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 ?

1 mol of CH_(4), 1 mole of CS_(2) and 2 mol of H_(2)S are 2 mol of H_(

1.	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 ?
Answer» Solution :`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.