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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_(2)` then the equilibrium constant `(K)` 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)`

Text Solution

Verified by Experts

The correct Answer is:
B
Forward reaction rate `(r_f)[A][B]=K_(1)[A][B]`
Backward reaction rate `(r_b) =K_2[C][D]=K_2[C][D]`
At equilibrium, `r_f=r_b`
`:. K_(1)[A][B]=K_2[C][D]`
The concentration of reactents & products of equilibrium are related by
`K=(K_(1))/(K_2)=([C][D])/([A][B])`
`:. (K_c)=(K_(1))/(K_2)`
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Knowledge Check

  • 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)`

  • The equilibrium constant (K) for the reaction. A+2BhArr2C+D is:

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

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