At 700 K, equilibrium constant for the reaction H_(2(g)) + I_(2(g)) hArr 2HI_((g))is 54.8. If 0.5 mol L^(-1) of НІ_((g)) is present at equilibrium at 700 K, what are the concentration of H_(2(g)) and I_(2(g)) assuming that we initially started with HI(g) and allowed it to reach equilibrium at 700 K ?

At 700 K, equilibrium constant for the reaction H_(2(g)) + I_(2(g)) hA

1.	At 700 K, equilibrium constant for the reaction H_(2(g)) + I_(2(g)) hArr 2HI_((g))is 54.8. If 0.5 mol L^(-1) of НІ_((g)) is present at equilibrium at 700 K, what are the concentration of H_(2(g)) and I_(2(g)) assuming that we initially started with HI(g) and allowed it to reach equilibrium at 700 K ?
Answer» Solution :`{:("EQUILIBRIUM reaction :", H_(2(G)) + I_(2(g)) hArr , 2HI_((g)) , K_c=54.8),("Reaction start with HI :",2HI_((g)) hArr, H_(2(g)) + , I_(2(g)) of K._c),("Concentration at equilibrium [M]:", 0.5,X,x "and" K._c=1/K_c=1/54.8):}` Expression of equilibrium constant of reaction which start with HI, `K._c=([H_2][I_2])/([HI]^2` `therefore 1/54.8=((x)(x))/(0.5)^2=(x/0.5)^2` `therefore (1/54.8)^(1/2)=x/0.5=1/7.403` `therefore` 7.403 x= 0.5 and `x=0.5/7.403` = 0.0675 M So, `[H_2]=[I_2]`= x = 0.0675 M