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.
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.
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)
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)
