The correct option is A 10 bar
H2+Br2⇔2HBr;Kp=1.6×105 at 1024K
when reaction will be reverse,
2HBr⇔H2+Br2;K′p=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.
K′p=P(H2)×P(Br2)/P2(HBr)
=(x/2).(x/2)/(10−x)2
1/1.6×105=x2/4(10−x)2
0.625×105=x2/4(10−x)2
because values of K'p is so small so, (10−x)≈10
0.625×10−5=x2/4×100
625×4×10−5=x2
x=0.05
x/2=0.025=2.5×10−2 bar
Hence, P(H2)=P(Br2)=2.5×10−2 bar
so, P(HBr)=10−x=10−0.05≈ 10bar