The equilibrium constant for the following reaction is `1.6 xx 10^(5)` at `1024 K`
`H_(2)(g) + Br_(2)(g) hArr 2HBr(g)`
Find the equilibrium pressure of all gases if 10.0 bar of HBr is introduced into a sealed container at 1024K.

Explanation

` {:(,2HBr(g) ,hArr, H_2(g) ,+, Br_2(g) ""),( " Initial pressure " , 10.0 ,, -,, - ),(" Pressure at eq ". , 10.0 rho , , (x)/(2),,(x)/(2)):}`

For the reactions
` " " K= (1)/( 1.6xx 10 ^(5)) = 6.25 xx 10 ^(-6)`
Further,` " " K_c =K_p = 6.25 xx 10^(-6)`
` because " " K_p = (rho_(H_2) -rho_(Br_2))/(rho_(HBr)^(2))`
Hence ` " " 6.25 xx 10^(-6) = ((rho)/(2) .(rho)/(2))/((10.0- rho)^(2))`
Taking square root of both sides, we have
` " " 2.5 xx 10 ^(-3) = ((rho)/(2))/(10.0 - rho) = (rho)/(2(10.0 - rho))`
or ` 0.05 - 5.0 xx 10 ^(-3) rho = rho `
` or " " rho ( 1+ 5.0 xx 10 ^(-3)) = 0.05`
` or " " (0.05)/(1+5.0 xx 10 ^(-3))= 4.98 xx 10^(-2) `bar
Hence, at equilibrium
` rho _(H_2) = rho_(Br_2) = (rho)/(2) = (4.98 xx 10 ^(-2))/(2) = 2.49 xx 10 ^(-2) ` bar
and ` rho_(HBr) = (10 .0 -rho) = 10.0 - 4.98 xx 10 ^(-2)`
` " " = 9.95 ` bar.