10.4: Using Molarity in Equilibrium Calculations
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.
- For the reaction shown below, if the molar concentrations of SO 3 , NO and SO 2 are all 0.100 M , what is the equilibrium concentration of NO 2 ?
- 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) + Cl 2 (g) ⇄ COCl 2 (g)