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A mixture of one mole of CO(2) and one m...

A mixture of one mole of `CO_(2)` and one mole of `H_(2)` attains equilibrium at a temperature of `250^(@)C` and a total pressure of 0.1 atm for the change `CO_(2(g)) +H_(2(g)) harr CO_((g))+H_(2)O_((g))`. Calculate `K_(P)` if the analysis of final reaction mixture shows 0.16 volume fraction of CO

A

0.46

B

0.63

C

0.22

D

0.82

Text Solution

Verified by Experts

The correct Answer is:
C

volume fraction = pressure fraction = mole fraction
`X_(CO)=X_(H_(2)O)=0.16`
`X_(CO_(2))=X_(H_(2))=(1-2 xx 0.16)/(2)=0.34`
`K_(P)=((X_(CO) xx P)(X_(H_(2)O) xx P))/((X_(CO_(2)) xx P)(X_(H_(2)O) xx P))`
`=(0.16 xx 0.16)/(0.34 xx 0.34)=0.22`
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