One mole of H_2O and one mole of CO are taken in 10 L vessel and heated to 725 K. At equilibrium 40% of water (by mass) reacts with CO according to the equation: H_2O_((g)) + CO_((g)) hArr H_(2(g)) + CO_(2(g)) Calculate the equilibrium constant for the reaction.

One mole of H_2O and one mole of CO are taken in 10 L vessel and heate

1.	One mole of H_2O and one mole of CO are taken in 10 L vessel and heated to 725 K. At equilibrium 40% of water (by mass) reacts with CO according to the equation: H_2O_((g)) + CO_((g)) hArr H_(2(g)) + CO_(2(g)) Calculate the equilibrium constant for the reaction.
Answer» Solution :In starting 1 MOL `H_2O` present. Then at equilibrium reaction is occurred by 40% `H_2O` by mass. `therefore` REACTED `H_2O` = 40% of 1 =0.4 mol `{:("Equilibrium reaction:", H_2O_((g)) + ,CO_((g)) hArr , H_(2(g)) + , CO_(2(g))),("Inital mol:", 1,1,0,0),("Change in mol :", -0.4, -0.4 , +0.4 , +0.4),("Mol of equilibrium :", (1-0.4) , (1-0.4) , 0.4 , 0.4),("10 L vessel",0.6,0.6,0.4,0.4),(therefore "mol L"^(-1) "(at equilibrium):",0.6/10,0.6/10,0.4/10,0.4/10),(,0.06,0.06,0.04,0.04):}` The expression of equilibrium CONSTANT `K_c` is , `K_c=([H_2][CO_2])/([H_2O][CO])=((0.04)(0.04))/((0.06)(0.06))=4/9`=0.4444