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In a reaction A+BhArrC+D the rate consta...

In a reaction `A+BhArrC+D` the rate constant of forward reaction & backward reaction is `K_(1)` and `K_(12)` then the equilibrium constant `(K)` for reaction is expressed as:

Text Solution

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The correct Answer is:
`K_(c)=(K_(1))/(K_(2))`
Forward reaction rate `(r_(f))=K_(1)[A][B]`
Bacward reaction rate `(r_(f))=K_(2)[C][D]`
At equilibrium, `R_(f)=R_(b)`
`therefore K_(1)[A][B]=K_(2)[C][D]`
The concentration of reactents & products at equlibrium are related by
`K=(K_(1))/(K_(2))=([C][D])/([A][B]) therefore K=(K_(1))/(K_(2))`
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Knowledge Check

  • In a reaction A + B harr C + D the rate constant of forward reaction & backward reaction is k_(1) and k_2 then the equilibrium constant ( K_c ) for reaction is expressed as:

    A
    `K_c=(K_(2))/(K_(1))`
    B
    `K_c=(K_(1))/(K_(2))`
    C
    `K_c=K_(1) xx K_(2)`
    D
    `K_c=K_(1) + K_(2)`

  • The equilibrium constant (K) of a reaction may be written as

    A
    `K=e^(-DeltaG//RT)`
    B
    `K=e^(-DeltaG^(@)//RT)`
    C
    `K=e^(-DeltaH//RT)`
    D
    `K=e^(-DeltaH^(@)//RT)`

  • The equilibrium constant (K) of a reaction may be written as

    A
    `K=e^(-DeltaG//RT)`
    B
    `K=e^(-DeltaG^(@)RT)`
    C
    `K=e^(-DeltaH//RT)`
    D
    `K=e^(-DeltaH^(@)RT)`

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