Question

Question 7.27 The equilibrium constant for the following reaction is 1.6 ×105 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.

Class XI Equilibrium Page 227

Answer 1

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Answer 2

H2 + Br2 <=> 2HBr ; Kp = 1.6 × 10^5 at 1024K
when reaction will be reverse ,
2HBr <=> H2 + Br2 ; K'p = 1/Kp = 1/1.6 × 10^5 at 1024K

at initial time , pressure of HBr is 10 bar
pressure of H2 and Br2 are 0 bar .

at equilibrium , pressure of HBr is (10-x) bar
pressure of H2 and Br2 are x/2 bar.

K'p = P(H2) × P(Br2)/P²(HBr)
= (x/2).(x/2)/(10-x)²
1/1.6 × 10^5 = x²/4(10-x)²
0.625 × 10^-5 = x²/4(10-x)²
because values of K'p is so small so, (10-x)≈10
0.625 × 10^-5 = x²/4×100
625 × 4 × 10^-5 = x²
x = 0.05
x/2 = 0.025 = 2.5 × 10^-2 bar
hence, P(H2) = P(Br2) = 2.5 × 10^-2 bar
so, P(HBr) =10-x = 10-0.05 ≈ 10bar