The equilibrium constant for the following reaction is 1.6×105 at 1024KH2(g)+Br2(g)⇌2HBr(g) Find the equilibrium pressure of H2 gas if 10.0bar of HBr is introduced into a sealed container at 1024K.
A
2.8×103
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B
5.2×10−2
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C
2.5×10−2
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D
3.4×10−3
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Solution
The correct option is C2.5×10−2 For the reaction 2HBr(g)⇌H2(g)+Br2(g) KC=11.6×105 , as the reaction is reverse of previous reaction.
As Δng=0, so KP=KC=11.6×105
Now decompostion, 2HBr(g)⇌Br2(g)+H2(g)initialpressure1000
Atequil.10−2ppp
So, Kp=p2(10−2p)2 putting the values we get, 11.6×105=p2(10−2p)2 ⇒400p=10−2p ⇒p=0.025bar so, PH2=p=2.5×10−2bar