Question

The equilibrium constant for the reaction `H_(2) (g) + Br_(2) (g) hArr 2 HBr (g) " at " 1024 K " is " 1*6 xx 10^(5)`
Find the equilibrium pressure of all gases if `10*0` bar of HBr is introduced into a sealed container at 1024 K.

Answer 1

` {:(,2HBr (g),hArr,H_(2)(g),+,Br_(2)(g),K=1/(1*6 xx10^(5))),("Intial",10"bar",,,,,),("At eqm.",10-p,,p//2,,p//2,):}`
` K_(p) = ((p//2)( p//2))/(10-p)^(2)= 1/(1*6 xx10^(5))(p^(2))/(4(10-p)^(2) )= 1/(1*6 xx 10^(5) )`
Taking square root of both sides, we get
` p/(2(10-p) )= 1/(4 xx10^(2)) or 4 xx 10^(2) p= 2 (10-p) or 402 p = 20 or p= 20/402 = 4* 98 xx 10^(-2) "bar "`
Hence , at equilibrium `p_(H_(2)) =pBr_(2) = p//2 = 2*5 xx 10^(-2) "bar" , p_(HBr) = 10-p ~~ 10 " bar" `