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.

Options

A. Forward
B. Backward
C. Won't shift
D. Unpredictable

Answer

A

Explanation

The initial concentrations are
`[HI]=(2.0xx10^(-2))/(2)=1.0xx10^(-2)"mol "L^(-1)`
`[H_(2)]=(1.0xx10^(-2))/(2)=0.5xx10^(-2)"mol "L^(-1)`
`[I_(2)]=(3.0xx10^(-2))/(2)=1.5xx10^(-2)"mol "L^(-1)`
Reaction quotient,
`Q=([HI]^(2))/([H_(2)][I_(2)])=((1.0xx10^(-2))^(2))/((0.5xx10^(-2))(1.5xx10^(-2)))=1.33`
Since `Q lt K_(c)`, the reaction will proceed in the forward direction to attain equilibrium, so that Q becomes equal to K.