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As far as I know, a large equilibrium constant shows that the a reaction like the one below wants to complete:

$$\ce{B <=> A}$$

If A and B are gases, then K is: $$K_1=\frac{[A]}{[B]}$$

but when I multiply the reaction by 20 , K is $$K_{20}=\frac{[A]^{20}}{[B]^{20}}$$ while the same number of moles of B are consumed. If $\small K_1 = 10$ then $\small K_{20} = $$10^{20}$ so the reaction must be approximately complete. Why does this happen while the reaction materials didn't change?

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First of all, you switched the numerator and denominator in the definition of $K$. If $\ce{A <=> B}$ , then $K=\frac{[B]}{[A]}$. Products over reactants.

Notice that, if $K$ is large, then mathematically it is necessary for either the numerator to be large and/or the denominator to be small. Therefore, qualitatively, if $K$ is large, then you either have a lot of products, or little left of reagents, both of which imply a reaction which occurs close to completion.

Now here's the problem - we were speaking qualitatively. Your first sentence ("a big equilibrium constant shows that the reaction wants to get complete") is only a rule of thumb. It happens to work well because in reactions where equilibria descriptions are quantitatively reliable/exist at all, the stoichiometric coefficients are relatively small when written with the lowest integers (using the lowest integer coefficients is almost always done). The rule of thumb is borne out of repeated calculations under these implied conditions. They are not the case for your second equilibrium example, so such a rule should not be expected to still be valid.

Indeed, though $10^{20}$ is an impressive number at first, when you actually go do the calculations to find the relative amounts of reactants and products, you will at some point have to take its twentieth root, promptly turning it back into a much smaller number. Another way to think about it is that with such large stoichiometric coefficients, even a small change in concentrations is amplified enormously by the large exponents and has a huge impact on the value of the reaction quotient; a small increase in products will go a long way to reaching the equilibrium condition.

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  • $\begingroup$ yes , edited that $\endgroup$ – user2561 Oct 26 '13 at 11:37
  • $\begingroup$ somebody said this example : A+B--->C+D then if k = 2 when we solve $$\frac{x^{2}}{(10 - x)^{2}}$$ then 5.85 moles of the reactants are consumed . for 100A+100B--->100C+100D , k = 2^100 and if you solve $$2^{100} = \frac{x^{200}}{(10 - x)^{200}}$$ , the answer will be the same . so the rule is false! $\endgroup$ – user2561 Oct 26 '13 at 12:08
  • $\begingroup$ forgot to say that moles of A and B are before the beginning 10 , I think it is wrong because they had multiply first moles by 100 so we would have 1000 moles of each reactant , Am I wrong? $\endgroup$ – user2561 Oct 26 '13 at 12:39
  • $\begingroup$ @user2561 When you multiply the global reaction by any number, you don't multiply the mole amounts of any species you're trying to solve the problem with; that's an entirely different piece of information related to the problem itself (a boundary condition, as they say), not the reaction . If you did everything correctly, the answer must be the same in either $A→B$ or $100A→100B$ if you use the same values of $[A]$ and $[B]$ in both situations. If the answer is not the same, then that's an indication you did not properly change the value of the equilibrium constant. $\endgroup$ – Nicolau Saker Neto Nov 4 '13 at 11:26
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You're a bit muddled up.
Let's say I have the following reaction: $$\ce{A <=> B}$$

This signifies that one molecule of $A$ reacts to give one molecule of $B$. The molecularity of this reaction is 1. The equilibrium constant (say $K_{eq}$) is defined as the ratio of concentration of products to that of reactants.

i.e. $K_{eq} = \frac{[B]}{[A]}$

If a reaction is written as: $$\ce{2A <=> 2B}$$

This signifies that two molecules of A react with each other to produce two molecules of B. It is not the same as our first reaction!
The equilibrium constant (say $K'_{eq}$) will be:

$$K'_{eq} = \frac{[B]\cdot[B]}{[A]\cdot[A]} = \frac{[B]^2}{[A]^2}$$

The equilibrium constant is used to get an idea of how much the reaction is favoured. If $K = 13$, then the concentrations of the products are thirteen times more than the concentrations of reactants. Therefore, any value of $K$ greater than one can signify that the reaction wants to proceed forward.

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  • $\begingroup$ There's an issue with your last sentence. The equilibrium ratio between products and reagents is unchanged after multiplying a reaction by any number. Also, technically, the reaction is exactly the same after multiplying the equation by a number. To say otherwise is to suggest that something changes in the kinetics/mechanism. Changing the relevant equilibrium constant is just a trick that ensures proper bookkeeping, and does not alter any parameter of the reactants/products/reaction. $\endgroup$ – Nicolau Saker Neto Nov 2 '13 at 20:38
  • $\begingroup$ I was told by my professor that saying 2A -> 2B would imply that 2 molecules of A react to form products, although multiplying a number to the equation can be used to compare it to other equations for adding two equations, or comparing stuff like $\Delta H$. Because it quantitatively doesn't make any difference :/ Is that wrong? $\endgroup$ – mikhailcazi Nov 3 '13 at 4:57
  • $\begingroup$ The implying part is a problem. The total chemical equation for a reaction actually does not in general convey any chemical information, in its purest form. All it does is count the atoms before and after reaction, making sure they're in the same amount and type. But we often try to add extra chemical meaning, and that can result in tricky situations. The reaction mechanism steps are what should tell you how a reaction works, and in general the mechanism has more than one step and does not involve all species interacting at once, as may be implied by the total chemical equation. $\endgroup$ – Nicolau Saker Neto Nov 3 '13 at 9:19
  • $\begingroup$ For a reaction $A→B$, we can imagine figuring out that it proceeds by a single step where two molecules of A react to produce two molecules of B. Thus, while $A→B$ is mathematically correct, it does not represent the true nature of the reaction. If you multiply the equation by two to get $2A→2B$, it is still mathematically valid, and you could say it now carries the actual mechanism. But the point is that this is not necessary, as the total equation usually isn't meant to represent the mechanism in the first place. The total equation is fundamentally just a mathematical equality, LHS = RHS. $\endgroup$ – Nicolau Saker Neto Nov 3 '13 at 9:33
  • $\begingroup$ @NicolauSakerNeto Oh, okay, I get what you mean. Does that mean that my answer is incorrect? $\endgroup$ – mikhailcazi Nov 4 '13 at 7:03

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