In the reaction, 2H2(g) = H2(g)+L(9), if the equilibrium concentration of Hy, 1, and HI is 2 M, 2 M and 4 M respectively at a particular temperature then the equilibrium constant of the reaction will be
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Answer
H
2(g)
+I
2(g)
⇌2HI
(g)
K
c
=
[H
2
][I
2
]
[HI]
2
=54.8
⇒[H
2
][I
2
]=
K
c
[HI
2
]
∵[HI]=0.5 mol litre
−1
;
∴[H
2
][I
2
]=
54.8
[0.5]
2
=4.56×10
−3
At equilibrium, equal concentration of H
2
and I
2
will exist if dissociation of HI is carried out. Thus,
[H
2
]=[I
2
]
⇒[H
2
]
2
=4.56×10
−3
⇒[H
2
]=0.068 mol litre
−1
⇒[I
2
]=0.068 mol litre
−1
Explanation:
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