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Calculating the Equilibrium Constant For A Reaction
CALCULATING THE EQUI L IBRIUM CONSTANT FOR A REACTION
Name Section
1 ’. Write the equilibrium expression, given the following general equation.
aA (g) + bB (g) cC (g) + dD (g)
K =
[C]
c [D]
d
[A]
a [B]
b
Distinguish between the Kp and Kc for the reaction.
K
c
[C]
c [D]
d
[A]
a [B]
b
K
p
P
c
C
P
d
D
P
a
A
P
b
B
K
c
is the equilibrium constant in which the concentration of reactant and
products are expressed in terms of molarity. K p
is the equilibrium
constant in which the amounts of reactant and products are expressed in
terms of partial pressures measured in atm or mm Hg.
Below are listed three reactions. Associated with each reaction is a 1.0 L
container with a particulate level representation of the reaction before the
reaction has occurred. To the right is the 1.0 L container with a particulate
level representation of the reaction after attaining equilibrium. In each case
indicate whether you think the equilibrium constant for the reaction is greater
than 1, less than 1, or equal to 1. In each case support your answer with a brief
explanation.
a. Reaction I: A 2
(g) + B 2
(g) 2AB (g) (where A 2
(g) and B 2
(g) is
Container before the
reaction started
Container after the reaction
achieves equilibrium
K =
[AB]
2
[A
]
1 [B ]
1
[ 8 ]
2
[ 3 ]
1 [ 1 ]
1
b. Reaction II: C 2
(g) + D 2
(g) 2CD (g) (where C 2
(g) and D 2
(g) is
Container before the
reaction started
Container after the reaction
achieves equilibrium
K =
[CD]
2
[C
2
]
1 [D 2
]
1
[2]
2
[3]
1 [3]
1
c. Reaction II: X 2
(g) + Y 2
(g) 2XY (g) (where X 2
(g) and Y 2
(g) is
Container before the
reaction started
Container after the reaction
achieves equilibrium
K =
[XY]
2
[X
2
]
1 [Y 2
]
1
[4]
2
[3]
1 [3]
1
d. If any of the cases (K > 1, K < 1, or K = 1) did not appear in the three
examples above, use the space below to draw the before container, and
the equilibrium container for the missing case.
Container before the
reaction started
Container after the reaction
achieves equilibrium
K
c
[HI]
2
[H
2
][I
2
]
When the reaction is reversed the new equilibrium constant expression
becomes:
K
c
[H
2
][I
2
]
[HI]
2
=
K
c
ii)
1
2
H
2
(g) +
1
2
I
2
(g) ä HI (g)
The equilibrium constant expression from part b is
K
c
[HI]
2
[H
2
][I
2
]
When the coefficients in the reaction are changed, the new equilibrium
constant expression becomes;
K
c
[HI]
[H
]
1/ [I ]
1/
= K
c
- The initial concentration of both H 2
and I 2
is 0.250 M. The reaction occurs as
shown below,
H
2
(g) + I 2
(g) 2HI (g)
When equilibrium is achieved the concentration of HI is 0.393 M. Calculate the
magnitude of K c
for the reaction.
H
2
(g) + I 2
(g) 2HI (g)
Initial 0.250 M 0.250 M 0
We know the reaction proceeds from left to right because there is no HI
present initially and the equilibrium amount of HI is 0.393 M. To form HI
some H 2
and some I 2
must react. How much? To determine how much we
must use the coefficients from the balanced chemical equation.
The amount of H 2
reacting is,
0.393 M HI
1 mol H 2
2 mol HI
= 0.196 M H
2
Note: that this is also the amount
of I 2
The equilibrium amount of H 2
and I 2
is determined by subtracting the
amount reacting from the initial amount.
[H
2
]
eq
= 0.250 M - 0.196 M = 0.054 M
[I
2
]
eq
= 0.250 M - 0.196 M = 0.054 M
K
c
[HI]
2
[H
2
][I
2
]
[0.393 M]
2
[0.054 M][0.054 M]
- A vessel initially has a partial pressure of NO equal to 0.526 atm and a partial
pressure of Br 2
equal to 0.329 atm. At equilibrium the partial pressure of Br 2
is 0.203 atm. Calculate K p
for the reaction (K p
2NO (g) + Br 2
(g) 2NOBr (g)
2NO (g) + Br 2
(g) 2NOBr (g)
Initial 0.526 atm 0.329 atm 0 let x = P Br 2
(reacting)
Change - 2x - x +2x
Equilibrium 0.526 - 2x 0.329 - x 0 + 2x
[Br 2
]
eq
= 0.203 atm = 0.329 atm - x
x = 0.329 atm - 0.203 atm = 0.126 atm
[NO]
eq
= 0.526 - 2x = 0.526 - 2(0.126 atm) = 0.274 atm
[NOBr] eq
= 2x = 2(0.126 atm) = 0.252 atm
K
p
P
2
NOBr
P
2
NO
·P
Br
2
2 ·(0.203)
