One mole of `H_(2)` and one mole of `CO` are taken in a `10 litre` vessel and heated to `725 K`. At equilibrium, `40 per cent` of water (by mass) reacts with carbon monoxide according to the equation, `H_(2)O_((g))+CO_((g))hArrH_(2(g))+CO_(2(g))` Calculate the equilibrium constant for the reaction.

One mole of `H_(2)` and one mole of `CO` are taken in a `10 litre` ves

1.	One mole of `H_(2)` and one mole of `CO` are taken in a `10 litre` vessel and heated to `725 K`. At equilibrium, `40 per cent` of water (by mass) reacts with carbon monoxide according to the equation, `H_(2)O_((g))+CO_((g))hArrH_(2(g))+CO_(2(g))` Calculate the equilibrium constant for the reaction.
Answer» `{:(,H_(2)O_((g)),+,CO_((g)),hArr,H_(2(g)),+,CO_(2(g))),("Initial moles",1,,1,,0,,0), ("Mole at eqm.",(1-.04),,(1-0.4),,0.4,,0.4):}` (`40%` of `H_(2)O` reacts at eqm.) `K_(c)=([CO_(2)][H_(2)])/([H_(2)O][CO])=(0.4xx0.4)/(0.6xx0.6)=0.44`

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One mole of `H_(2)` and one mole of `CO` are taken in a `10 litre` vessel and heated to `725 K`. At equilibrium, `40 per cent` of water (by mass) reacts with carbon monoxide according to the equation, `H_(2)O_((g))+CO_((g))hArrH_(2(g))+CO_(2(g))` Calculate the equilibrium constant for the reaction.

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