net direction and whether Kcis greater or less

than 1 for each of the following reactions (assume equal

initial concentrations of all species):

(a) H2PO4-(aq) + NH3(aq) HPO42-(aq) + NH4+(aq)

(b) H2O(l) + HS-(aq) OH-(aq) + H2S(aq)

18-36

Direction of an Acid-Base Reaction

PROBLEM: Given that 0.10 M of HX (blue and green) has a pH of 2.88,

and 0.10 M HY (blue and orange) has a pH 3.52, which

scene best represents the final mixture after equimolar solutions of HX and Y - are mixed?

1 Write a balanced equation.

2. Write an expression for Ka

3. Define x as the change in concentration that

occurs during the reaction.

4. Construct a reaction table in terms of x.

5 Make assumptions that simplify the calculation.

6. Substitute values into the Ka expression and

The notation system

• Molar concentrations are indicated by [ ].

• A bracketed formula with no subscript indicates an equilibrium concentration.

HPAc) builds up in the blood of persons with

phenylketonuria, an inherited disorder that, if untreated,

causes mental retardation and death A study of the acid

shows that the pH of 0.12 M HPAc is 2.62 What is the Ka

of phenylacetic acid?

18-40

Initial [HA]

PROBLEM: Propanoic acid (CH3CH2COOH, which we simplify as

HPr) is a carboxylic acid whose salts are used to retard mold growth in foods What is the [H 3 O +] of 0.10 M HPr (Ka = 1.3x10 −5 )?

18-41

Concentration and Extent of Dissociation

Percent HA dissociated = [HA]dissoc

[HA]init

x 100

As the initial acid concentration decreases, the percent

dissociation of the acid increases.

HA(aq) + H2O(l) H3O+(aq) + A-(aq)

A decrease in [HA]init means a

decrease in [HA]dissoc= [H3O + ] = [A - ],

causing a shift towards the products.

The fraction of ions present increases, even though the

actual [HA] dissoc decreases.

18-42

Extent of HA Dissociation

PROBLEM: A 0.15 M solution of acid HA (blue and green) is 33%

dissociated Which scene best represents a sample of the solution after it is diluted with water?

18-43

A polyprotic acid is an acid with more than one ionizable

proton In solution, each dissociation step has a different

value for Ka:

Ka1> Ka2 > Ka3

Polyprotic Acids

Ka1 = [H3 O + ][H 2 PO 4-] [H 3 PO 4 ] = 7.2x10

-3

H 3 PO 4(aq) + H2O(l) H2 PO 4-(aq) + H3 O +(aq)

Ka2 = [H3 O + ][HPO 42-] [H 2 PO 4-] = 6.3x10

-8

H 2 PO 4-(aq) + H2O(l) HPO42-(aq) + H3 O +(aq)

Ka3 =[H3 O + ][PO 43-] [HPO 42-] = 4.2x10

-13 HPO 42-(aq) + H2O(l) PO43-(aq) + H3 O +(aq)

We usually neglect [H3O+] produced after the first dissociation

18-44

Table 18.5 Successive Ka values for Some Polyprotic Acids at

25°C

18-45

for a Polyprotic Acid PROBLEM: Ascorbic acid (H2C6H6O6; H2Asc for this problem),

known as vitamin C, is a diprotic acid (Ka1 = 1.0x10 -5 and

Ka2 = 5x10 -12 ) found in citrus fruit Calculate [H 2 Asc],

[HAsc - ], [Asc 2-], and the pH of 0.050 M H2 Asc.

18-46

Weak Bases

A Brønsted-Lowry base is a species that accepts an H+ For a weak base that dissolves in water:

B(aq) + H2O(l) BH+(aq) + OH-(aq)

The base-dissociation or base-ionization constant is

given by:

[BH + ][OH - ] [B]

Kb =

Note that no base actually dissociates in solution, but ions are

produced when the base reacts with H 2 O.

18-49

PROBLEM: Dimethylamine, (CH3 ) 2 NH, a key intermediate in

detergent manufacture, has a Kb of 5.9x10 -4 What is the

pH of 1.5 M (CH3 ) 2 NH?

18-50

Anions of Weak Acids as Weak Bases

The anions of weak acids often function as weak bases.

A-(aq) + H2O(l) HA(aq) + OH-(aq) K

b= [HA][OH-][A-]

A solution of HA is acidic, while a solution of A-is basic.

HF(aq) + H2O(l) H3O+(aq) + F-(aq)

HF is a weak acid, so this equilibrium lies to the left

[HF] >> [F-], and [H3O+]from HF>> [OH-] ;

the solution is therefore acidic.

from H 2 O

18-51

F-(aq) + H2O(l) HF(aq) + OH-(aq)

If NaF is dissolved in H2O, it dissolves completely, and F

-can act as a weak base:

HF is a weak acid, so this equilibrium also lies to the left

[F-] >> [HF], and [OH-] >> [H3O+] ;

the solution is therefore basic.

from H 2 O from F -

18-52

Kaand Kb for a Conjugate Acid-Base Pair

HA + H2O H3O++ A

-A-+ H2O HA + OH2H2O H3O++ OH-

This relationship is true for any conjugate acid-base pair.

18-53

has applications in photographic development and textile

dyeing What is the pH of 0.25 M NaAc? Kaof acetic

acid (HAc) is 1.8x10 -5

18-54

Acid Strength of Nonmetal Hydrides

For nonmetal hydrides (E-H), acid strength depends on:

• the electronegativity of the central nonmetal (E), and

• the strength of the E-H bond

Across a period, acid strength increases.

Electronegativity increases across a period, so the acidity of E-H increases.

Down a group, acid strength increases.

The length of the E-H bond increases down a group and its bond strength therefore decreases.

HBr

HI

Electronegativity increases, so acidity increases

nonmetal hydride acidity.

18-56

Acid Strength of Oxoacids

All oxoacids have the acidic H bonded to an O atom.

Acid strength of oxoacids depends on:

• the electronegativity of the central nonmetal (E), and

• the number of O atoms around E

For oxoacids with the same number of O atoms, acid

strength increases as the electronegativity of E increases

For oxoacids with different numbers of O atoms, acid

strength increases with the number of O atoms

Hydrated Metal Ions

Some hydrated metal ions are able to transfer an H+to

H2O These metal ions will form acidic solutions.

M(H2O)x n+ (aq) + H2O(l) M(H2O)x-1OH(n-1)

(aq)+ H3O+(aq)

Mn+ (aq) + H2O(l) → M(H2O)x n+ (aq)

Consider a metal ion in solution, Mn+:

If Mn+ is small and highly charged, it will withdraw enough

e-density from the O-H bonds of the bound H2O molecules

to release H+:

18-59

Figure 18.12 The acidic behavior of the hydrated Al 3+ ion.

18-60

Salts that Yield Neutral Solutions

A salt that consists of the anion of a strong acid and the cation of a strong base yields a neutral solution.

NaNO3

Na + is the cation of NaOH, a strong base.

NO 3 - is the anion of HNO 3 , a strong acid.

This solution will be neutral, because neither Na+nor

NO3-will react with H2O to any great extent

18-61

A salt that consists of the anion of a strong acid and

the cation of a weak base yields an acidic solution.

NH4Cl

NH 4 + is the cation of

NH 3 , a weak base.

Cl - is the anion of HCl, a strong acid.

This solution will be acidic, because NH4 will react with

H2O to produce H3O+:

NH4(aq) + H2O(l) NH3(aq) + H3O+(aq)

Salts that Yield Acidic Solutions

18-62

A salt that consists of the anion of a weak acid and the

cation of a strong base yields a basic solution.

CH 3 COO - is the anion of

CH 3 COOH, a weak acid.

Na + is the cation of NaOH, a strong base.

This solution will be basic, because CH3COO-will react with H2O to produce OH-:

CH3COO-(aq) + H2O(l) CH3COOH(aq) + OH-(aq)

Salts that Yield Basic Solutions

18-63

from Reactions of the Ions with Water PROBLEM: Predict whether aqueous solutions of the following are

acidic, basic, or neutral, and write an equation for the

reaction of any ion with water:

(a) Potassium perchlorate, KClO4 (b) Sodium benzoate, C6 H 5 COONa

(c) Chromium(III) nitrate, Cr(NO3 ) 3

18-64

If a salt that consists of the anion of a weak acid and the

cation of a weak base, the pH of the solution will

depend on the relative acid strength or base strength of the ions

NH4CN

NH 4 + is the cation of a weak base, NH 3

CN - is the anion of a weak acid, HCN.

Salts of Weakly Acidic Cations and Weakly Basic Anions

CN-(aq) + H2O(l) HCN(aq) + OH-(aq)

NH4(aq) + H2O(l) NH3(aq) + H3O+(aq)

18-65

Kaof NH4 = Kw

Kbof NH3

1.0x10-141.76x10-5

Kbof CN-= Kw

Kaof HCN

1.0x10-146.2x10-10

The reaction that proceeds farther to the right determines the

pH of the solution, so we need to compare the Kaof NH4

with the Kbof CN-

Since Kbof CN-> Kaof NH4, CN-is a stronger base than

NH4 is an acid A solution of NH4CN will be basic

18-66

Table 18.7 The Acid-Base Behavior of Salts in Water

18-67

Solutions from Kaand Kb of the Ions PROBLEM: Determine whether an aqueous solution of zinc formate,

Zn(HCOO) 2 , at 25°C is acidic, basic, or neutral.

18-68

The Leveling Effect

H2O exerts a leveling effect on any strong acid or base.

All strong acids and bases are equally strong in water.

All strong acids dissociate completely to form H3O+, while all strong bases dissociate completely to form OH-

In water, the strongest acid possible is H 3 O +and the

strongest base possible is OH -

18-69

The Lewis Acid-Base Definition

A Lewis base is any species that donates an electron

pair to form a bond.

A Lewis acid is any species that accepts an electron pair

to form a bond

The Lewis definition views an acid-base reaction as the

donation and acceptance of an electron pair to form

a covalent bond.

18-70

Lewis Acids and Bases

The Lewis definition expands the classes of acids.

A Lewis base must have a lone pair of electrons to

donate

Any substance that is a Brønsted-Lowry base is also a Lewis base.

A Lewis acid must have a vacant orbital (or be able to

rearrange its bonds to form one) to accept a lone pair and form a new bond

Many substances that are not Brønsted-Lowry acids are Lewis

acids.

18-71

Electron-Deficient Molecules as Lewis Acids

B and Al often form electron-deficient molecules, and

these atoms have an unoccupied p orbital that can accept

a pair of electrons:

BF3accepts an electron pair from ammonia to form a covalent bond

18-72

Lewis Acids with Polar Multiple Bonds

Molecules that contain a polar multiple bond often function

as Lewis acids:

The O atom of an H 2 O molecule donates a lone pair to the S of SO 2 , forming a new S‒O σ bond and breaking one of the S‒O p bonds.

18-73

Metal Cations as Lewis Acids

A metal cation acts as a Lewis acid when it dissolves in

water to form a hydrated ion:

The O atom of an H2O molecule donates a lone pair to an available

orbital on the metal cation.

18-74

Figure 18.13 The Mg 2+ ion as a Lewis acid in chlorophyll.

18-75

PROBLEM: Identify the Lewis acids and Lewis bases in the following

reactions:

(a) H+ + OH - H2O (b) Cl- + BCl3 BCl4- (c) K+ + 6H2O K(H2O6) +