Chemical equilibrium and stoichiometry

In reactions that go until the end the stoichiometric number indicates the exact amounts of reactants and products you will get. However, this is not the case in equilibrium reactions and you will get a different amounts (concentrations) than what exhibits the stoichiometric numbers. $$\ce{Na + Cl -> NaCl}$$ You will get in the end 1 mole of NaCl from 1 mole of Na and one mole of Cl. $$\ce{2SO3<=>2SO2 + O2}$$ In equilibrium you will not get 1 molar of $\ce{O2}$ and 2 molar of $\ce{SO2}$ from 2 molar of $\ce{SO3}$. Then what do the stoichiometric numbers indicate in equilibrium?

• They indicate that for each mole of O2 you will get two moles of SO2. Mar 7 '18 at 11:29
• How is it true? You will get totally different mollar concentrations based on Kc. I can't catch the issue exactly. Mar 7 '18 at 13:15
• This is true because the stoichiometric coefficients tell you so. Yes, you might get totally different molar concentrations. But whatever they might be, the concentration of SO2 will be twice that of O2. Mar 7 '18 at 14:14

Let us take the example of your equation: $$\ce{2SO3<=>2SO2 + O2}$$
1. If 2 mole $\ce{SO3}$ reacts, it will give 2 mole $\ce{SO2}$ and 1 mole $\ce{O2}$ but one needs to remember here that all of the 2 moles will not react. But the number of moles that do dissociate, give products in the above ratio i.e. if initially have 2 moles of $\ce{SO3}$ and degree of dissociation is 50%, then we will have 1 mole of $\ce{SO3}$ reacting to give 1 mole of $\ce{SO2}$ and half mole of $\ce{O2}$
2. The stoichiometric coefficients also provide the powers to which the concentrations of each compound will be raised to in the expression for equilibrium constant $$K_\mathrm c = \frac{[\ce{SO2}]^2[\ce{O2}]}{[\ce{SO3}]^2}$$