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  • Calculating Equilibrium Pressures: A Step‑by‑Step Guide

    By Lee Johnson • Updated Mar 24, 2022

    Chemical equilibrium describes the steady state of a reversible reaction where reactants convert to products and vice versa at equal rates. In practice, chemists quantify this balance using the equilibrium constant, Kp, which links the partial pressures of the gases involved.

    Understanding the Equilibrium Constant

    For a generic gas‑phase reaction:

    \(aA(g)+bB(g)\rightleftharpoons cC(g)+dD(g)\)

    the equilibrium constant is defined as:

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

    When all stoichiometric coefficients equal one, the expression simplifies to “products over reactants.” This form is valid only at equilibrium.

    Sometimes you’ll see the equilibrium constant expressed in terms of molar concentrations, Kc, related to Kp by:

    \(K_p = K_c (RT)^{\Delta n}\)

    where Δn is the change in the number of gas moles between products and reactants.

    Rearranging the Equation to Find Equilibrium Pressure

    The key step is introducing the variable x, which represents the change in pressure from the initial value to equilibrium. Suppose the initial pressure of each reactant is P_i and the products start at zero pressure. Then each equilibrium pressure can be expressed in terms of x.

    With all coefficients set to one, the Kp expression becomes:

    \(\begin{aligned} K_p &= \frac{x^2}{(P_i - x)^2} \end{aligned}\)

    Solving for x gives:

    \(x = \frac{\sqrt{K_p}\,P_i}{1 + \sqrt{K_p}}\)

    The equilibrium partial pressure of a reactant is P_i - x, while that of a product is simply x.

    Example: Producing Chloromethane

    Consider the reaction:

    \(\text{CH}_3\text{OH(g)} + \text{HCl(g)} \rightleftharpoons \text{CH}_3\text{Cl(g)} + \text{H}_2\text{O(g)}\)

    With Kp = 5{}900 and an initial pressure P_i = 0.75 atm for each reactant, compute x:

    \(\begin{aligned} x &= \frac{\sqrt{K_p}\,P_i}{1 + \sqrt{K_p}} \\ & = \frac{\sqrt{5900}\times 0.75\;\text{atm}}{1 + \sqrt{5900}} \\ &\approx 0.74\;\text{atm} \end{aligned}\)

    Thus, the equilibrium pressure of each product is about 0.74 atm, and that of each reactant is 0.75 - 0.74 = 0.01 atm.

    By following this systematic approach, you can accurately determine equilibrium pressures for any gas‑phase reaction.

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