The concentration of H_2, I_2 and HI at 731K respectively 0.92 xx 10^(-2), 0.20 xx 10^(-2) and 2.96 xx 10^(-2) "mol L"^(-1), calculate equilibrium constant. H_(2(g)) + I_(2(g)) hArr 2HI_((g))

The concentration of H_2, I_2 and HI at 731K respectively 0.92 xx 10^(

1.	The concentration of H_2, I_2 and HI at 731K respectively 0.92 xx 10^(-2), 0.20 xx 10^(-2) and 2.96 xx 10^(-2) "mol L"^(-1), calculate equilibrium constant. H_(2(g)) + I_(2(g)) hArr 2HI_((g))
Answer» Solution :The given REACTION of SYNTHESIS of HI is as under. `H_(2(g)) + I_(2(g)) HARR 2HI_((g))` and its `K_c` can represent as under `K_c=([HI]^2)/([H_2][I_2])` where, [HI] =`0.92xx10^(-2) "mol L"^(-1)` `[I_2]=0.20xx10^(-2) "mol L"^(-1)` `[HI]=2.96xx10^(-2) "mol L"^(-1)` `=((2.96xx10^(-2))^2("mol L"^(-1))^2)/((0.92xx10^(-2))(0.20xx10^(-2))("mol L"^(-1))( mol L^(-1)))` =47.62