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

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

1.	One mole of `H_(2)O` and one mole of CO are taken in a 10 litre vessel and heated at 725 K. At equilibrium 40% of water (by mass) reacts with CO according to the equation : `H_(2) O (g) + CO(g) hArr H_(2)(g) + CO_(2) (g) ` calculate the equilibrium constant for the reaction.
Answer» ` [H_(2) O ] = (1 - 040 )/10 "mol"L^(-1) = 006 "mol"L^(-1) , [CO] = 0 06 "mol" L^(-1)` ` [H_(2)] = (04)/10 "mol" L^(-1) = 004 "mol" L^(-1) , [CO_(2) ] = 004 "mol" L^(-1) ` ` K = ([H_(2) ] [ CO_(2)] )/([H_(2)O][CO])=(004 xx004 )/(006 xx 006)= 0*444 .`

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

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