First Law of Thermodynamics

draweryaleMechanics

Oct 27, 2013 (3 years and 11 months ago)

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Mullis

1

First Law of Thermodynamics

(Law of Conservation of Energy)


The combined amount of matter and energy in
the universe is constant.


Potential energy


Kinetic energy


Enthalpy = heat gained or lost by system:
Δ
H


Δ
H < 0: Exothermic process




Δ
H > 0: Endothermic Process





Endothermic vs. Exothermic


Chemistry,
Raymond Chang, 11
th

edition

2

Mullis

3

q and
Δ
H: Signs

Sign is determined by the experience of the
system:

Heat in

System

(The reaction

occurs in here!)

Δ
H > 0

Δ

H
is positive.

Endothermic
Rxn

System

(The reaction

occurs in here!)

Δ
H < 0

Δ

H
is negative.

Exothermic
Rxn

Heat out

Mullis

4

Calorimetry


Calorimeter = Measures temp change in a process



Heat capacity = amount of heat to raise temp by 1K (or
1º C)



Specific heat = heat capacity for 1 g of a substance

(Symbol for specific heat is usually C)



Amount of heat absorbed by a substance calculated
using mass, specific heat and temperature change:


q =
Δ
E = mc
Δ

T



Mullis

5

Heat Measurements Using Calorimeter

50.0 mL of 0.400 M CuSO
4

at 23.35 ºC is mixed with 50.0 mL of
0.600 M NaOH at 23.35 ºC in a coffee
-
cup calorimeter with heat
capacity of 24.0 J/ ºC. After reaction, the temp is 25.23 ºC. The
density of the final solution is 1.02 g/mL. Calculate the amount
of heat evolved. Specific heat of water is 4.184J/g ºC .

q =
Δ
E = mc
Δ

T

Mass = (50 mL+50 mL)(1.02g/mL) = 102 g

Δ

T = 25.23
-
23.35 = 1.88
ºC

Heat absorbed by solution = (102g)(
4.18J
)(1.88 ºC) = 801J








g
ºC

Add this to heat absorbed by calorimeter

= (
24.0J
)(1.88 ºC) = 45.1J







g
ºC

Total heat liberated by this reaction = 846 J




Mullis

6

Hess’s Law Example

C
2
H
5
OH + 3O
2



2CO
2

+ 3H
2
O
Δ
H =
-
1367 kJ/mol


C
2
H
4

+ 3O
2



2 CO
2

+ 2H
2
O
Δ
H =
-
1411 kJ/mol

Find
Δ
H for C
2
H
4

+ H
2
O


C
2
H
5
OH


2CO
2

+ 3H
2
O


C
2
H
5
OH + 3O
2

Δ
H =

1367

kJ/mol

C
2
H
4

+ 3O
2



2 CO
2

+ 2H
2
O
Δ
H =
-
1411 kJ/mol

___________________________________________

C
2
H
4

+ H
2
O


C
2
H
5
OH

Δ
H =
-
44 kJ/mol


Mullis

7

Heating Curve at Constant Pressure

Curve is flat during phase
changes.

Area A: Temperature
remains constant until
all the solid has become
liquid because melting
requires energy.

Once the energy is no
longer required for
phase change, kinetic
energy again increases.

Mullis

8

Heating Curve at Constant Pressure,
cont.

Length of horizontal
line A is proportional
to the heat of fusion.

The higher the heat of
fusion, the longer the
line.

Line B is longer than A
because heat of
vaporization is higher
than heat of fusion.

Ex: For water:

Heat of fusion

= 334
J/g

Heat of vaporization

=2260J/g

Δ
H
fus

= 6.02 kJ/mol

Δ
H
vap

= 40.7 kJ/mol


Mullis

9

Heating Curve at Constant Pressure,
cont.

Gas and solid warming
slopes are steeper than
that for liquid warming.

The specific heat of the
liquid phase is usually
greater than that of the
solid or gas phase.

Ex: for water:

a.
Solid = 2.09 J/g
-
ºC

b.
Liquid = 4.18
J/g
-
ºC

c.
Gas =
2.03 J/g
-
ºC