10.4: Using Molarity in Equilibrium Calculations

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As we have pointed out several times in the preceding sections, the Ideal Gas Laws (Chapter 10) tell us that the partial pressure of a gas and the molar concentration of that gas are directly proportional. We can show this simply by beginning with the combined gas law:

\[P_{gas}V=nRT \nonumber \]

If we divide both sides by the volume, V, and state that V must be expressed in liters, the right side of the equation now contains the term ${\$ . Realizing that the number of moles of gas (n) divided by the volume in liters is equal to molarity, M, this expression can be re-written as:

\[P_{gas}=MRT \nonumber \]

Using this expression, molar concentrations can easily be substituted for partial pressures, and visa versa.

Exercise $\PageIndex{1}$

For the reaction shown below, if the molar concentrations of SO₃, NO and SO₂ are all 0.100 M, what is the equilibrium concentration of NO₂?
For the reaction between carbon monoxide and chlorine to form phosgene, the equilibrium constant calculated from partial pressures is K = 0.20. How does this value relate to the equilibrium constant, K_C, under the same conditions, calculated from molar concentrations?

CO (g) + Cl₂ (g) ⇄ COCl₂ (g)