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1
Water – Ice Problem
Calculate the change in entropy of water,
S
system
, when 2 mol of water freezes at
0
o
C (273 K).
For ice,
H
fusion
= 6.01 kJ/mol
S = S
final
–S
initial
= q/T
A) + 44.0 J/K
B) – 44.0 J/K
C) + 0.022 J/K
D) – 0.022 J/K
2
Water (the system)
Heat flow: system (water) to surroundings (freezer)
During the freezing process T
water
is constant (0°C)
S = q/T
and q = n
H
fusion
H
fusion
= 6.01 kJ/mol
q
H2O
is negative since energy flows out
of the
system.
q
H2O
= 2 x 6.01 x 10
3
J = 1.202 x 10
4
J
S
H2O
= q
H2O
/T
H2O
= 1.202x10
4
/
273
= 44.0 J K
1
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When 2 mol of water freezes,
S
system
=  44 J/K.
The entropy decreases!
Makes sense liquid
solid.
Water freezes spontaneously, but we said
S > 0 for a spontaneous process.
What gives here?
Need to consider
the total entropy
not just the
entropy of the system.
4
Total Entropy Change
When water freezes, energy is transferred to
the surroundings.
Which means that the energy is dispersed over
more states in the surroundings.
So the entropy of the surroundings increases.
S
total
=
S
system
+
S
surroundings
S
total
is also called
S
universe
3
5
Calculate the change in entropy of the freezer
(surroundings),
S
surroundings
, when 2 mol of water
freezes at 0
o
C (273 K).
The temperature of the
freezer is – 15
o
C (258 K)
For ice,
H
fusion
= 6.01 kJ/mol
S = S
final
–S
initial
= q/T
A) + 46.6 J/K
B) – 46.6 J/K
C) + 44.0 J/K
D) – 44.0 J/K
6
Freezer (the surroundings)
The freezer absorbs energy at its temp. (15°C).
Absorbs energy so q
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This note was uploaded on 02/13/2011 for the course CHE 132 taught by Professor Hanson during the Spring '08 term at SUNY Stony Brook.
 Spring '08
 HANSON
 Chemistry

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