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The concentration of a pure solid or liq...

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.

Text Solution

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The correct Answer is:
a
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Concentration of pure solid and liquid is not included in the expression of equilibrium constant because

Explain why pure liquids and solids can ignored while writing the equilibrium constant expression?

Knowledge Check

  • For the system 3A+2BhArrC, the expression for equilibrium constant is

    A
    `([3A][2B])/(C)`
    B
    `([C])/([3A][2B])`
    C
    `([A]^(2)[B]^(2))/([C])`
    D
    `([C])/([3A]^(3)[2B]^(2))`

  • 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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    For the reaction A+B hArr C+D , the initial concentrations of A and B are equal. The equilibrium concentration of C is two times the equilibrium concentration of A. The value of equilibrium constant is ………..

    For the reaction, A+BhArr C+D, the initial concentration of A and B are equal, but the equilibrium concentration of C is twice that of equlibrium concentration of A. The equlibrium constant is

    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:

    statement-1 : No tem in the concentration of a pure solid ora pure liquid appears in an equilibrium constant expression. statement-2 : Each pure solid or pure liquid is in a phase by itself, and has a constant concentration at constant temperature.

    Assertion: No term in the concentration of a pure solid or a pure liquid apperas in an equilibrium constant expression. Reason: Each pure solid or pure liquid is in a phase by itself and has a constant concentration at constant temperature.

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