Question

At `448^(@)C`, the equilibrium constant `(K_(c))` for the reaction
`H_(2)(g)+I_(2)(g) hArr 2HI(g)`
is `50.5`. Presict the direction in which the reaction will proceed to reach equilibrium at `448^(@)C`, if we start with `2.0xx10^(-2)` mol of HI, `1.0xx10^(-2)` mol of `H_(2)` and `3.0xx10^(-2)` mol of `I_(2)` in a `2.0 L` constainer.

Answer 1

The initial concentration are
`[HI]=(2.0xx10^(-2))/(2) "mol" L^(-1) =1.0xx10^(-2) "mol" L^(-1)`
`[H_(2)]=(1.0xx10^(-2))/(2) "mol" L^(-1) =0.5xx10^(-2) "mol" L^(-1)`
`[I_(2)]=(3.0xx10^(-2))/(2) "mol" L^(-1) =1.5xx10^(-2) "mol" L^(-1)`
Concentration quotient.
`Q=([HI]^(2))/([H_(2)][I_(2)])`
`=((1.0xx10^(-2) "mol" L^(-1))^(2))/((0.5xx10^(-2) "mol" L^(-1))xx(1.5xx10^(-2) "mol" L^(-1)))=1.3`
Since `Q lt K`, the reaction will proceed in the forward direction to attain equilibrium so that `Q` becomes equal to `K`.