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Calculating an Equilibrium Constant Using Partial Pressures

  • Page ID
    1366
  • The equilibrium constant is known as \(K_{eq}\). A common example of \(K_{eq}\) is with the reaction: 

    \[aA + bB \rightleftharpoons cC + dD\]

    \[K_{eq} = \dfrac{[C]^c[D]^d}{[A]^a[B]^b}\]

    where:

    • At equilibrium, [A], [B], [C], and [D] are either the molar concentrations or partial pressures.
    • Products are in the numerator. Reactants are in the denominator.
    • The exponents are the coefficients (a,b,c,d) in the balanced equation.
    • Solids and pure liquids are omitted. This is because the activities of pure liquids and solids are equal to one, therefore the numerical value of equilibrium constant is the same with and without the values for pure solids and liquids.
    • \(K_{eq}\) does not have units. This is because when calculating activity for a specific reactant or product, the units cancel. So when calculating \(K_{eq}\), one is working with activity values with no units, which will bring about a \(K_{eq}\) value with no units.

    Various \(K_{eq}\)

    All the equilibrium constants tell the relative amounts of products and reactants at equilibrium. For any reversible reaction, there can be constructed an equilibrium constant to describe the equilibrium conditions for that reaction. Since there are many different types of reversible reactions, there are many different types of equilibrium constants:

    • \(K_{c}\): constant for molar concentrations
    • \(K_{p}\): constant for partial pressures
    • \(K_{sp}\): solubility product
    • \(K_{a}\): acid dissociation constant for weak acids
    • \(K_{b}\): base dissociation constant for weak bases
    • \(K_{w}\): describes the ionization of water (\(K_{w} = 1 \times 10^{-14}\))

    Calculating Kp

    Referring to equation: 

    \[aA + bB \rightleftharpoons cC + dD\]

    \[K_p = \dfrac{(P_C)^c(P_D)^d}{(P_A)^a(P_B)^b}\]

    Partial Pressures: In a mixture of gases, it is the pressure an individual gas exerts. The partial pressure is independent of other gases that may be present in a mixture. According to the ideal gas law, partial pressure is inversely proportional to volume. It is also directly proportional to moles and temperature.

    Example \(\PageIndex{1}\)

    At equilibrium in the following reaction at room temperature, the partial pressures of the gases are found to be \(P_{N_2}\) = 0.094 atm, \(P_{H_2}\) = 0.039 atm, and \(P_{NH_3}\) = 0.003 atm.

    \[\ce{N_2 (g) + 3 H_2 (g) \rightleftharpoons 2 NH_3 (g)} \nonumber \]

    What is the \(K_p\) for the reaction?

    SOLUTION

    First, write \(K_{eq}\) (equilibrium constant expression) in terms of activities.

    \[K = \dfrac{(a_{NH_3})^2}{(a_{N_2})(a_{H_2})^3} \nonumber\]

    Then, replace the activities with the partial pressures in the equilibrium constant expression.

    \[K_p = \dfrac{(P_{NH_3})^2}{(P_{N_2})(P_{H_2})^3} \nonumber\]

    Finally, substitute the given partial pressures into the equation.

    \[K_p = \dfrac{(0.003)^2}{(0.094)(0.039)^3} = 1.61 \nonumber\]

    Example \(\PageIndex{2}\)

    At equilibrium in the following reaction at 303 K, the total pressure is 0.016 atm while the partial pressure of \(P_{H_2}\) is found to be 0.013 atm.

    \[\ce{3 Fe_2O_3 (s) + H_2 (g) \rightleftharpoons 2 Fe_3O_4 (s) + H_2O (g)} \nonumber\]

    What is the \(K_p\) for the reaction?

    SOLUTION

    First, calculate the partial pressure for \(\ce{H2O}\) by subtracting the partial pressure of \(\ce{H2}\) from the total pressure.

    \[ \begin{align*} P_{H_2O} &= {P_{total}-P_{H_2}} \\[5pt] &= (0.016-0.013) \; atm \\[5pt] &= 0.003 \; atm \end{align*}\]

    Then, write K (equilibrium constant expression) in terms of activities. Remember that solids and pure liquids are ignored.

    \[K = \dfrac{(a_{H_2O})}{(a_{H_2})}\nonumber\]

    Then, replace the activities with the partial pressures in the equilibrium constant expression.

    \[K_p = \dfrac{(P_{H_2O})}{(P_{H_2})}\nonumber\]

    Finally, substitute the given partial pressures into the equation.

    \[K_p = \dfrac{(0.003)}{(0.013)} = 0.23 \nonumber\]

    Example \(\PageIndex{3}\)

    A flask initially contained hydrogen sulfide at a pressure of 5.00 atm at 313 K. When the reaction reached equilibrium, the partial pressure of sulfur vapor was found to be 0.15 atm.

    \[\ce{2 H_2S (g) \rightleftharpoons 2 H_2 (g) + S_2 (g) } \nonumber\]

    What is the \(K_p\) for the reaction?

    SOLUTION

    For this kind of problem, ICE Tables are used.

      \(\ce{2H2S (g)}\)  \( \rightleftharpoons \) \(\ce{2H2(g)}\) + \(\ce{S2(g)}\)
    Initial Amounts 5.00 atm   0 atm   0 atm
    Change in Amounts -0.3 atm   +0.3 atm   +0.15 atm
    Equilibrium Amounts 4.7 atm   0.3 atm   0.15 atm

    Now, set up the equilibrium constant expression, \(K_p\).

    \[K_p = \dfrac{(P_{H_2})^2(P_{S_2})}{(P_{H_2S})^2} \nonumber\]

    Finally, substitute the calculated partial pressures into the equation.

    \[ \begin{align*} K_p &= \dfrac{(0.3)^2(0.15)}{(4.7)^2} \\[5pt] &= 6.11 \times 10^{-4} \end{align*} \]

    References

    1. Petrucci, et al. General Chemistry: Principles & Modern Applications; Ninth Edition. Pearson/Prentice Hall; Upper Saddle River, New Jersey 07.

    Contributors

    • Bianca Yau