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Question

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 H2 gas if 10.0 bar of HBr is introduced into a sealed container at 1024 K.

A
2.8×103
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B
5.2×102
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C
2.5×102
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D
3.4×103
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Solution

The correct option is C 2.5×102
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)initial pressure1000

At equil. 102ppp

So,
Kp=p2(102p)2
putting the values we get,
11.6×105=p2(102p)2
400p=102p
p=0.025 bar
so, PH2=p=2.5×102 bar

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