7.27 The equilibrium constant for the following reaction is 1.6 ×105 at 1024K

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 1024K.

NEET SOLUTION:

Given, 
Kp for the reaction i.e., H2g+Br2g  2HBr(g) is 1.6×105.
Therefore, for the reaction, the equilibrium constant will be,
K'p=1Kp=11.6×105=6.25×10-6
Given,
Now, let p be the pressure of both H2 and Br2 a equilibrium.
                       2HBrg           H2g    +     BrgInitial conc.      10                       0                   0At equilibrium  10-2p               p                   p
Now, we can write,
pH2×pBr2p2HBr=K'pp×p10-2p2=6.25×10-6p10-2p=2.5×10-3p=2.5×10-2-5.0×10-3pp+5.0×10-3p=2.5×10-21005×10-3p=2.5×10-2p=2.49×10-2bar=2.5×10-2 barapprox.
Therefore, at equilibrium,
H2=Br2=2.49×10-2barHBr=10-2×2.49×10-2bar=9.95bar=10barapprox.