# Thermodynamics - Fort Bend Independent School District

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27 Οκτ 2013 (πριν από 4 χρόνια και 10 μήνες)

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Thermodynamics

Thermochemistry

Carol Brown

Saint Mary’s Hall

The 0th Law

Heat flows from hot to cold

The 1st Law

In all changes, energy is conserved

The energy of the universe is constant

∆E=q + w

(both q and w are from the system’s point of
view)

P∆V Work

Using the MKS system, determine the
units for P∆V.

Calorimetry
-
Basics

Heat: The total amount of thermal energy
contained in a sample of matter

Measured in calories or joules

Temperature: The average kinetic energy
of the molecules in a sample of mater.

Measured in Kelvins or degrees Celsius

More basics

Specific heat capacity:

The amount of heat
it takes to raise one gram of a substance
one degree Celsius.

Units: J/g
o
C

Heat capacity:

The amount of heat it takes
to raise a system one degree Celsius.

Units
--
J/
o
C

Coffee Cup Calorimeter

q = mc∆T

How much heat, in kJ, is
necessary to take 20 g of ice
at
-
5
o

C to steam at 100
o

C?
The specific heat of ice is 2.1
J/g
o
C; of liquid water is 4.2
J/g
o
C. The latent heat of
fusion is 334 J/g. The latent
heat of vaporization is 2268
J/g.

60.7 kJ

Potential Energy Diagrams
--
Exothermic Reactions

Potential Energy Diagrams
--
Endothermic Reactions

Important Terms

Enthalpy

(heat of reaction)
--
The amount of
heat lost or absorbed during the course of a
reaction when the only work done is
expansion or contraction at a constant
pressure (P∆V). Change in enthalpy is a
state function and is symbolized by ∆H.

What drives a reaction?

Enthalpy

∆H

Entropy

S

What is a State Function?

Examples of State Functions

Enthalpy

Entropy

Internal Energy

Temperature

Pressure

Thermodynamics

Defining a system

System

Surroundings

Universe

Stoichiometric Thermochemistry

When 2.50 g of methane burns in oxygen,
125 kJ of heat is produced. What is the
molar heat of combustion of methane
under these conditions?

Stoichiometric Thermochemistry:

When 2.50 g of methane burns in oxygen,
125 kJ of heat is produced. What is the
molar heat of combustion of methane
under these conditions?

-
802 kJ/mol

The oxidation of glucose is described by
the following equation:

C
6
H
12
O
6

+ 6O
2

--
> 6CO
2

+ 6H
2
O ∆H
o

=
-
2816 kJ

How much heat in kJ is produced by the
oxidation of 1.0 g of glucose?

The oxidation of glucose is described by
the following equation:

C
6
H
12
O
6

+ 6O
2

--
> 6CO
2

+ 6H
2
O ∆H
o

=
-
2816 kJ

How much heat in kJ is produced by the
oxidation of 1.0 g of glucose?

16 kJ

Another term

Standard molar heat of combustion:

The
amount of heat released when one mole of
a substance is burned in oxygen. The
measurements must be taken at standard
thermodynamic conditions. i.e. 298 K and
1.00 atm pressure. Symbolized by
∆H
o
comb
.

Still another important term

Standard molar heat of formation:

The
amount of heat lost or absorbed when one
mole of product is formed from its
elements in their most stable state. Again,
the measurements must be taken at
thermodynamic standard conditions.
Symbolized by ∆H
o
f
.

Is this an example of ∆H
o
comb
,
∆H
o
f
, neither, or both?

C + 1/2 O
2

--
> CO

CO + 1/2 O
2

--
> CO
2

CH
3
OH + 3/2 O
2

--
> CO
2

+ 2H
2
O

8 C + 9H
2

--
> C
8
H
18

CH
4

+ 2O
2

--
> CO
2

+ 2H
2
O

FeCl
2

+ 1/2Cl
2
--
> FeCl
3

S + O
2

--
> SO
2

Hess’s Law

Since enthalpy is a state function, the change in
enthalpy in going from some initial state to some
final state is independent of the pathway. This
means that in going from a particular set of
reactants to a particular set of products, the
change in enthalpy is the same whether the
reaction takes place in a single step or in a series
of steps. This principle is known as Hess’s Law.

Hess’s Law

Hess’s Law Problem #1

Given the following data:

H
2

+ 1/2O
2

--
> H
2
O(l)

∆H
o
=
-
285.8 kJ

N
2
O
5

+ H
2
O
--
> 2HNO
3

∆H
o
=
-
76.6 kJ

1
/
2
N
2

+

3
/
2
O
2

+

1
/
2
H
2
_
--
>

HNO
3

∆H
o
=
-
174
.
1

kJ

Calculate

∆H
o

for

the

reaction

2
N
2

+

5
O
2

--
>

2
N
2
O
5

Given the following data:

H
2

+ 1/2O
2

--
> H
2
O(l)

∆H
o
=
-
285.8 kJ

N
2
O
5

+ H
2
O
--
> 2HNO
3

∆H
o
=
-
76.6 kJ

1/2N
2

+ 3/2O
2

+ 1/2H
2
--
>HNO
3

∆H
o
=
-
174.1 kJ

Calculate ∆H
o

for the reaction

2N
2

+ 5O
2

--
> 2N
2
O
5

28.4 kJ

Another Hess’s Law Problem

Calculate ∆H
o

for the process

Sb(s) + 5/2 Cl
2
(g)
--
> SbCl
5
(g)

from the following information.

Sb(s) + 3/2 Cl
2
(g)
--
> SbCl
3
(g) ∆H
o

-
314 kJ

SbCl
3
(g) + Cl
2
(g)
--
> SbCl
5
(g) ∆H
o

-
80 kJ

Another Hess’s Law Problem:

Calculate ∆H
o

for the process

Sb(s) + 5/2 Cl
2
(g)
--
> SbCl
5
(g)

from the following information.

Sb(s) + 3/2 Cl
2
(g)
--
> SbCl
3
(g) ∆H
o

-
314 kJ

SbCl
3
(g) + Cl
2
(g)
--
> SbCl
5
(g) ∆H
o

-
80 kJ

-
394 kJ

Spontaneity

The driving force of reactions.

Enthalpy

Entropy

Free Energy ∆G

∆G = ∆H
-

T∆S