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 direcition 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 lit"^(-2)`
`"Volume = 500 ml"=1/2L`
`{:([CH_(4)]_("in")=("1 mol")/(1)" "[CS_(2)]_("in")=("1 mol")/(1/2L)),(" = 2 mol L"^(-1)" = 2 mol L"^(-1)),([H_(2)S]_("in")=("2 mol")/(1/2L)" "[H_(2)]=("2 mol")/(1/2L)),(" = 4 mol L"^(-1)" = 4 mol L"^(-1)):}`
`Q=([CS_(2)][H_(2)]^(4))/([CH_(4)][H_(2)S]^(2))`
`:.Q=(2xx(4)^(4))/((2)xx(2)^(2))=64`
`QgtKC`
The reaction will proceed in the reverse direation to reach the equilibrium.