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Chemistry Question on Law Of Chemical Equilibrium And Equilibrium Constant

The equilibrium constant for the following reaction is 1.6 ×105 at 1024 K.
H2(g)+Br2(g)⇋2HBr(g)H_2(g) + Br_2(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.

Answer

Given, Kp for the reaction i.e.,
H2(g)+Br2(g)↔2HBr(g)H_2(g) + Br_2(g) ↔ 2HBr(g) is 1.6×1051.6×10^55.
Therefore, for the reaction 2HBr(g)↔H2(g)+Br(g)2HBr(g) ↔ H_2(g) + Br(g), the equilibrium constant will be,
K′p=1KpK'p = \frac {1}{K_p}
K′p=11.6×105K'p= \frac {1}{1.6×10^5}
K′p=6.25×10−6K'p= 6.25×10^{-6}
Now, let p be the pressure of both H2 and Br2 at equilibrium.
2HBr(g)↔H2(g)+Br2(g)2HBr(g) ↔ H_2(g) + Br_2(g)
Initial conc. 1010 00 00
At equilibrium 10−2p10 - 2p pp pp
Now, we can write,
PHBr×P2PHBr2=K′p\frac {P_{HBr} × P_2}{P^2_{HBr}} = K'p

p×p(10−2p)2=6.25×10−6\frac {p × p }{(10 - 2p)^2} = 6.25×10^{- 6 }

p10−2p=2.5×10−3\frac {p}{10 - 2 p} = 2.5 × 10^{- 3}
P=2.5×10−2−(5.0×10−3)PP = 2.5 × 10^{-2} - (5.0 × 10^{-3})P
P+(5.0×10−3)P=2.5×10−2P + (5.0 × 10^{- 3})P = 2.5 × 10^{- 2}
(1005×10−3)P=2.5×10−2(1005×10^{- 3}) P = 2.5 × 10^{- 2}
P=2.49×10−2 barP = 2.49 × 10^{- 2}\ bar
P=2.5×10−2 barP= 2.5×10^{- 2}\ bar (approximately)
Therefore, at equilibrium,
[H2]=[Br2]=2.49×10−2 bar[H_2] = [Br_2] = 2.49 × 10^{-2}\ bar
[HBr]=10−2×(2.49×10−2) bar=9.95 bar=10 bar[HBr] = 10-2×(2.49×10^{-2})\ bar = 9.95 \ bar = 10\ bar (approximately)