At 700K, equilibrium constant for the reaction: `H_(2)(g)+I_(2)(g)hArr 2HI(g)` is 54.8. if 0.5 `mol*L^(-1)` of HI(g) is present at equilibrium at 700 K, what are the concentrations 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

`[H_2]=[I_2]` = 0.068 M

Explanation

`{:(,2HI(g) hArr, H_2(g) + ,I_2),("At equi.",0.5,x,x):}`
`K_c=([H_2][I_2])/"[HI]"^2=1/54.8`
`x^2/(0.5)^2=1/54.8`
`x^2=0.25/54.8=4.56xx10^(-3)`
x=0.068
`[H_2]=[I_2]=0.068 mol L^(-1)`