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One mole of water and one mole of carbon...

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

A

0.444

B

0.044

C

4.44

D

444

Text Solution

Verified by Experts

The correct Answer is:
A
At equilibrium,
`[H_(2)O]=(1-0.40)/10"mol"L^(_1)`
= `0.06"mol"L^(-1)`
`[CO]=0.06"mol"L^(-1)`
`[H_(2)]=0.4/10=0.04"mol"L^(-1)=[CO_(2)]`
`K=([H_(2)][CO_(2)])/([H_(2)O][CO])=(0.04xx0.04)/(0.06xx0.06)=0.444`
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