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

Lecture PowerPoint
Chapter 16

Chemistry
Kinetics: Rates and Mechanisms
of Chemical Reactions

The Molecular Nature of Matter and Change

Martin S. Silberberg

16-1

Copyright  The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

16-2

Figure 16.1

A faster reaction (top) and a slower reaction (bottom).

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
16.5 Theories of Chemical Kinetics
16.6 Catalysis: Speeding Up a Reaction

16-3

Figure 16.2

16-4

The wide range of reaction rates.

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.

16-5

16-6

1


Figure 16.3

The effect of surface area on reaction rate.

A hot steel nail glows feebly
when placed in O2.

Figure 16.4

Sufficient collision energy is required for a
reaction to occur.

The same mass of steel wool
bursts into flame.

16-7

16-8

Table 16.1

Expressing the Reaction Rate

Concentration of O3 at Various Times in its
Reaction with C2H4 at 303 K

Reaction rate is measured in terms of the changes in
concentrations of reactants or products per unit time.
Time (s)

For the general reaction A → B, we measure the

concentration of A at t1 and at t2:
Rate =

change in concentration of A
change in time

=-

conc A2 - conc A1
t2 - t1

=-

[A]
t

Square brackets indicate a concentration in moles per liter.
The negative sign is used because the concentration of A
is decreasing. This gives the rate a positive value.

16-9

Figure 16.5

Concentration of O3
(mol/L)

0.0

3.20x10-5

10.0

2.42x10-5

20.0

1.95x10-5

30.0

1.63x10-5

40.0

1.40x10-5

50.0

1.23x10-5

60.0

1.10x10-5

rate = =-

 [C2H4]
t
 [O3]

t

16-10

Three types of reaction rates for the reaction of O3
and C2H4.

Figure 16.6A

Plots of [reactant] and [product] vs. time.

C2H4 + O3 → C2H4O + O2

[O2] increases just as fast as
[C2H4] decreases.
Rate = -

=

16-11

[C2H4]
t
[C2H4O]
t

=-

=

[O3]
t
[O2]
t

16-12

2


Figure 16.6B

Plots of [reactant] and [product] vs. time.

In general, for the reaction

H2 + I2 → 2HI

[HI] increases twice as
fast as [H2] decreases.

Rate = -

Rate =

[H2]

=-

t

[IH]
t

= -2

[I2]
t
[H2]
t

aA

[HI]
= 1
t
2
= -2

+

bB → cC

+

dD

where a, b, c, and d are the coefficients for the balanced
equation, the rate is expressed as:

[I2]

rate = -

t

1
a

[A]
1 [B] = 1 [C]
=b t
c t
t

= 1 [D]
d t

The expression for the rate of a reaction
and its numerical value depend on which
substance serves as the reference.

16-13

16-14

Sample Problem 16.1

Expressing Rate in Terms of Changes in
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:
2H2(g) + O2(g) → 2H2O(g)

The Rate Law
For any general reaction occurring at a fixed temperature
aA

(b) When [O2] is decreasing at 0.23 mol/L·s, at what rate is
[H2O] increasing?
PLAN: We choose O2 as the reference because its coefficient is 1. For
every molecule of O2 that disappears, two molecules of H2
disappear, so the rate of [O2] decrease is ½ the rate of [H2]
decrease. Similarly, the rate at which [O2] decreases is ½ the rate
at which [H2O] increases.

16-15

+

bB → cC

+

dD

Rate = k[A]m[B]n

(a) Express the rate in terms of changes in [H2], [O2], and

[H2O] with time.

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.

16-16

Reaction Orders

Figure 16.7 Plots of reactant concentration, [A], vs. time for
first-, second-, and zero-order reactions.

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]1 and the reaction is first order with respect to A.
If the rate quadruples when [A] doubles, the rate depends
on [A]2 and 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.
16-17

16-18

3


Figure 16.8 Plots of rate vs. reactant concentration, [A], for
first-, second-, and zero-order reactions.

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 H2 and third order overall.
Note that the reaction is first order with respect to H2
even though the coefficient for H2 in the balanced
equation is 2.
Reaction orders must be determined from experimental
data and cannot be deduced from the balanced
equation.

16-19

16-20

Sample Problem 16.2

Determining Reaction Orders from Rate
Laws
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.
k[NO]2[O

(a) 2NO(g) + O2(g) → 2NO2(g); rate =
2]
(b) CH3CHO(g) → CH4(g) + CO(g); rate = k[CH3CHO]3/2
+
(c) H2O2(aq) + 3I (aq) + 2H (aq) →I3 (aq) + 2H2O(l); rate = k[H2O2][I-]
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.

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.

SOLUTION:

16-21

16-22

Table 16.2

Initial Rates for the Reaction between A and B

Experiment

Initial Rate
(mol/L·s)

Initial [A]
(mol/L)

Initial [B]
(mol/L)

1

1.75x10-3

2.50x10-2

3.00x10-2

2

3.50x10-3

5.00x10-2

3.00x10-2

3

3.50x10-3

2.50x10-2

6.00x10-2

4

7.00x10-3

5.00x10-2

6.00x10-2

Finding m, the order with respect to A:
We compare experiments 1 and 2, where [B] is kept
constant but [A] doubles:
Rate 2
Rate 1

k[A] m [B] n
=

2
k[A] m
1

16-23

m

=

[A] 2

m
[A]1

=

[A]2 m
[A]1
m

3.50x10-3 mol/L·s
1.75x10-3mol/L·s

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

2
[B] n1

=

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

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

16-24

4


Table 16.3

Finding n, the order with respect to B:

O2(g) + 2NO(g) → 2NO2(g)

We compare experiments 3 and 1, where [A] is kept
constant but [B] doubles:
Rate 3
Rate 1

k[A] m [B] n
=

3

3

k[A] m
[B] n1
1

n

=

[B] 3

[B]1

Experiment

Initial Rate
(mol/L·s)

[O2]

[NO]

1

3.21x10-3

1.10x10-2

1.30x10-2

m

3.50x10-3 mol/L·s

=

1.75x10-3mol/L·s

Rate = k[O2]m[NO]n
Initial Reactant

Concentrations (mol/L)

[B]3 n

=

n
[B]1

Initial Rates for the Reaction between O2 and NO

6.00x10-2 mol/L

2

6.40x10-3

2.20x10-2

1.30x10-2

3.00x10-2

3

12.48x10-3

1.10x10-2

2.60x10-2

4

9.60x10-3

3.30x10-2

1.30x10-2

5

28.8x10-3

1.10x10-2

3.90x10-2

mol/L

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

16-25

16-26

Sample Problem 16.3

Determining Reaction Orders from Rate Data

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

system. One of these is
NO2(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)

Sample Problem 16.3
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:

Initial [NO2] Initial [CO]
(mol/L)
(mol/L)

1

0.0050

0.10

0.10

2

0.080

0.40

0.10

3

0.0050

0.10

0.20

16-27

16-28

Sample Problem 16.4

Determining Reaction Orders from Molecular
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:

Table 16.4

Overall
Reaction Units of k
Order
(t in seconds)
0

Expt no:
1
Initial rate (mol/L·s) 0.50x10-4

2
1.0x10-4

3
2.0x10-4

4
?

(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.

16-29

Units of the Rate Constant k for Several Overall

Reaction Orders

mol/L·s
(or mol L-1s-1)

1

1/s (or s-1)

2

L/mol·s
(or L mol-1s-1)

3

L2/mol2·s
(or L2 mol-2s-1)

General formula:
L
Units of k =

order-1

mol
unit of t

The value of k is easily determined from experimental rate
data. The units of k depend on the overall reaction order.

16-30

5


Integrated Rate Laws

Figure 16.9 Information sequence to determine the kinetic
parameters of a reaction.

An integrated rate law includes time as a variable.

Series of plots of
concentration vs.
time

First-order rate equation:
rate = - [A] = k [A]
Determine slope of tangent at t0 for
each plot.

Initial rates

Compare initial rates when [A] changes and
[B] is held constant (and vice versa).
Reaction
orders

Substitute initial rates, orders, and

concentrations into rate = k[A]m[B]n, and
solve for k.

Rate constant (k)
and actual rate law

ln

t
Second-order rate equation:
[A]
rate = = k [A]2
t
Zero-order rate equation:
rate = - [A] = k [A]0

[A]0
[A]t

= - kt

1
1
= kt
[A]t [A]0
[A]t - [A]0 = - kt

t

16-31

16-32

Sample Problem 16.5

Determining the Reactant Concentration
after a Given Time

Figure 16.10A Graphical method for finding the reaction order
from the integrated rate law.

PROBLEM: At 1000oC, 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 C4H8 concentration is 2.00 M, what is the
concentration after 0.010 s?
(b) What fraction of C4H8 has decomposed in this time?

First-order reaction
integrated rate law
ln [A]0 = - kt
[A]t

PLAN: We must find the concentration of cyclobutane at time t, [C4H8]t.
The problem tells us the reaction is first-order, so we use the
integrated first-order rate law:
ln

[C4H8 ]0
[C4H8 ]t

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

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

16-33

16-34

Figure 16.10B Graphical method for finding the reaction order
from the integrated rate law.

Figure 16.10C Graphical method for finding the reaction order
from the integrated rate law.

Second-order reaction

Zero-order reaction
integrated rate law
1
[A]t

-

1
[A]0

integrated rate law

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

= kt

straight-line form
1
1
= kt +
[A]t
[A]0

A plot of

16-35

1
vs. time gives a straight line for a second-order reaction.
[A]

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

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

16-36

6


Figure 16.11 Graphical determination of the reaction order for the

decomposition of N2O5.

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, t1/2 does not depend on the
starting concentration.
ln 2 0.693
t1/2 =
=
k
k

The concentration data is used to
construct three different plots. Since the
plot of ln [N2O5] vs. time gives a straight
line, the reaction is first order.

16-37

Figure 16.12

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.

16-38

A plot of [N2O5] vs. time for three reaction half-lives.
for a first-order process

t1/2 =

ln 2
k

=

0.693
k

Sample Problem 16.6

Using Molecular Scenes to Find Quantities
at Various Times
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)?

16-39

Sample Problem 16.7

16-40

Determining the Half-Life of a First-Order
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 firstorder 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 t1/2 using the half-life
equation for a first order reaction. Once we know t1/2 we can
calculate the time taken for the concentration to drop to 0.25 of
its initial value.

16-41

Half-life Equations
For a first-order reaction, t1/2 does not depend on the
initial concentration.
For a second-order reaction, t1/2 is inversely proportional
to the initial concentration:
t1/2 =

1
k[A]0

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

For a zero-order reaction, t1/2 is directly proportional to
the initial concentration:
t1/2 =

[A]0
2k0

(zero order process; rate = k)

16-42

7


Table 16.5

An Overview of Zero-Order, First-Order, and
Simple Second-Order Reactions

Collision Theory and Concentration

Zero Order

First Order

Second Order

Rate law

rate = k

rate = k[A]

rate = k[A]2

Units for k

mol/L·s
[A]0
2k

1/s
ln 2
k

Integrated rate law
in straight-line form

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

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

L/mol·s
1
k[A]0
1
= kt +
[A]

Plot for straight line

[A]t vs. t

ln[A]t vs. t

Slope, y intercept

-k, [A]0

-k, ln[A]0

Half-life

t

The basic principle of collision theory is that particles
must collide in order to react.
1
[A]0

1
vs. t
[A]t
k, 1
[A]0

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.

16-43

16-44

Temperature and the Rate Constant

Figure 16.13 The number of possible collisions is the product,
not the sum, of reactant concentrations.

Temperature has a dramatic effect on reaction rate.
For many reactions, an increase of 10°C will double or triple the rate.

add another
6 collisions

Experimental data shows that k increases exponentially
as T increases. This is expressed in the Arrhenius
equation:

4 collisions
add another

k = rate constant
A = frequency factor
Ea = activation energy

k = Ae-Ea/RT

Higher T

larger k

increased rate

9 collisions

16-45

16-46

Figure 16.14 Increase of the rate constant with temperature for
the hydrolysis of an ester.

Activation Energy
In order to be effective, collisions between particles
must exceed a certain energy threshold.

Expt [Ester] [H2O] T (K)
1
2
3
4

0.100
0.100
0.100
0.100

0.200
0.200
0.200
0.200

288
298
308
318

Rate
k
(mol/L·s) (L/mol·s)
1.04x10-3 0.0521
2.20x10-3
0101
3.68x10-3
0.184
6.64x10-3
0.332

When particles collide effectively, they reach an
activated state. The energy difference between the
reactants and the activated state is the activation
energy (Ea) for the reaction.
The lower the activation energy, the faster the reaction.
Smaller Ea

larger f

larger k

increased rate

Reaction rate and k increase exponentially as T increases.

16-47

16-48

8


Figure 16.15 Energy-level diagram for a reaction.

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 Ea increases.
Reaction rate therefore increases.

Collisions must occur with
sufficient energy to reach an
activated state.

This particular reaction is reversible
and is exothermic in the forward

direction.

16-49

16-50

Figure 16.16 The effect of temperature on the distribution of
collision energies.

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.
The term A in the Arrhenius equation is the frequency
factor for the reaction.
k = Ae -Ea/RT

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.

16-51

16-52

Figure 16.23

Catalysis: Speeding up a Reaction

Reaction energy diagram for a catalyzed (green) and
uncatalyzed (red) process.

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 reaction.
A catalyst will speed up both the forward and the reverse
reactions.
A catalyst does not affect either H or the overall yield for
a reaction.
16-53

16-54

9


Figure 16.24

The catalyzed decomposition of H2O2.

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

A small amount of NaBr is
added to a solution of H2O2.

Chemical Connections
Figure B16.1

Satellite images of the increasing size of the
Antarctic ozone hole (purple).

Oxygen gas forms quickly as Br-(aq)
catalyzes the H2O2 decomposition; the
intermediate Br2 turns the solution orange.

16-55

16-56

Chemical Connections
Figure B16.2 Reaction energy diagram for breakdown of O2
by Cl atoms. (Not drawn to scale.)

16-57

10


Lecture PowerPoint
Chapter 17

Chemistry
The Molecular Nature of
Matter and Change

Equilibrium: The Extent of Chemical Reactions

Martin S. Silberberg

17-1

Copyright  The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

17-2

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 Kc and Kp

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.

17.4 Comparing Q and K to Determine Reaction Direction

At equilibrium: rateforward = ratereverse

17.5 How to Solve Equilibrium Problems
17.6 Reaction Conditions and Equilibrium:

Le Châtelier’s Principle

17-3

Figure 17.1

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.
17-4

Reaching equilibrium on the macroscopic and
molecular levels.

The Equilibrium Constant
Consider the reaction N2O4(g)

2NO2(g)

At equilibrium ratefwd = raterev
2

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

kfwd
krev

2
= [NO2] eq

[N2O4]eq

The ratio of constants gives a new constant, the equilibrium
constant:
K=
17-5

2
kfwd
= [NO2]eq
krev [N2O4]eq

17-6

1


Figure 17.2

The range of equilibrium constants.

K and the extent of reaction
K reflects a particular ratio of product concentrations to
reactant concentrations for a reaction.
K therefore indicates the extent of a reaction, i.e., how
far a reaction proceeds towards the products at a
given temperature.
A small value for K indicates that the reaction yields little
product before reaching equilibrium. The reaction favors

the reactants.

small K
The reaction mixture
contains mostly
reactants.
large K
The reaction mixture
contains mostly
products.

A large value for K indicates that the reaction reaches
equilibrium with very little reactant remaining. The
reaction favors the products.
17-7

intermediate K

17-8

Figure 17.3

The Reaction Quotient Q

The change in Q during the N2O4-NO2 reaction.

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.
17-9

17-10

Table 17.1

Initial and Equilibrium Concentration Ratios for
the N2O4-NO2 System at 200°C (473 K)

Initial
[N2O4]

[NO2]

Q, [NO2]2
[N2O4]

1

0.1000

0.0000

0.0000

2

0.0000

0.1000

∞

3

0.0500

0.0500

0.0500

4

0.0750

0.0250

0.0833

17-11

[NO2]eq

K, [NO2] 2eq
[N2O4]eq

0.0357

0.193

10.4

0.000924

0.0982

10.4

0.00204

0.146

10.4

0.00275

0.170

10.5

[N2O4]eq

Writing the Reaction Quotient from the
Balanced Equation

PROBLEM: Write the reaction quotient, Qc, for each of the following
reactions:
(a) The decomposition of dinitrogen pentaoxide,

Equilibrium

Expt

Sample Problem 17.1

(b) The combustion of propane,

17-12

2


Forms of K and Q
For an overall reaction that is the sum of two more
individual reactions:
Qoverall = Q1 x Q2 x Q3 x ….. and
Koverall = K1 x K2 x K3 x ……

Forms of K and Q
If the coefficients of a balanced equation are multiplied by

a common factor,
[C]c[D]d
Q' = Qn = [A]a[B]b

n

and K' = Kn

The form of Q and K depend on the direction in which the
balanced equation is written:
Qc(rev) =

1
Qc(fwd)

Kc(rev) =

1
Kc(fwd)

17-13

17-14

Sample Problem 17.2

Writing the Reaction Quotient and Finding
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:

Sample Problem 17.3

Finding the Equilibrium Constant for an
Equation Multiplied by a Common Factor

PROBLEM: For the ammonia formation reaction, the reference
equation is
Kc is 2.4x10-3 at 1000 K. What are the values of Kc for
the following balanced equations:

(a) Show that the overall Qc for 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.

17-15

(a)

(b)

17-16

K and Q for hetereogeneous equilibrium

Figure 17.4

The reaction quotient for a heterogeneous system
depends only on concentrations that change.

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]
17-17

solids do not change
their concentrations

17-18

3


Table 17.2

Expressing Equilibria with Pressure Terms
Kc and Kp

Ways of Expressing Q and Calculating K

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.
2

Qp =

P NO2
P 2NO

x PO

Qc =

[NO2]2
[NO]2 [O2]

2

Kp = Kc (RT)Dn(gas)
If the amount (mol) of gas does not change in the reaction,
Dngas = 0 and Kp = Kc.

17-19

17-20

Sample Problem 17.4

Determining the Direction of Reaction

Converting Between Kc and Kp

PROBLEM: A chemical engineer injects limestone (CaCO3) into the hot
flue gas of a coal-burning power plant for form lime (CaO),
which scrubs SO2 from the gas and forms gypsum
(CaSO4·2H2O). Find Kc for the following reaction, if CO2
pressure is in atmospheres.

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.

17-21

Figure 17.5

17-22

Reaction direction and the relative sizes of Q and K.

Q>K
Q

Q=K

Sample Problem 17.5

Using Molecular Scenes to Determine
Reaction Direction

PROBLEM: For the reaction A(g)
B(g), the equilibrium mixture at
175°C is [A] = 2.8x10-4 M and [B] = 1.2x10-4 M. 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?

PLAN: We must compare Qc with Kc to determine the reaction direction,
so we use the given equilibrium concentrations to find Kc. Then we
count spheres and calculate Q for each mixture.

17-23

17-24

4


Using Concentrations to Determine Reaction
Direction
PROBLEM: For the reaction N2O4(g)

2NO2(g), Kc = 0.21 at
100°C. At a point during the reaction, [N2O4] = 0.12 M
and [NO2] = 0.55 M. Is the reaction at equilibrium? If not,
in which direction is it progressing?

Solving Equilibrium Problems

Sample Problem 17.6

If equilibrium quantities are given, we simply substitute
these into the expression for Kc to 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.

17-25

17-26

Example: In a study of carbon oxidation, an evacuated vessel
containing a small amount of powdered graphite is heated to 1080 K.
Gaseous CO2 is added to a pressure of 0.458 atm and CO forms. At
equilibrium, the total pressure is 0.757 atm. Calculate Kp.
CO2(g) + C(graphite)

Once we have the equilibrium amounts expressed in terms of x, we

use the other information given in the problem to solve for x.
The total pressure at equilibrium is 0.757 atm = PCO2(eq) + PCO(eq)

2CO(g)
0.757 atm = 0.458 – x + 2x

Pressure (atm)

Initial
Change
Equilibrium

CO2(g) +

C(graphite)

0.458

-

0

-x

-

+2x

0.458 -x

-

2x

0.757 atm = 0.458 + x
x = 0.757 – 0.458 = 0.299 atm
At equilibrium PCO2(eq) = 0.458 – x = 0.458 – 0.299 = 0.159 atm
PCO = 2x = 2(0.299) = 0.598 atm

The amount of CO2 will decrease. If we let the decrease in CO2 be x,
then the increase in CO will be +2x.

2

Kp =

Equilibrium amounts are calculated by adding the change to the
initial amount.

17-27

PCO2(eq)
PCO(eq)

2
= 0.598
0.159

= 2.25

17-28

Sample Problem 17.7
PROBLEM:

(from reaction table)

2CO(g)

Calculating Kc from Concentration Data

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)

Sample Problem 17.8
PROBLEM:

H2(g) + I2(g)

At equilibrium, [HI] = 0.078 M. Calculate Kc.

Determining Equilibrium Concentrations
from Kc

In a study of the conversion of methane to other fuels, a
chemical engineer mixes gaseous CH4 and H2O in a 0.32-L
flask at 1200 K. At equilibrium the flask contains 0.26 mol
of CO, 0.091 mol of H2, and 0.041 mol of CH4. What is the

[H2O] at equilibrium? Kc = 0.26 for this process at 1200 K.

PLAN: To calculate Kc we 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 H2 and x mol of I2 form. We
use this information to set up a reaction table.

17-29

17-30

5


Sample Problem 17.9

Determining Equilibrium Concentrations from
Initial Concentrations and Kc

PROBLEM: Fuel engineers use the extent of the change from CO and
H2O to CO2 and H2 to regulate the proportions of synthetic
fuel mixtures. If 0.250 mol of CO and 0.250 mol of H2O 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.

Sample Problem 17.10

Making a Simplifying Assumption to
Calculate Equilibrium Concentrations

PROBLEM: Phosgene is a potent chemical warfare agent that is now
outlawed by international agreement. It decomposes by the
reaction COCl2(g)
CO(g) + Cl2(g); Kc = 8.3x10-4 at
360°C. Calculate [CO], [Cl2], and [COCl2] when the following
amounts of phosgene decompose and reach equilibrium in a
10.0-L flask.
(a) 5.00 mol COCl2

17-31

17-32

The Simplifying Assumption
We assume that x([A]reacting can be neglected if
• Kc is relatively small and/or
• [A]init is relatively large.
If [A]init > 400, the assumption is justified; neglecting x
Kc
introduces an error < 5%.
If [A]init < 400, the assumption is not justified; neglecting x
Kc
introduces an error > 5%.

17-33

Figure 17.6

(b) 0.100 mol COCl2

Sample Problem 17.11

Predicting Reaction Direction and
Calculating Equilibrium Concentrations

PROBLEM: The research and development unit of a chemical company is
studying the reaction of CH4 and H2S, two components of
natural gas: CH4(g) + 2H2S(g)
CS2(g) + 4H2(g)
In one experiment, 1.00 mol of CH4, 1.00 mol of CS2, 2.00 mol
of H2S, and 2.00 mol of H2 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.

17-34

Steps in solving equilibrium problems.

Figure 17.6

PRELIMINARY SETTING UP

continued
SOLVING FOR x AND EQUILIBRIUM
QUANTITIES

1. Write the balanced equation.
2. Write the reaction quotient, Q.

6. Substitute the quantities into Q.

3. Convert all amounts into the
correct units (M or atm).

7. To simplify the math, assume that x is
negligible:
([A]init – x = [A]eq ≈ [A]init)

WORKING ON THE REACTION
TABLE

8. Solve for x.
 Check that assumption is
justified (<5% error). If not, solve
quadratic equation for x.

4. When reaction direction is not
known, compare Q with K.
5. Construct a reaction table.

9. Find the equilibrium quantities.

 Check the sign of x, the
change in the concentration
(or pressure).

17-35

 Check to see that calculated
values give the known K.

17-36

6


Le Châtelier’s Principle

The Effect of a Change in Concentration

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.
The system responds to a disturbance by a shift in the
equilibrium position.
A shift to the left is a net reaction from product to reactant.
A shift to the right is a net reaction from reactant to product.

17-37

Figure 17.7

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
have an effect.
A change in concentration has no effect on the value of K.

17-38

The effect of a change in concentration.
Table 17.3 The Effect of Added Cl2 on the PCl3-Cl2-PCl5 System

Concentration (M)

PCl3(g) +

Original equilibrium

0.200

Disturbance

Change
New equilibrium

17-39

Figure 17.8

PCl3(g)
0.600

0.200

0.200

-x

-x

0.600
+x

0.200 - x

0.200 - x

0.600 + x
(0.637)*

determined value

17-40

The effect of added Cl2 on the PCl3-Cl2-PCl5 system.

PCl3(g) + Cl2(g)

PCl5(g)

Sample Problem 17.12

Predicting the Effect of a Change in
Concentration on the Equilibrium Position

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 O2:
2H2S(g) + O2(g)

When Cl2 (yellow curve) is added, its
concentration increases instantly (vertical
part of yellow curve) and then falls
gradually as it reacts with PCl3 to form
more PCl5. Equilibrium is re-established at
new concentrations but with the same
value of K.

17-41

0.125
+0.075

New initial

*Experimentally

Cl2(g)

What happens to
(a) [H2O] if O2 is added?

2S(s) + 2H2O(g)
(b) [H2S] if O2 is added?

(c) [O2] if H2S is removed? (d) [H2S] 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.

17-42

7


The Effect of a Change in Pressure (Volume)

Figure 17.9

The effect of a change in pressure (volume) on a
system at equilibrium.

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.
17-43

17-44

Sample Problem 17.13

The Effect of a Change in Temperature

Predicting the Effect of a Change in Volume
(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?
(a) CaCO3(s)

CaO(s) + CO2(g)

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

SF6(g)

(c) Cl2(g) + I2(g)

2ICl(g)

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.

17-45

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
endothermic reaction.
A decrease in temperature removes heat, which favors the
exothermic reaction.

17-46

Temperature and K
The only factor that affects the value of K for a given
equilibrium system is temperature.

Sample Problem 17.14

Predicting the Effect of a Change in
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?
(a) CaO(s) + H2O(l)

For a reaction with DH°rxn > 0, an increase in temperature
will cause K to increase.

(b) CaCO3(s)
(c) SO2(g)

For a reaction with DH°rxn < 0, an increase in temperature
will cause K to decrease.
The van’t Hoff equation shows this relationship:
ln

K2
K1

17-47

o
= - DH rxn
R

1
T2

-

1
T1

R = 8.314 J/mol·K
K1 is the equilibrium constant at T1

Ca(OH)2(aq) DH° = -82 kJ
CaO(s) + CO2(g) DH° = 178 kJ

S(s) + O2(g) DH° = 297 kJ

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.

17-48

8

Catalysts and Equilibrium

Table 17.4 Effects of Various Disturbances on a System at
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.

17-49

17-50

Sample Problem 17.15

Determining Equilibrium Parameters from
Molecular Scenes

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):

The Synthesis of Ammonia
Ammonia is synthesized industrially via the Haber process:
N2(g) + 3H2(g)

(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
[Y2]?

17-51

There are three ways to maximize the yield of NH3:
• Decrease [NH3] by removing NH3 as it forms.
• Decrease the volume (increase the pressure).
• Decrease the temperature.

17-52

Table 17.5 Effect of Temperature on Kc for Ammonia Synthesis
N2(g) + 3H2(g)

2NH3(g)

Figure 17.11 Percent yield of ammonia vs. temperature at five
different pressures.

DH°rxn = -91.8 kJ

T (K)

Kc

200.

7.17x1015

300.

2.69x108

400.

3.94x104

500.

1.72x102

600.

4.53x100

700.

2.96x10-1

800.

3.96x10-2

An increase in temperature causes the equilibrium to shift towards
the reactants, since the reaction is exothermic.

17-53

2NH3(g) DH°rxn = -91.8 kJ

17-54

At very high P and low T (top left), the yield is high, but the
rate is low. Industrial conditions (circle) are between 200 and
300 atm at about 400°C.

9


Figure 17.12 Key stages in the Haber process for
synthesizing ammonia.

Chemical Connections
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.

17-55

17-56

Chemical Connections
Figure B17.2

The effect of inhibitor binding on the shape of an
active site.

17-57

10


03-May-20

Lecture PowerPoint
Chapter 18

Chemistry

Acid-Base Equilibria

The Molecular Nature of
Matter and Change

Martin S. Silberberg

18-1

Copyright  The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

Acid-Base Equilibria

18-2

Table 18.1

Some Common Acids and Bases and their
Household Uses.

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
Effect
18.9 Electron-Pair Donation and the Lewis Acid-Base Definition

18-3

18-4

Strong and Weak Acids

Arrhenius Acid-Base Definition
This is the earliest acid-base definition, which classifies
these substances in terms of their behavior in water.

A strong acid dissociates completely into ions in water:

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

An acid is a substance with H in its formula that dissociates
to yield H3O+.

A dilute solution of a strong acid contains no HA molecules.
A weak acid dissociates slightly to form ions in water:

A base is a substance with OH in its formula that
dissociates to yield OH-.
When an acid reacts with a base, they undergo
neutralization:
H+(aq) + OH-(aq) → H2O(l) DH°rxn = -55.9 kJ

18-5

HA(aq) + H2O(l)

H3O+(aq) + A-(aq)

In a dilute solution of a weak acid, most HA molecules are
undissociated.
Kc =

[H3O+][A-]
has a very small value.
[HA][H2O]

18-6

1


03-May-20

Figure 18.1A

The extent of dissociation for strong acids.

Figure 18.1B

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

The extent of dissociation for weak acids.

H3O+(aq) + A-(aq)

Weak acid: HA(aq) + H2O(l)

There are no HA molecules in solution.
Most HA molecules are undissociated.

18-7

18-8

Figure 18.2

Reaction of zinc with a strong acid (left) and a
weak acid (right).

The Acid Dissociation Constant, Ka
H3O+(aq) + A-(aq)

HA(aq) + H2O(l)
Kc =
Zinc reacts rapidly with
the strong acid, since
[H3O+] is much higher.

1 M HCl(aq)

Kc[H2O] = Ka =

[H3O+][A-]
[HA]

The value of Ka is an indication of acid strength.
Stronger acid

higher [H3O+]

Weaker acid

lower % dissociation of HA

larger Ka
smaller Ka

1 M CH3COOH(aq)

18-9

Table 18.2

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

18-10

Ka Values for some Monoprotic Acids at 25°C

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.)

18-11

18-12

2


03-May-20

Classifying the Relative Strengths of Bases

Sample Problem 18.1

Classifying Acid and Base Strength from
the Chemical Formula

PROBLEM: Classify each of the following compounds as a strong
acid, weak acid, strong base, or weak base.

• Strong bases include
– water-soluble compounds containing O2- or OH- ions.

(a) KOH

(b) (CH3)2CHCOOH

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

(c) H2SeO4

(d) (CH3)2CHNH2

• M2O or MOH, where M = Group 1A(1) metal (Li, Na, K, Rb, Cs)
• MO or M(OH)2 where 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.

18-13

18-14

Autoionization of Water

The Ion-Product Constant for Water (Kw)
2H2O (l)
Kc =

H3O+ (aq) + OH- (aq)
[H3O+][A-]
[H2O]2

Kc[H2O]2 = Kw = [H3O+][OH-] = 1.0x10-14 (at 25°C)
Water dissociates very slightly into ions in an equilibrium
process known as autoionization or self-ionization.
2H2O (l)

H3

O+

(aq) +

OH-

(aq)

18-15

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

= 1.0x10-7 (at 25°C)

Both ions are present in all aqueous systems.

18-16

Figure 18.3

A change in [H3O+] causes an inverse change in [OH-],
and vice versa.
Higher [H3O+]

lower [OH-]

Higher [OH-]

lower [H3O+]

The relationship between [H3O+] and [OH-] and the

relative acidity of solutions.

We can define the terms “acidic” and “basic” in terms of
the relative concentrations of H3O+ and OH- ions:
In an acidic solution,
In a neutral solution,
In basic solution,

18-17

[H3O+] > [OH-]
[H3O+] = [OH-]
[H3O+] < [OH-]

18-18

3


03-May-20

Sample Problem 18.2

Calculating [H3O+] 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 [H3O+] =
3.0x10-4 M. Calculate [OH-]. Is the solution neutral,
acidic, or basic?

The pH Scale
pH = -log[H3O+]
The pH of a solution indicates its relative acidity:
In an acidic solution,
In a neutral solution,
In basic solution,

pH < 7.00
pH = 7.00
pH > 7.00

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

18-19

18-20

Table 18.3

The Relationship between Ka and pKa

Figure 18.4
Acid Name (Formula)

The pH values of some
familiar aqueous
solutions.

Ka at 25°C

pKa

1.0x10-2

1.99

Nitrous acid (HNO2)

7.1x10-4

3.15

Acetic acid (CH3COOH)

1.8x10-5

4.75

Hypobromous acid (HBrO)

2.3x10-9

8.64

Phenol (C6H5OH)

1.0x10-10

10.00

Hydrogen sulfate ion (HSO4

pH = -log [H3O+]

-)

pKa = -logKa
A low pKa corresponds to a high Ka.

18-21

18-22

Figure 18.5

pH, pOH, and pKw

The relations among [H3O+], pH, [OH-], and pOH.

Kw = [H3O+][OH-] = 1.0x10-14 at 25°C
pH = -log[H3O+]
pOH = -log[OH-]

pKw = pH + pOH = 14.00 at 25°C

pH + pOH = pKw for any aqueous solution at any temperature.

Since Kw is 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).

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03-May-20

Sample Problem 18.3

Calculating [H3O+], pH, [OH-], and pOH

Figure 18.6

Methods for measuring the pH of an aqueous solution.

PROBLEM: In an art restoration project, a conservator prepares
copper-plate etching solutions by diluting concentrated
HNO3 to 2.0 M, 0.30 M, and 0.0063 M HNO3. Calculate
[H3O+], pH, [OH-], and pOH of the three solutions at 25°C.

pH paper

pH meter

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Brønsted-Lowry Acid-Base Definition

Figure 18.7

Lone pair
binds H+

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
an H+ ion.
• A base must contain a lone pair of electrons to bond
to H+.

Dissolving of an acid or base in water as a BrønstedLowry acid-base reaction.

(acid, H+ donor)

(base, H+ acceptor)

Lone pair
binds H+

An acid-base reaction is a proton-transfer process.
(base, H+ acceptor)

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(acid, H+ donor)

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Conjugate Acid-Base Pairs

Conjugate Acid-Base Pairs

In the forward reaction:
NH3 accepts a H+ to form NH4+.

H2S + NH3

HS- + NH4+

H2S donates a H+ to form HS-.
In the reverse reaction:
NH4+ donates a H+ to form NH3.

H2S + NH3

H2S + NH3

HS- + NH4+

H2S and HS- are a conjugate acid-base pair:
HS- is the conjugate base of the acid H2S.

NH3 and NH4+ are a conjugate acid-base pair:
NH4+ is the conjugate acid of the base NH3.
A Brønsted-Lowry acid-base reaction occurs when an
acid and a base react to form their conjugate base
and conjugate acid, respectively.

HS- + NH4+

acid1 + base2

base1 + acid2

HS- accepts a H+ to form H2S.
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