1 mol of `CH_(4)`, 1 mole of `CS_(2)` and 2 mol of `H_(2)S` are 2 mol of `H_(2)` are mixed in a 500 ml flask. The equilibrium constant for the reaction `K_(c)=4xx10^(-2)"mol"^(2)"lit"^(-2)`. In which direction will the reaction proceed to reach equilibrium?

Explanation

`CH_(4)(g)+2H_(2)S(g)hArrCS_(2)(g)+4H_(2)(g)`
`K_(c)=4xx10^(-2)"mol"^(2)"lit"^(-2)`
Volume = 500 ml = `1/2L`
`[CH_(4)]_("initial")=(1"mol")/(1/2L)`
`[CS_(2)]_("initial")=(1"mol")/(1/2L)`
= `2"mol"L^(-1)=2"mol"L^(-1)`
`[H_(2)S]_("initial")=(2"mol")/(1/2L)=4"mol"L^(-1)`
`[H_(2)]=(2"mol")/(1/2L)=4"mol"L^(-1)`
`Q=([CS_(2)][H_(2)]^(4))/([CH_(4)][H_(2)S]^(2))`
`thereforeQ=(2xx(4)^(4))/((2)xx(4)^(4))=16`
`QgtK_(c)`
`therefore` The reaction will proceed in the reverse direction to reach the equilibrium.