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Help with natural logarithm question? Please :) Watch

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    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.

    Answers:

    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]
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    (Original post by Princess Kawaii)
    ...
    If I understand correctly:

    K=\prod A_i^{\nu_i}

    which in the case of your particular equation becomes:

    K=A^{-a}B^{-b}C^{c}D^{d}


    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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