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Question

The equilibrium constant for the reaction is H2+Br22HBr at 1024 k is 1.60×105. Find the equilibrium pressure of all gases if 10 bar of HBr is introduced into a sealed container at 1024 K.

A
10 bar
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B
20 bar
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C
15 bar
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D
17 bar
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Solution

The correct option is A 10 bar
H2+Br22HBr;Kp=1.6×105 at 1024K
when reaction will be reverse,
2HBrH2+Br2;Kp=1/Kp=1/1.6×105 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.

Kp=P(H2)×P(Br2)/P2(HBr)
=(x/2).(x/2)/(10x)2
1/1.6×105=x2/4(10x)2
0.625×105=x2/4(10x)2
because values of K'p is so small so, (10x)10
0.625×105=x2/4×100
625×4×105=x2
x=0.05
x/2=0.025=2.5×102 bar
Hence, P(H2)=P(Br2)=2.5×102 bar
so, P(HBr)=10x=100.05 10bar

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