The equilibrium constant for the following reaction is `1.6 xx10^5` 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 1024 K.

Explanation

`{:("Equilibrium reaction :",H_(2(g)) + , Br_(2(g)) hArr, 2HBr_((g))),("Stoichiometry :",1,1,2):}`
So, `Deltan_((g))="(mol of product - mol of reactant)"_g`
=2-1-1=0
`Deltan=0`, So, `K_p = K_c`
`{:("Reverse reaction :",2HBr + , H_(2(g)) hArr,Br_(2(g))),("Initial pressure :","10 bar","zero","zero"),("Change in pressure :","-2x",x,x),("Partial pressure at equilibrium :",(10-2x)"bar",x "bar" , x "bar"):}`
`K_(p_r)=((p_(H_2))(p_(Br_2)))/(p_(HBr))^2="(x bar)(x bar)"/(10-2x)^2`
but for reserve reaction, `K_p =1/(K_p "of forward")`
`1/(1.6xx10^5)=x^2/(10-2x)^2`
`1/(16xx10^4)=(x/(10-2x))` (take square root both side)
`1/(4xx10^2)=x/(10-2x)`
`therefore` 10-2x=400x
`therefore` 402x=10
`therefore x=10/402`=0.02488 bar `approx` 0.025 bar
`therefore` At equilibrium `p_(H_2)=p_(Br_2)`=x=0.025 bar and `p_(HBr)`= 10-2x=10-2 (0.025)
=10-0.05 = 9.95 bar