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One mole of H2O and one mole of CO are t...

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.

Text Solution

Verified by Experts

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
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