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(1)

Chapter 5

Thermochemistry

Chemistry, The Central Science, 11th edition Theodore L. Brown; H. Eugene LeMay, Jr.;

(2)

Energy

• Energy is the ability to do work or transfer

heat.

– Energy used to cause an object that has mass to move is called work.

(3)

Potential Energy

(4)

Kinetic Energy

Kinetic energy is energy an object possesses by virtue of its motion.

1 2

(5)

Units of Energy

• The SI unit of energy is the

joule (J)

.

• An older, non-SI unit is still in widespread

use: the

calorie (cal)

.

1 cal = 4.184 J

1 J = 1

kg m



2

(6)

Definitions:

System and Surroundings

• The system includes the molecules we want to

study (here, the hydrogen and oxygen molecules). • The surroundings are

(7)

Definitions: Work

• Energy used to move an object over some distance is work.

w = Fd

(8)

Heat

• Energy can also be transferred as heat. • Heat flows from

(9)

Conversion of Energy

(10)

Conversion of Energy

(11)

First Law of Thermodynamics

• Energy is neither created nor destroyed.
(12)

Internal Energy

(13)

Internal Energy

By definition, the change in internal energy, E, is the final energy of the system minus the initial energy of the system:

(14)

Changes in Internal Energy

• If

E

> 0,

E

final

>

E

initial

– Therefore, the system

absorbed energy from

the surroundings.

(15)

Changes in Internal Energy

• If

E

< 0,

E

final

<

E

initial

– Therefore, the system

released energy to the

surroundings.

(16)

Changes in Internal Energy

• When energy is

(17)
(18)

Exchange of Heat between

System and Surroundings

(19)

Exchange of Heat between

System and Surroundings

• When heat is absorbed by the system from the surroundings, the process is endothermic.

(20)

State Functions

Usually we have no way of knowing the

(21)

State Functions

• However, we do know that the internal energy of a system is independent of the path by

which the system achieved that state.

(22)

State Functions

• Therefore, internal energy is a state function. • It depends only on the present state of the

system, not on the path by which the system arrived at that state.

(23)

State Functions

• However, q and w are

not state functions.

• Whether the battery is shorted out or is

(24)

Work

Usually in an open

container the only work done is by a gas

pushing on the

(25)

Work

We can measure the work done by the gas if the reaction is done in a vessel that has been fitted with a piston.

(26)

Enthalpy

• If a process takes place at constant

pressure (as the majority of processes we

study do) and the only work done is this

pressure-volume work, we can account

for heat flow during the process by

(27)

Enthalpy

• When the system changes at constant

pressure, the change in enthalpy,

H

, is

H

=

(

E

+

PV

)

• This can be written

(28)

Enthalpy

• Since

E

=

q

+

w

and

w

= -

P

V

, we can

substitute these into the enthalpy

expression:

H

=

E

+

P

V

H

= (

q

+

w

) −

w

(29)

Endothermicity and

Exothermicity

• A process is

endothermic when

(30)

Endothermicity and

Exothermicity

• A process is

endothermic when

H is positive. • A process is

exothermic when

(31)

Enthalpy of Reaction

The change in

enthalpy, H, is the enthalpy of the

products minus the enthalpy of the

(32)

Enthalpy of Reaction

This quantity, H, is called the enthalpy of

(33)

The Truth about Enthalpy

1. Enthalpy is an extensive property.

2. 

H

for a reaction in the forward direction

is equal in size, but opposite in sign, to

H

for the reverse reaction.

(34)

Calorimetry

Since we cannot know the exact enthalpy of the reactants and

products, we measure

H through

calorimetry, the

(35)

Heat Capacity and Specific Heat

The amount of energy required to raise the

temperature of a substance by 1 K (1C) is its

(36)

Heat Capacity and Specific Heat

We define specific heat capacity (or simply

specific heat) as the amount of energy required

(37)

Heat Capacity and Specific Heat

Specific heat, then, is

Specific heat =

heat transferred

mass

temperature change

s

=

q

(38)

Constant Pressure Calorimetry

By carrying out a

reaction in aqueous solution in a simple

calorimeter such as this one, one can indirectly measure the heat

(39)

Constant Pressure Calorimetry

Because the specific heat for water is well

known (4.184 J/g-K), we can measure H for the reaction with this

equation:

(40)

Bomb Calorimetry

• Reactions can be

carried out in a sealed “bomb” such as this one.

(41)

Bomb Calorimetry

• Because the volume in the bomb

calorimeter is

constant, what is

measured is really the change in internal

(42)

Hess’s Law

H

is well known for many reactions,

and it is inconvenient to measure

H

for

every reaction in which we are

interested.

• However, we can estimate

H

using

(43)

Hess’s Law

Hess’s law states that “[i]f a reaction is

carried out in a series of steps, H for the

overall reaction will be equal to the sum of

(44)

Hess’s Law

Because H is a state function, the total

enthalpy change

depends only on the initial state of the

(45)

Enthalpies of Formation

An enthalpy of formation,

H

f

, is defined

as the enthalpy change for the reaction in

which a compound is made from its

(46)

Standard Enthalpies of Formation

Standard enthalpies of formation, Hf°, are
(47)

Calculation of

H

• Imagine this as occurring in three steps:

C

3

H

8 (g)

+ 5 O

2 (g)



3 CO

2 (g)

+ 4 H

2

O

(l)
(48)

Calculation of

H

• Imagine this as occurring in three steps:

C

3

H

8 (g)

+ 5 O

2 (g)



3 CO

2 (g)

+ 4 H

2

O

(l)
(49)

Calculation of

H

• Imagine this as occurring in three steps:

C

3

H

8 (g)

+ 5 O

2 (g)



3 CO

2 (g)

+ 4 H

2

O

(l)
(50)

C

3

H

8 (g)

+ 5 O

2 (g)



3 CO

2 (g)

+ 4 H

2

O

(l)

C3H8 (g)  3 C (graphite) + 4 H2 (g) 3 C (graphite) + 3 O2 (g)  3 CO2 (g) 4 H2 (g) + 2 O2 (g)  4 H2O (l)

Calculation of

H

(51)

Calculation of

H

We can use Hess’s law in this way:

H

=

n



H

f

°

products

m



H

f

°

reactants
(52)

H = [3(-393.5 kJ) + 4(-285.8 kJ)] – [1(-103.85 kJ) + 5(0 kJ)]

= [(-1180.5 kJ) + (-1143.2 kJ)] – [(-103.85 kJ) + (0 kJ)] = (-2323.7 kJ) – (-103.85 kJ) = -2219.9 kJ

C

3

H

8 (g)

+ 5 O

2 (g)



3 CO

2 (g)

+ 4 H

2

O

(l)
(53)

Energy in Foods

(54)

Energy in Fuels

The vast

majority of the energy

References

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