how to solve it for a reaction N2+3H -----> 2NH3 the equilibrium mixture contains 0.25M nitrogen and 0.15M hydrogen gas at 25 degree centigrade calculate the concentration of NH3 given kc=9.6 The volume of container is 1dm^3
Answers
Answer:
Answer
N
2
(g)+3H
2
(g)⇌2NH
3
(g)
At equilibrium, (1−x) (3−3x) 2x
Given, x=0.0025
Active conc.
4
(1−0.0025)
4
(3−0.0075)
4
(0.0050)
Applying law of mass action,
K
c
=
[N
2
][H
2
]
3
[NH
3
]
2
=
(
4
0.9975
)(
4
2.9925
)
3
(
4
(0.0050)
)
2
=1.49×10
−5
litre
2
mol
−2
K for the reaction
2
1
N
2
(g)+
2
3
H
2
(g)⇌NH
3
(g) is equal to
K
c
.
K=
K
c
=
1.49×10
−5
=3.86×10
−3
litre mol
−1
Explanation:
1 mole of nitrogen and 3 moles of hydrogen are mixed in a 4 L container. If 0.25 percent of nitrogen is converted to ammonia by the following reaction:
N
2
(g)+3H
2
(g)⇌2NH
3
(g).
Calculate the equilibrium constant (K
c
) in concentration units. What will be the value of K for the following equilibrium?
2
1
N
2
(g)+
2
3
H
2
(g)⇌NH
3
(g)