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Equilibrium Law Calculations
(with RICE charts)
Example 14.7  pg. 567
H
2
+ I
2
↔
2HI
R
I
C
E
H
2
I
2
HI
R
atio,
I
nitial,
C
hange,
E
quilibrium
Q  Try
PE 9 on
pg. 568
• Read 566 (from “Calculating Kc…”) to 568.
Follow the sample calculation carefully.
PE 9  pg. 568
PCl
3
+ Cl
2
↔
PCl
5
Q  Try 14.38, 14.39 pg. 589
14.9  pg. 570
CO + H
2
O
↔
CO
2
+ H
2
CO
H
2
OC
O
2
11
1
0.100
0.100
0
x
x
+x
0.10  x 0.10  x
x
R
I
C
E
H
2
1
0
+x
x
= 4.06
=
[0.10 x]
2
[x]
2
Kc =
[CO
2
][H
2
]
[CO][H
2
O]
x/[0.10x] = 2.01, x = 0.2012.01x, 3.01x = 0.201
x=0.0668
Read 5701. Follow sample calculation carefully.
PE 11, 14.40, 14.41 pg. 589 (notice [ ] for 14.41)
Equilibrium calculations when Kc is very small
• Thus far, problems have been designed so
that the solution for x is straightforward
• If the problems were not so carefully designed
we might have to use quadratic equation (or
calculus) to solve the problem.
• If Kc is very large or very small we can use a
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This note was uploaded on 04/12/2011 for the course CHEM 1010 taught by Professor Marshall during the Spring '11 term at North Texas.
 Spring '11
 Marshall
 Chemistry, Equilibrium

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