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One mole of H(2)O and one mole of CO are...

One mole of `H_(2)O` 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.

Text Solution

Verified by Experts

The reaction is
`H_2O(g)+CO(g) hArr H_2(g) + CO_2(g)`
Number of moles of water initially present =1 mol
Percentage of water reacted =40%
`therefore` Moles of water reacted `=(1xx40)/100`=0.4 mol
Moles of water left unreacted =1.0-0.4=0.6 mol
According to the above equation, for every one mole of `H_2O`, one mole of CO reacts and forms one mole each of `CO_2` and `H_2`. Therefore , 0.4 mole of `H_2O` will react with 0.4 mole of CO to give 0.4 mole of `H_2` and 0.4 mole of `CO_2`. Since the total volume is 10 L, the molar concentrations at equilibrium are :
`[H_2]=0.4/10 "mol L"^(-1) , [CO_2]=0.4/10 "mol L"^(-1)`
`[H_2O]=0.6/10 "mol L"^(-1) , [CO]=0.6/10 "mol L"^(-1)`
Now , `K=([H_2][CO_2])/([H_2O][CO])=((0.4/10)xx(0.4/10))/((0.6/10)xx(0.6/10))`=0.444
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