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S40.
Using the thermodynamic data from your textbook, calculate
Δ
S° at 298K for each of the following reactions.
Explain
the sign of each of the entropy changes by qualitatively comparing the molecular disorder in the reactants and
products.
To find
Δ
S°, we need to take the product of the number of moles and the
standard molar entropies and sum up those results for both the reactants and the
product, then take the difference of the two sums according to the equation
Δ
S
° =
nS
m
°
(products)
∑
−
nS
m
°
(reactants)
∑
a)
Cl
2
(g)
→
2Cl(g)
Δ
S
° =
(2 moles)
×
(165.2 J/mole K)
−
(1 mole)
×
(223.07 J/mole K)
=107.33 J/K
The sign is positive because chlorine atoms are more randomly distributed (they
have a larger number of microstates and hence greater disorder) than the paired atoms
or Cl
2
.
Therefore going from Cl
2
to 2 Cl increases the entropy of the system.
b)
H
2
O(l)
→
H
2
O(g)
Δ
S
° =
(1 mole)
×
(188.83 J/mole K)
−
(1 mole)
×
(69.91 J/mole K)
=118.92 J/K
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This note was uploaded on 02/01/2011 for the course CHEM 111 taught by Professor Owens,g during the Fall '08 term at University of Utah.
 Fall '08
 Owens,G
 Mole, Reaction

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