The value of `K_(c)` for the reaction `2HI(g)hArrH_(2)+I_(2)(g)` is `1xx10^(-4)`. At a given time, t he composition of reaction mixture is
`[HI]=2xx10^(-5) mol`,`[H_(2)]=1xx10^(-5)` mol and `[l_(2)]=1xx10^(-5)` mol
In which direction will the reaction proceed ?

Options

A. forward
B. backward
C. equilibrium
D. none

Answer

B

Explanation

Given that,
`[HI]=2 xx 10^(-5)" mol" [H_(2)=1 xx 10^(-5)" mol"]`
`[I_(2)]=1 xx 10^(-5)"mol"`
At a given time, the reaction quotient Q for the reaction will be given by the expression `Q=([H_(2)] [I_(2)])/([HI]^(2))=(1 xx 10^(-5) xx 1 xx 10^(-5))/((2 xx 10^(-5))^(2))=1/4`
`=0.25 =2.5 xx 10^(-1)`
As the value of reaction quotient is greater than the value of `K_(c), i.e, 1 xx 10^(-4)` the reaction will proceed in the reverse reaction.