I'm wondering if this is possible because could there not be a situation in which the reactants were pure solids/liquids and would therefore not contribute to the value of k but still contain more total moles that the product side?


Yes, even if the equilibrium constant is much greater than 1, there can be almost all reactants and very little products.

A wood pile can just sit outside (in an oxygen containing atmosphere) all winter, but until you stick the wood in the fireplace and supply activation energy almost no reactants will be converted to product.

If all the reactants are mixed in the same phase, the same is true. Mix methane and air in a Bunsen burner or gas stove, and nothing noticeable happens until you provide a spark or flame.

  • $\begingroup$ Ok. If however, the reaction is allowed to go through to equilibrium when K is greater than 1, there will always be more products than reactants? $\endgroup$ – Emu27 Nov 12 '14 at 16:25
  • $\begingroup$ If the sum of the exponents of the product activities (concentrations) does not equal the sum of the exponents of the reactant activities, then it is arbitrary when K is 1. In other words, it depends upon the choice of units and the definition of the "concentration constant". So, at least for this reason there could be more reactants than products at equilibrium even though K is greater than 1. $\endgroup$ – DavePhD Nov 12 '14 at 17:14
  • $\begingroup$ @Emu27 for example, if $K = 10 = ([B]/[A]^3) \times M^2$ and [B] = 0.01M and [A] = 0.1M then there are more moles of reactants than products. $\endgroup$ – DavePhD Nov 12 '14 at 17:42

The equilibrium constant does only contain the activities/concentrations of the species that are actually available to the reaction, e.g. in solution it would only contain species that are actually dissolved in the solvent or that are present at the interface between the solution and another phase. So, solids would not enter the equation with all there moles but only with the small percentage that is acutally available to the reaction: if some of the solid is dissolved and those dissolved species take part in the reaction then it is those (and only those) molecules that enter into the equation for the equilibrium constant; in the case of a heterogeneous reaction the solid can only enter the equation via its available surface sites. So, the bulk of a solid would not contribute to the equilibrium constant in a reaction taking part in a liquid phase that shares and interface with the solid.

Now, concerning the main question: If the equilibrium constant in a reaction is greater than 1 is it possible that there are more moles of reactants than products? This can certainly be the case if you already have some of the products present and keep their concentrations constant. Look at the example reaction

\begin{equation} \ce{A + B <=>> C + D} \end{equation}

which shall have an equilibrium constant of $K_{c} = 10$. Let's assume you keep the products at a concentration of 1 mol/L each, so $[\ce{C}] = 1 \, \mathrm{mol}/\mathrm{L}$ and $[\ce{D}] = 1 \, \mathrm{mol}/\mathrm{L}$. The equation for the equilibrium constant is

\begin{equation} K_{c} = \frac{[\ce{C}] [\ce{D}]}{[\ce{A}] [\ce{B}]} \end{equation}

Since you've already fixed the product $[\ce{C}] [\ce{D}]$ at $1 \, \mathrm{mol}^{2}/\mathrm{L}^{2}$ the product of your reactants must fulfill

\begin{equation} [\ce{A}] [\ce{B}] = \frac{[\ce{C}] [\ce{D}]}{K_{c}} = 0.1 \, \mathrm{mol}^{2}/\mathrm{L}^{2} \end{equation}

So, if you use only very little $\ce{B}$, say $[\ce{B}] = 0.01 \, \mathrm{mol}/\mathrm{L}$, your concentration of $\ce{A}$ in the solution will be

\begin{equation} [\ce{A}] = \frac{[\ce{C}] [\ce{D}]}{K_{c} [\ce{B}]} = 10 \, \mathrm{mol}/\mathrm{L} \end{equation}

Now, if you conduct this reaction in a vessel with a volume of 1 L you'll find a total of 2 mol of products and 10.01 mol of reactants in your reaction mixture, i.e. more reactants than products.

  • $\begingroup$ Does that mean that it is not possible? $\endgroup$ – Emu27 Nov 12 '14 at 4:04
  • $\begingroup$ It does mean that once equilibrium is reached the product of the concentrations of the available reactant species raised to the power of their respective stoichiometric coefficients will be smaller than the product for the product species. $\endgroup$ – Philipp Nov 12 '14 at 7:28
  • $\begingroup$ @Emu27 I've extended my answer. $\endgroup$ – Philipp Nov 12 '14 at 10:57
  • $\begingroup$ If we do not cheat as you put it, your answer is at odds with the one below. $\endgroup$ – Emu27 Nov 12 '14 at 17:21
  • $\begingroup$ @Emu27 Ups, sorry, it was quite early and I didn't really consider cases with different stoichiometric coefficients. Forget the last sentence of my answer. $\endgroup$ – Philipp Nov 12 '14 at 18:10

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