Homework 6
Solutions
1.
2.25
K
3NO
NO
O
N
P
(g)
2(g)
(g)
2
=
↔
+
The equilibrium constant expression in terms of pressure is
25
.
2
)
)(P
(P
)
(P
K
2
2
NO
O
N
3
NO
P
=
=
In the problem, we are given P
NO
= 2.00 atm
2
NO
P
= 3.00 atm
Thus,
)
(3.00)(P
(2.00)
2.25
NO2
3
=
2
NO
P
= 1.19 atm
2. Since the reactants and products are given in moles, first calculate K
C
The general expression for the relationship between K
P
and K
C
is.
Δ
n
C
P
(RT)
K
K
=
5
P
10
*
3
.
1
K
−
=
1
1
.K
m.mol
0.0821L.at
R
=
500K
T
=
∆
n = 1
Thus, 1.3 * 10
5
= K
C
(0.0821 * 500)
K
C
=
3.2 * 10
7
[SO
3
] = 0.200M
[SO
2
] = 0.100M
[O
2
] = 0.150M
The Q
C
expression for the reaction is
2(g)
2(g)
3(g)
O
2SO
2SO
+
↔
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2
3
1
2
2
2
C
]
[SO
]
[O
]
[SO
Q
=
Now you all should be able to take it from here,
3.
3(g)
2(g)
2(g)
2NH
3H
N
↔
+
According to this reaction 1 mol N
2
, reacts with 3 mols H
2
to
produce 2 mols NH
3
.
However, at equilibrium there are
1.5mol
3.0mol
2.0mol
3(g)
2(g)
2(g)
2NH
3H
N
↔
+
Therefore in order to calculate the moles of H2O required to produce
1.5 mol NH3 we write the mole ratio.
3
2
2molNH
3molH
ratio
mole
=
2
3
3
2
2.25molH
1.5molNH
*
2molNH
3molH
=
Thus, in order to produce 1.5 mol NH
3
, 2.25 mol H
2
are required.
Now at equilibrium, we have 3.0 mol H
2
. Hence, initially there are
3.0+2.25 = 5.25 mol H
2
. In correct numbers of significant figures,
initially there are 5.3 mol H
2
.
4.
(g)
(g)
B
A
↔
Initial
1.00 atm
0.00 atm
If “x” atm of A goes to form B, then
At equilibrium
x
x
00
.
1
x
x
−
−
Thus, K
P
A
B
P
P
Substituting the respective values.
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 Spring '08
 AMATEIS
 Chemistry, Equilibrium, pH, H+

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