K_c and K_p

For a reaction involving gases, we should use the equilibrium constant with partial pressures if we want to relate this constant to the thermodynamic energies.  However, we can solve equilibrium problems for gases with a constant written in terms of concentrations.   To keep these two straight (since they are unit less) we give them different notations.  The equilibrium constant with pressures is K_p and the equilibrium constant with concentrations is K_c.  These two are related since both depend on the number of moles of each chemical species.  This is easiest to see for a given reaction.  Look at the reaction

\[\rm 2H_2(g) + O_2(g) \rightleftharpoons 2H_2O(g)\]

\[K_p = {P_{H_2O}^2 \over {P_{H_2}^2 \; P_{O_2}}}\]

writing the partial pressure in terms of the number of moles of each species we get

\[K_p = {{n_{H_2O}RT \over V}^2 \over {n_{H_2}RT \over V}^2 \; {n_{O_2}RT \over V}}\]

Since the concentration for each species is given by the number of moles divided by the volume, you can see this appears in the given expression.  The partial pressure is simply the molar concentration times a factor of RT.   For this reaction, if we gather all the concentrations together you can see we have

\[K_p = {{{n_{H_2O} \over V}^2 \over {n_{H_2} \over V}^2 \; {n_{O_2} \over V}} \times {1 \over RT}} = {K_c \times {1 \over RT}}\]

So you can see that K_p is related to K_c by a factor of RT.   This factor depends on the number of moles of gas in the products compared to the reactants.  If they have the same number of moles of gas then K_p = K_c.  Otherwise

\[K_p = K_c(RT)^{\Delta n}\]

where \(\Delta n\) is the change in the number of moles of gas (products - reactants). For the example reaction, there are three moles of gas in the reactants and two moles of gas in the products so \(\Delta n\) is negative one.

© 2013 mccord/vandenbout/labrake

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