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    • Thread Starter

    Help with this question please? And you can explain the principles of how to work it out?

    Q) The most general form of an equilibrium constant is given by:

    K = Π[concentration]u
    where Π (‘pi’) refers to the product of all the concentration terms of both reactants (e.g. [A] and [B]) and their products (e.g. [C] and [D]). u (‘nu’) is the stoichiometric coefficient (positive for products and negative for reactants).

    Write the equilibrium constant K for the general reaction:

    aA + bB → cC + dD
    where ‘A’ refers to a reactant A and a to its stoichiometric coefficient, ‘C’ refers to a product C and c to its stoichiometric coefficient, etc.

    Give the expression of the natural logarithm of K.


    a) ln(K) = [A]-a × [B]-b × [C]c × [D]d

    b) ln(K) = - a ln [A] ×(- b ln [B]) × c ln [C] × d ln[D]

    c) ln(K) = - ln [A]/a - ln [B]/b + ln [C]/c + ln[D]/d

    d) ln(K) = - a ln [A] - b ln [B] + c ln [C] + d ln[D]
    • Study Helper

    Study Helper
    (Original post by Princess Kawaii)
    If I understand correctly:

    K=\prod A_i^{\nu_i}

    which in the case of your particular equation becomes:


    and taking logs, we have

    \ln K=\ln\left(A^{-a}B^{-b}C^{c}D^{d})

    There are two rules of manipulating logs that you need to know:

    \ln xy = \ln x + \ln y

    \ln x^y = y\ln x

    Try and have a go yourself. Start by splitting the product into a sum.
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Updated: December 31, 2012

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