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Question

The equilibrium constant for the following reaction is1.6\times 10^5 at 1024 K.
H2(g)+Br2(g)2HBr(g)
Find the equilibrium pressure of all gases if 10.0 bar of HBr is introduced into a sealed container at 1024 K.


Solution

Given,
Kp for the reaction i.e., H2(g)+Br2g2HBr(g)is1.6×105
Therefore, for the reaction 2HBr(g)H2(g)+Br2(G), the equilibrium constant will be,
Kp=1Kp=11.6×105=6.25×106
Now, let p be the pressure of both H2 and Br2 at equilibrium
 2HBr(g)H(2(g))+Br2(g)Initial cone.1000At equlibrium102ppp
Now, we can write,

PHBr×PP2HBr=KpP×p(102p)2=6.25×106P102p=2.5×103p=2.5×02(5.0×103)pp+(5.0×103)p=2.5×102(0005×103)p=2.5×102
p=2.49×102bar=2.5×102bar (approximately)
Therefore, at equilibrium,

[H2]=[br2]=2.49×102 bar
[HBr]=102×(2.49×102) bar
=9.95bar=10 bar (approximately)

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