Skip to main content
Chemistry LibreTexts

2.2 The Meaning of Keq

  • Page ID
    32233
  • \( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \) \( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)\(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\) \(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\)\(\newcommand{\AA}{\unicode[.8,0]{x212B}}\)

    What can the value of Keq tell us about a reaction?

    • If Keq is very large, the concentration of the products is much greater than the concentration of the reactants. The reaction essentially "goes to completion"; all - or most of - of the reactants are used up to form the products.
    • If Keq is very small, the concentration of the reactants is much greater than the concentration of the products. The reaction does not occur to any great extent - most of the reactants remain unchanged, and there are few products produced.
    • When Keq is not very large or very small (close to a value of 1) then roughly equal amounts of reactants and products are present at equilibrium.

    Here are some examples to consider:


    the decomposition
    of ozone, O3

    \(\ce{2O_3(g) \leftrightarrow 3O_2(g)}\)

    \(\mathrm{K_{eq}=2.0 \times 10^{57}}\)

    Keq is very large, indicating that mostly O2 is present in an equilibrium system, with very little O3

    production of
    nitrogen monoxide
    \(\ce{N_2(g) + O_2(g) \leftrightarrow 2NO(g)}\)

    \(\mathrm{K_{eq}=1.0 \times 10^{-25}}\)

    Very little NO is produced by this reaction; N2 and O2 do not react readily to produce NO (lucky for us - otherwise we would have little oxygen to breath in our atmosphere!)

    reaction of carbon monoxide and water \(\ce{CO(g) + H_2O(g) \leftrightarrow CO_2(g) + H_2(g)}\)

    \(\mathrm{K_{eq}=5.09}\)

    (at 700 K)

    The concentrations of the reactants are very close to the concentrations of the products at equilibrium

    2.2 The Meaning of Keq is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by LibreTexts.

    • Was this article helpful?