10.4: Using Molarity in Equilibrium Calculations
- Page ID
- 79594
\( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \) \( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)\(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\) \(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\)\(\newcommand{\AA}{\unicode[.8,0]{x212B}}\)
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 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, KC, under the same conditions, calculated from molar concentrations?
CO (g) + Cl2 (g) ⇄ COCl2 (g)