Skip to main content
Chemistry LibreTexts

15.3: Determining an Equilibrium Constant

  • Page ID
  • Overview

    In general chemistry 1 we calculated % Yield based on the complete consumption of the limiting reagent. We know now that only product favored reactions with large equilibrium constants go to completion, and in this section we will learn how to calculate the equilibrium constant for a reaction that does not go to completion. In the next section we will show how we can use the equilibrium constant to calculate the equilibrium concentration of all species if we mix reactants, products, or a mixture of the two.

    Extent of Reaction & RICE Diagram

    To calculate the equilibrium constant we need to know what we started with, and the concentration of just one species at equilibrium. In section 14.1 we looked at relative reaction rates and saw that the rates of consumption and production of the various species were related to each other by the stoichiometric coefficients. If we had pure ammonia it must be reactant loaded (Q<K), and we know that for every two ammonia consumed, we would make one nitrogen and three hydrogen, and the reaction would proceed until Q=K. If we consider x to be the "extent of reaction", then the amount of ammonia consumed would be -2x, while the amount of nitrogen and hydrogen produced would be x and 3x respectively. So x, the extent of reaction is a measure of how far the system is from equilibrium, and if K was a large number, the reaction would proceed and create lots of products, but if K was a small fraction, very few products would be produced before Q=K and equilibrium is achieved.

    \[2NH_3\rightarrow N_2 + 3H_2 \]

    On the other hand, if we had no ammonia and a mixture of nitrogen and hydrogen, then the system is "product loaded" (Q>K) and they would react to produce two ammonia for every three hydrogen and one nitrogen consumed, with the reaction proceeding until equilibrium is achieved (Q=K). If we had a mixture of all three species it could be either reactant loaded, product loaded, or at equilibrium. If it is not at equilibrium, a reaction occurs until equilibrium is achieved. Since this reaction can go in the forward or reverse direction, we need a convention, and we will always call the species on the left the reactant, and the species on the right the product, and if the reverse reaction occurs and we make reactants, the value of x, the extent of reaction, will be a negative number.

    Generic RICE Diagram

    For the generic reaction

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

    the RICE diagram is:

    Reactants aA + bB cC dD
    Initial [A]Initial [B]Initial [C]Initial [D]Initial
    Change -ax -bx +cx +dx
    Equilibrium [A]Eq [B]Eq [C]Eq [D]Eq

    noting that:

    \[ \left [ A\right ]_{Eq}=\left [ A\right ]_{Initial}-ax \\ \left [B\right ]_{Eq}=\left [ B\right ]_{Initial}-bx \\ \left [ C\right ]_{Eq}=\left [C\right ]_{Initial}+cx \\ \left [D\right ]_{Eq}=\left [D\right ]_{Initial}+dx \label{15.3.3} \]

    So if we know both the initial and equlibrium concentrations of one species, we can calculate x from the above equation for that species, and then use that value of x to calculate the equilibrium concentration of the other species

    Once we know the equilibrium concentration of all species, we can calculate K.

    \[K=\frac{\left [C\right ]_{eq}^{c}\left [D\right ]_{eq}^{d}}{\left [A\right ]_{eq}^{a}\left [B\right ]_{eq}^{b}}\]

    Solved Problem

    Calculate the equilibrium constant for the following equation if 0.063 mol NO(g) is produced after 0.100 mol of NO, 0.050 mol of H2 and 0.100 mol H2O(g) are mixed in a 1 L container.

    \[2NO(g)\: +\: 2H_2 \rightleftharpoons\: N_2\: +\: 2H_2O\]

    Before proceeding to watch the video, think about what you have been given. First, you know the balanced equation and have the initial concentration of all species and the equilibrium concentration of one. This means you have the data you need. Second, you note there is initially no nitrogen, and so the reaction must move in the forward direction and x is positive. Note, if you wrote the reaction backwards, so the Nitrogen was a reactant, then the back reaction would have occurred and x would be negative.

    Problem Sets

    Question 4 of Interactive Quiz 16.2 and Quiz 16.4

    Robert E. Belford (University of Arkansas Little Rock; Department of Chemistry). The breadth, depth and veracity of this work is the responsibility of Robert E. Belford, You should contact him if you have any concerns. This material has both original contributions, and content built upon prior contributions of the LibreTexts Community and other resources, including but not limited to:

    • Was this article helpful?