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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:


    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:


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

    Exercise \(\PageIndex{1}\)

    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?
    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, KC, under the same conditions, calculated from molar concentrations?

    CO (g) + Cl2 (g) ⇄ COCl2 (g)


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