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.
Given,kp for the reaction i.e., H2(g) + Br2(g) ↔ 2HBr(g) is 1.6 ×105
Therefore, for the reaction
2HBr(g) ↔ H2(g) + Br2(g)
the equilibrium constant will be,
K'p = 1/Kp
= 1/1.6 ×105
= 6.25x10-6
Now, let p be the pressure of both H2 and Br2 at equilibrium.
2HBr(g) ↔ H2(g) + Br2(g)
Initial Conc. 10 0 0
at equilibrium 10-2p p p
Now, we can write,
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Welcome to the NCERT Solutions for Class 11 Chemistry - Chapter . This page offers a step-by-step solution to the specific question from Excercise 1 , Question 27: The equilibrium constant for the following reaction is 1.6 ×105 at 1024K H2(g) + Br2(g) &ha....
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