The reaction, CO_((g)) + 3H_(2(g)) hArr CH_(4(g)) + H_2O_((g))is at equilibrium at 1300K in a 1L flask. It also contain 0.30 mol of CO, 0.10 mol of H_2 and 0.02 mol of H_2O and an unknown amount of CH_4 in the flask. Determine the concentration of CH_4 in the mixture. The equilibrium constant, K_c for the reaction at the given temperature is 3.90.

The reaction, CO_((g)) + 3H_(2(g)) hArr CH_(4(g)) + H_2O_((g))is at eq

1.	The reaction, CO_((g)) + 3H_(2(g)) hArr CH_(4(g)) + H_2O_((g))is at equilibrium at 1300K in a 1L flask. It also contain 0.30 mol of CO, 0.10 mol of H_2 and 0.02 mol of H_2O and an unknown amount of CH_4 in the flask. Determine the concentration of CH_4 in the mixture. The equilibrium constant, K_c for the reaction at the given temperature is 3.90.
Answer» Solution :`{:("EQUILIBRIUM reaction :",CO_((g))+,3H_(2(g)) HARR , CH_(4(g))+ , H_2O_((g))),("Mol at equilibrium :",0.30,0.10,x ,0.02),("Volume = 1L So, mol L"^(-1):,0.30,0.10,x,0.02):}` `K_c`=3.90, where, x=Concentration of `CH_4` equilibrium According to chemical equilibrium constant, `K_c=([CH_4][H_2O])/([CO][H_2]^3)` `therefore 3.90=((x)(0.02))/((0.30)(0.10)^3)` `therefore x=((3.9)xx(0.30)(1xx10^(-3)))/0.02` `=58.5xx10^(-3)` `=5.85xx10^(-2) "mol L"^(-1)`