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Concentration of pure solid and liquid i...

Concentration of pure solid and liquid is not included in the expression of equilibrium constant because

Text Solution

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[Pure liquid] or [Pure solid]=`("Number of moles")/("Volume of L")`
`=("Mass"//"Molecular mass")/("Volume")`
`=("Mass")/("Volume")xx1/("Molecular mass")`
`=("Density")/("Molecular mass")`
As density of a pure liquid or pure solid is constant at constant temperature and molecular mass is also constant, therefore, their molar concentration are constant and not included into the equilibrium constant.
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If more than one phase is present in the reversible reaction then it is said to be heterogenous system. Example: CaO(s)+CO_(2)(g) Expression of equilibrium constant for the above reaction can be taken as : K=([CaO(s)][CO_(2)(g)])/([CaO(s)]) ." ".....(i) Now concentration of CaO(s)=[CaO(s)] =("moles of CaO")/("volume of CaO") as density of CaO[rho_(CaO(s))] and molar mass of CaO[M_(CaO(s))] are a fixed quantity therefore concentration of pure solid and liquid term is uncharge with respect to time. Hence, equilibrium constant for the equation (i) can be written as : K_(C)=[CO_(2)(g))] K_(P)=P_(CO_2) As K_(p) and K_(c) is not containing solid terms therefore, addition or removel of pure solid and pure liquid has no effect on the equilibrium process. CaCO_(3)(s)hArr+CaO(s)+CO_(2)(s) At equilibrium in the above case, 'a' moles of CaCO_(3) , 'b' moles of CaO and 'c' moles of CO_(2) are found then identify the wrong statement:

Concentrates include :

Knowledge Check

  • The concentration of a pure solid or liquid phase is not include in the expression of equilibrium constant becase :

    A
    density of solid and liquid are independent of their quantities .
    B
    solids and liquids react slowly.
    C
    solids and liquids at equilibrium do not interact with gaseous phase.
    D
    the molecules of solids and liquids cannot migrate to the gaseous phose.

  • If more than one phase is present in the reversible reaction then it is said to be heterogenous system. Example: CaO(s)+CO_(2)(g) Expression of equilibrium constant for the above reaction can be taken as : K=([CaO(s)][CO_(2)(g)])/([CaO(s)]) ." ".....(i) Now concentration of CaO(s)=[CaO(s)] =("moles of CaO")/("volume of CaO") as density of CaO[rho_(CaO(s))] and molar mass of CaO[M_(CaO(s))] are a fixed quantity therefore concentration of pure solid and liquid term is uncharge with respect to time. Hence, equilibrium constant for the equation (i) can be written as : K_(C)=[CO_(2)(g))] K_(P)=P_(CO_2) As K_(p) and K_(c) is not containing solid terms therefore, addition or removel of pure solid and pure liquid has no effect on the equilibrium process. K_(p) for the reaction NH_(4)I(s)hArrNH_(3)(g)+HI(g) is 1//4 at 300K .If above equilibrium is established by taking 4 moles of NH_(4)I(s) in 100 litre contanier, then moles of NH_(4)I(s) left in the container at equilibrium is ["Taken R=1/12Lt.atm mol"^(-1)K^(-1)] .

    A
    1
    B
    2
    C
    3
    D
    4

  • If more than one phase is present in the reversible reaction then it is said to be heterogenous system. Example: CaO(s)+CO_(2)(g) Expression of equilibrium constant for the above reaction can be taken as : K=([CaO(s)][CO_(2)(g)])/([CaO(s)]) ." ".....(i) Now concentration of CaO(s)=[CaO(s)] =("moles of CaO")/("volume of CaO") as density of CaO[rho_(CaO(s))] and molar mass of CaO[M_(CaO(s))] are a fixed quantity therefore concentration of pure solid and liquid term is uncharge with respect to time. Hence, equilibrium constant for the equation (i) can be written as : K_(C)=[CO_(2)(g))] K_(P)=P_(CO_2) As K_(p) and K_(c) is not containing solid terms therefore, addition or removel of pure solid and pure liquid has no effect on the equilibrium process. 200g of CaCO_(3)(g) taken in 4Ltr container at a certain temperature. K_(c) for the dissociation of CaCO_(3) at this temperature is found to be 1//4 mole Ltr^(-1) then the concentration of CaO in mole/litre is : [Given : rho_(CaO)=1.12gcm^(-3)][Ca=40,O=16]

    A
    `(1)/(2)`
    B
    `(1)/(4)`
    C
    `0.02`
    D
    `20`

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