Stoichiometric Coefficients and Relative Concentrations

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For a simple equilibrium or non-equilibrium chemical reaction, while I understand that the ratio of the coefficients of each reactant/product is mathematically equal to the molar ratio, I am not quite sure whether or not it is equal to the ratio of concentrations of each component in the reaction mixture (assuming the reaction is taking place in an aqueous environment, for instance). Can someone please clarify this for me?
 

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  • #2
Borek
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I am not sure what your question is. In general stoichiometric coefficients describe the reaction (in what ratio substances react and are being produced) not composition of mixtures.
 
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My question arose because of so-called ICE tables that are commonly used while solving equilibrium problems. My doubt is with regard to the "C" (change) part. I am given to understand that if we are to use the stoichiometric coefficients of each reactant/product in the equilibrium to compute in what proportion "change" is occurring in the equilibrium, it would be change in molar AMOUNT, not molar concentration (that is, of course, since the mole ratio and the stoichiometric coefficient ratio are exactly the same). But why do we assume that stoichiometric ratios indicate the change in concentrations of reactants/products in the equilibrium?
 
  • #4
Borek
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In general you are right that we should stick to moles, not to concentration. However, in the case of a mixture volume is identical for all substances involved, so the change in number of moles is equivalent to the change in concentrations (in other words: it is the same because volume cancels out).
 
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Does this apply even if we're given initial molar amounts of reactants/products?
Consider the Haber Process:
N2 + 3H2 <=> 2NH3
Kc = [NH3]2/[N2][H2]3
Suppose we're initially given 2 mol of N2 and 1 mol of H2, in a system of unknown volume, V.
Now, I understand that change is going to be -x for Nitrogen, -3x for Hydrogen, and +2x for Ammonia (x has dimensions mol/L).
While computing Kc for this equilibrium, can we substitute (2-x) for [N2] as though the initial concentration was 2 mol/L?
If we substitute the given data without glossing over any mathematical detail, we are left with:
Kc = (2x)2/(2/V mol - x)(1/V mol - 3x)3
If we do the binomial expansion, we would be left with Vs all over the place. I don't mean to be pedantic, it's just that I can't get my head around it and it's really bugging me.
 
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  • #6
Borek
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Suppose we're initially given 2 mol of N2 and 1 mol of H2, in a system of unknown volume, V.

If the V is unknown, and you are given Kc, problem can't be solved.
 
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Unless the total number of molecules on the right is the same as the total number of molecules on the left of the equilibrium, am I getting this right?
 
  • #8
Borek
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Definitely in some cases volume doesn't matter and cancels out, yes.

Note you can always rescale the problem to V=1L, in which case volume again stops to matter.
 

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