Skip to main content
Chemistry LibreTexts

2.7 Stability of Compounds

  • Page ID
    32253
  • \( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \) \( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)\(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\) \(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\)\(\newcommand{\AA}{\unicode[.8,0]{x212B}}\)

    The stability of a compound concerns how readily it breaks down. You can predict stability by knowing the heat of formation, ΔHfo of the compound if you keep in mind that the reverse of the formation reaction describes how readily the compound decomposes into its elements.
    Follow along carefully with these examples:
    1.

    If ΔHfo is large and negative, energy is released when the compound is formed.

    For example, ΔHfo for MgO(s) is -601.7 kJ. Thus:

    Mg(s) + ½O2 → MgO(s) + 601.7 kJ

    Conversely, energy is required to decompose the compound. The decomposition reaction is the reverse of the formation reaction, and thus will be:

    MgO(s) + 601.7 kJ → Mg(s) + ½O2

    Notice that when we reverse an equation, the energy term moves to the other side of the equation. This should make sense to you, and will be important to understand.

    Since a large amount of energy must be supplied for MgO to break down into its elements (601.7 kJ), we conclude that MgO is a stable compound. If a sample of MgO were sitting on your desk, it would remain as MgO and not break down into solid magnesium and oxygen gas.

    2.

    If ΔHfo is small and negative (only slightly exothermic), little energy is required to decompose the compound. These compounds are often unstable & decompose easily.

    Let's look at HBr(g) as an example. ΔHfo for this compound is -36.4 kJ. Our equation looks like this:

    ½ H2(g) + ½ Br2(g) → HBr(g) + 36.4 kJ

    Again, stability of a compound refers to the reverse of the formation reaction, or

    HBr(g) + 36.4 kJ → ½ H2(g) + ½ Br2(g)

    Since only a small amount of energy (36.4 kJ) is required for this reaction to occur, it is possible that this compound will break down spontaneously on its own. We therefore conclude that HBr is likely an unstable compound.

    3.

    Finally, if ΔHfo is positive (endothermic), the compound is likely to be unstable.

    ΔHfo for CS2 is +117.4 kJ. The heat of formation reaction:

    C(s) + 2 S(s) + 117.4 kJ → CS2(g)

    Reversing this to see the decomposition reaction:

    CS2(g) → C(s) + 2 S(s) + 117.4 kJ

    we see that energy is released when this compound breaks down. Since we now know that exothermic reactions tend to occur spontaneously (Section 2-3), we conclude that CS2 is an unstable compound.


    2.7 Stability of Compounds is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by LibreTexts.

    • Was this article helpful?