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Consider a general, single-step reaction...

Consider a general, single-step reaction of the type `A + BhArrC`. Show that the equilibrium constant is equal to the ratio of the rate constant for the forward and reverse reaction, `K_(c)=(k_(f))/(k_(r))`

Text Solution

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`k_(f)[A][B]=k_(r)[C],(k_(f))/(k_(r))=([C])/([A][B])=k_(c)`
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Knowledge Check

  • For a reversible reaction A + B iff C + D the equilibrium constant, K_(C) is equal to

    A
    `([A][B])/([C][D])`
    B
    `([C][D])/([A][B])`
    C
    `([A][C])/([B][D])`
    D
    None of these
  • For a chemical reaction of the type AhArrB, K=2.0 and BhArrC, K=0.01 . Equilibrium constant for the reaction 2ChArr 2A is

    A
    25
    B
    50
    C
    2500
    D
    `4xx10^(-4)`
  • The unit of equilibrium constant K_(c) for the reaction A+B hArr C would be

    A
    `mol^(-1) L`
    B
    `mol L^(-1)`
    C
    `mol L`
    D
    No unit
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