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In our chemistry textbook, the equilibrium constant $K_c$ is derived this way:

$$\ce{aA + bB <=> cC + dD}$$

in this reaction, rate of forward reaction is $k_\mathrm f= k_1[A]^a[B]^b$

rate of backward reaction is $k_\mathrm b = k_2[C]^c[D]^d$

in equilibrium, equalizing the two rates we get $$ K_c = \frac{k_1}{k_2}=\frac{[A]^a[B]^b}{[C]^c[D]^d}$$

Now, there is a problem... we know the $K_c$ depends on how we equalize the reaction... like we can write $$\ce{H2 + 0.5O2 -> H2O}$$ or $$\ce{2H2 + O2 -> 2H2O}$$, etc ways. But how the rate of reaction depends on the concentration? to what power?

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    $\begingroup$ To make your equations nicer, have a look at some other questions with formatted equations to see how to make your equations look like those in a textbook. Its quite simple, you will be able to copy examples. To answer your question, you reduce the equation to its simplest form by dividing through by any common numbers. You must also check that the equation is balanced, i.e. that there are the same $whole$ number of each type of atoms on both sides. Thus the second equation is the one to use 2H$_2$+O$_2$=2H$_2$O $\endgroup$
    – porphyrin
    Commented Jul 10, 2016 at 8:47
  • $\begingroup$ what is special in the second equation? Any will be fine from a clear point of view considering we are dealing with molar units. $\endgroup$ Commented Jul 10, 2016 at 11:43
  • $\begingroup$ @TahmidHossain what do you mean by equalize the reaction.Do you mean equalizing a reaction and its reverse reaction with same stoichiometric coefficients(of course it has to be)? $\endgroup$ Commented Jul 10, 2016 at 12:46
  • $\begingroup$ yeah. that;s what i meant $\endgroup$ Commented Jul 10, 2016 at 13:21
  • $\begingroup$ Equalize it any way you want, your $K_c$ will be just right. See also chemistry.stackexchange.com/questions/55926/… $\endgroup$ Commented Sep 9, 2016 at 20:49

1 Answer 1

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Well, there are two cases:

  1. $a + b = c + d$; and
  2. $a + b \neq c + d$

For case 2, it's fairly simple to tell how the equation was balanced because your $K_C$ will come in a different unit, but for case 1, because $K_C$ is dimensionless we depend on the convention that Chemical Reactions are balanced on the lowest whole number. So in the case of the example given, we would balance it as:

$$\ce{2H2 + O2 <=> H2O}$$

and $K_C$ would be defined on that basis.

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