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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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Knowledge Check

  • For the reaction CO_((g)) + (1)/(2) O_(2(g)), K_(1)//K_(c) is

    A
    RT
    B
    `(RT)^(-1)`
    C
    `(RT)^(-1//2)`
    D
    `(RT)^(1//2)`

  • For the reaction CO_((g)) + Cl_(2(g)) hArr COCl_(2(g)). The K_(p)//K_(c) is equal to

    A
    `(1)/(RT)`
    B
    1
    C
    `sqrt(RT)`
    D
    RT

  • In what manner the increase of pressure will affect the equation. C_((s))+H_2O_((g)) harr CO_((g))+H_(2(g))

    A
    Shift in the forward direction
    B
    Shift in the reverse reaction
    C
    Increase in the yield of `H_2`
    D
    from high motes to low gaseous moles

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