SELECT * FROM question_mgmt as q WHERE id=3051 AND status=1 SELECT id,question_no,question,chapter FROM question_mgmt as q WHERE courseId=2 AND subId=9 AND chapterId=48 and ex_no='1' AND status=1 ORDER BY CAST(question_no AS UNSIGNED) CBSE Free NCERT Solution of 11th chemistry Equilibrium the equilibrium constant for the following reactio

Question:

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.

Answer:

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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