1 2 7 The Stability of Compounds Stability

1. 2. 7 The Stability of Compounds

Stability of Compounds ¥ The stability of a compound is how readily it breaks down. ¥ You can predict stability by knowing the heat of formation, ΔHfo except it is the reverse of the formation reaction

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 Mg. O(s) is -601. 7 k. J. Thus: Mg(s) + ½O 2 → Mg. O(s) + 601. 7 k. J ¥ But energy is required to break down the compound. This reaction is the reverse of formation, and will be: Mg. O(s) + 601. 7 k. J → Mg(s) + ½O 2 ¥ When we reverse an equation, the energy term moves to the other side of the equation.

Stability of Compounds ¥ ¥ Since a large amount of energy must be supplied to break down Mg. O it is a stable compound. Stable means: If a sample of Mg. O were sitting on your desk, it would remain as Mg. O and not break down into solid magnesium and oxygen gas.

Example 2 ¥ If ΔHfo is small and negative (only slightly exothermic), little energy is required to break down the compound. These compounds are often unstable.

¥ Let's look at HBr(g) as an example. ΔHfo for this compound is -36. 4 k. J. Our equation looks like this: ½ H 2(g) + ½ Br 2(g) → HBr(g) + 36. 4 k. J ¥ Again, stability refers to the reverse of the formation reaction, or HBr(g) + 36. 4 k. J → ½ H 2(g) + ½ Br 2(g) ¥ Only a small amount of energy (36. 4 k. J) is required, so this compound will break down spontaneously. Therefore, HBr is unstable compound.

Example 3 ¥ Finally, if ΔHfo is positive (endothermic), the compound is likely to be unstable.

¥ ΔHfo for CS 2 is +117. 4 k. J. Reversing this to see the decomposition reaction: CS 2(g) → C(s) + 2 S(s) + 117. 4 k. J ¥ we see that energy is released when this compound breaks down. We know that exothermic reactions tend to occur spontaneously, so CS 2 is an unstable compound.