Question

# The equilibrium constant for the following reaction is1.6\times 10^5 at 1024 K. 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 1024 K.

Solution

## Given, Kp for the reaction i.e., H2(g)+Br2g↔2HBr(g)is1.6×105 Therefore, for the reaction 2HBr(g)↔H2(g)+Br2(G), the equilibrium constant will be, K′p=1Kp=11.6×105=6.25×10−6 Now, let p be the pressure of both H2 and Br2 at equilibrium  2HBr(g)↔H(2(g))+Br2(g)Initial cone.1000At equlibrium10−2ppp Now, we can write, PHBr×PP2HBr=K′pP×p(10−2p)2=6.25×10−6P10−2p=2.5×10−3p=2.5×0−2−(5.0×10−3)pp+(5.0×10−3)p=2.5×10−2(0005×10−3)p=2.5×10−2 p=2.49×10−2bar=2.5×10−2bar (approximately) Therefore, at equilibrium, [H2]=[br2]=2.49×10−2 bar [HBr]=10−2×(2.49×10−2) bar =9.95bar=10 bar (approximately)

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