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Here’s a HARD problem…
Determine the equilibrium pressures for all species of the
reaction:
N
2
(g) + 3 H
2
(g)
!
2 NH
3
(g)
K = 1.64 x 10
6
Initially: P
N2
= 4.8 atm, P
H2
= 0.5 atm; P
NH3
= 2.0 atm
55
A) P
N2
= 4.8 atm, P
H2
= 3.50 atm, P
NH3
= 0.018 atm
B) P
N2
= 3.0 atm, P
H2
= 0.40 atm, P
NH3
= 5.6
x 10
4
atm
C) P
N2
= 5.79 atm, P
H2
= 3.47atm, P
NH3
= 0.020atm
D) P
N2
= 5.3 atm, P
H2
= 2.0 atm, P
NH3
= 1.0 atm
E) P
N2
= 3.3 atm, P
H2
= 0.0 atm, P
NH3
= 2.75 atm
HINT:
You should be able to rationalize the correct
answer without a full equilibrium calculation (if you tried
the calculation over the weekend, good work)
Reaction goes to the left (towards reactants)
56
Step 1
. Calculate Q
Step 2
. Make up reaction I.C.E. table
I
C
E
nitial
hange
quilibrium
4.8
0.5
+3x
+x
4.8+x
N
2
(g)
+
3 H
2
(g)
!
2 NH
3
(g)
2.0
 2x
2.02x
0.5+3x
Q
=
P
NH3
2
P
N2
P
H2
3
=
(2.0)
2
(4.8)(0.5)
3
= 6.67 > K
PROBLEM:
Looks like small K problem BUT
57
Step 3
. Substitute into K and solve for x
K
=
1.64
!
10
"
6
=
P
NH3
2
P
N2
P
H2
3
=
(2.0  2x)
2
(4.8 + x)(0.5 + 3x)
3
A quartic equation!
Option 1:
Option 2:
Solve exactly using Maple, Mathematica,
MATLAB…
Make an approximation
DIRECTION of reaction (to left) is a large K problem
K
reverse
=
1
K
forward
=
1
1.64
!
10
"
6
=
6.098
!
10
5
58
Step 2a
. Make up table for complete reverse reaction
I
C
C
nitial
hange
ompletion
2.0
4.8
+1.0
2.0
0.0
5.8
2 NH
3
(g)
!
N
2
(g)
+
3 H
2
(g)
0.5
+ 3.0
3.5
Step 2b
. Make up reaction I.C.E. table using results
from Step 2a as initial conditions.
I
C
E
nitial
hange
quilibrium
5.8
3.5
3x
x
5.8x
N
2
(g)
+
3 H
2
(g)
!
2 NH
3
(g)
0.0
+ 2x
+2x
3.53x
59
x = 0.0101
P
N2
= 5.79 atm; P
H2
= 3.47 atm; P
NH3
= 0.020 atm
Step 3
. Substitute into K and solve for x
K
=
1.64
!
10
"
6
=
P
NH3
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 Winter '10
 Brown
 Chemistry, Equilibrium, pH

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