The Law of Mass Action explained

  1. Hi!

    I wondered if there is possible to get an explanation of the "Law of mass action." Here is a simple derivation:

    eq.1 aA + bB = cC + dD
    eq.2 R(f)= k(f)[A]^a*^b
    eq.3 R(b)= k(b)[C]^c*[D]^d
    eq.4 At equilibrium R(f)=R(b)
    eq.5 k(e)=k(f)/f(b)=([C]^c*[D]^d)/([A]^a*^b)

    Well, if a=4 and b=3, the left side of eq.1 becomes:
    eq.6 {left side}= 4A + 3B
    eq.7 {left side}= A + A + A + A + B + B + B

    Now (this is probably a pretty stupid question), why is it that all the factors in eq.2-5 are *multiplied* together, and not *added*?
     
    Last edited: Jun 11, 2008
  2. jcsd
  3. not stupid. 98% people dont realize what it is and ask the same.
    But theres one valid, very valid and very strongly reasonable answer to this.

    I give you the explaination, no crap.
    Consider 3A + 2B --> Z
    fine?

    Now, I let the concentration of A be (0.5 moles/liter)
    then I 'define' the (0.5) molarity or 'moles per litre', ("Implicity") as follows:
    "You have 50% chance to find exactly 1 mole of A in 1 litre of given volume of the solution."

    so when you say its 0.3 molars or moles/litre, I say "Its as good as a 30% chance to find exactly one mole of it in 1 litre of given volume of the solution."

    And for the reaction to happen you need 3 moles of A to be present, and each of them has a 50% probability to be present (as defined by 'definition') so... its as good as saying you have three coins and you want heads in all of them as the probability to get heads in all of them (possibility of head in each is 1/2 so possibility to get heads at a single time in all of them is 1/2 * 1/2 * 1/2 or (1/2)^3. Because here, you want all of em to be present at the same time and each has 1/2 chance to be present (molarity = 0.5 = 1/2) so.... if you want n moles of A its as good as the probability of exactly n moles of A to be present at the same time, and since each has a probability of k to be present (let the molarity of A be 'k' moles per litre) so... to get the probability for all of em to be present you do [k]^n;
    but similarly you also need m moles of B at the same time, so (let the molarity of B be 'j' moles per litre, then...) you have probability of exactly m moles of B to be present at the same time = [j]^m

    and since you need both A and B
    you do [k]^n * [j]^n

    get it?
    reference: http://www.khanacademy.org/video/ke...-not-necessary-to-progress?playlist=Chemistry
     
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