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Discussion in 'MCAT Study Question Q&A' started by SonhosDaVida, Sep 14, 2018.

  1. SonhosDaVida

    2+ Year Member

    May 23, 2014
    Likes Received:
    For this questions, why is it that in the Keq equation, you can ignore the numerator x? The Keq is a large value so the x will play a large role in it's value.

    Question 12

    1536926393409535006157622724926.jpg 15369264209218806008261655785290.jpg
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  3. Lawper

    Lawper In 3D
    Rocket Scientist Gold Donor 2+ Year Member

    Jun 17, 2014
    Likes Received:
    As far as the MCAT is concerned, when you are dealing with complicated equations to solve for unknown concentrations, you can safely assume x is significantly smaller than the initial concentrations and basically ignore it completely when solving for it. This is important to save time on test day.

    For this problem, if you want to be exact and not want to assume x is small, you're stuck with a scary equation of:

    (1-2x)^2/((2x)^2*x) = 2.1*10^4 -->
    (1-2x)^2/(4x^3) = 2.1*10^4 -->
    (1-2x)^2 = 2.1*10^4 * 4x^3

    This is not an easy equation to solve by hand but it can be solved by using a software. This is why for MCAT purposes, you can safely reduce 1-2x expression to 1 by assuming that x will be significantly smaller than the initial concentrations at equilibrium. This makes sense because in equilibrium, you'll have a mix of both reactants and products and generally, all the reactants or all products won't be entirely consumed. Not to mention, x can't be larger than 0.5 because the ICE table in the problem would suggest the final amount of product would be negative, which is not possible.

    Note that this small x approximation shortcut won't always work but you won't see this on the MCAT.
    SonhosDaVida likes this.

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