Answer:
For the equilibrium reaction 2HBr(g)⇌H
2
(g)+Br
2
(g) , the equilibrium constant is K=
1.6×10
5
1
.
Let the decease in the equilibrium pressure of HBr be p bar.
The initial pressures of HBr, H
2
,Br
2
are 10 bar, 0 bar and 0 bar respectively.
The equilibrium pressures are 10-p, p/2 and p/2 respectively.
The equilibrium constant expression is
K
p
=
(P
HBr
)
2
P
H
2
P
Br
2
=
(10−p)
2
(p/2)×(p/2)
=
1.6×10
5
1
4(10−p)
2
p
2
=
1.6×10
5
1
400p=20−2p
p=0.0498 bar
The equilibium pressures are
P
H
2
=P
Br
2
=
2
0.0498
=0.0249 bar
P
HBr
=10−0.0498=10 bar
Explanation:
