Finding equilibrium concentrations: A mixture of `0.50` mol `H_2` and `0.50` mol `I_2` is placed in a `1.00L` stainless steel container at `400^@C`. The equilibrium constant `K_c` for the reaction
`H_2(g)+I_2(g)hArr 2HI(g)`
is `54.3` at this temperature. Calculate the equilibrium concentrations of `H_2`, `I_2`, and `HI`.

Options

A. `{:(,,[H_(2)] [M],[I_(2)] [M],[HI] [M]),(,1,0.200,0.200,0.0200):}`
B. `{:(,,[H_(2)] [M],[I_(2)] [M],[HI] [M]),(,2,0.00427,0.00427,0.0315):}`
C. `{:(,,[H_(2)] [M],[I_(2)] [M],[HI] [M]),(,3,0.315,0.0315,0.00850):}`
D. `{:(,,[H_(2)] [M],[I_(2)] [M],[HI] [M]),(,4,0.00478,0.00478,0.0352):}`

Answer

B

Explanation

`H_(2)+I_(2) Leftrightarrow 2HI, K_(1)=54.4`
`2HI Leftrightarrow H_(2)+I_(2)`
Eq. mol (0.04-2x) x x
`K_(2)=1/K_(1), (1)/(54.4)=(x^(2))/((0.04-2x)^(2))`
on solving, x=0.00427