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which of the following compounds would be most likely to decompose into…

Question

which of the following compounds would be most likely to decompose into its elements if it were left out on a lab bench? note the values given are for $delta g^circ_{\text{formation}}$ for each compound. nabr(s), $delta g^circ$f = -349.0 kj/mol hbr(g), $delta g^circ$f = -53.4 kj/mol cdf₂(s), $delta g^circ$f = -647.7 kj/mol baco₃(s), $delta g^circ$f = -1134.4 kj/mol

Explanation:

Response

To determine which compound is most likely to decompose into its elements, we use the concept of Gibbs free energy of formation ($\Delta G_f^\circ$). A compound is more likely to decompose if the reverse reaction (decomposition into elements) has a negative $\Delta G$ (spontaneous). The decomposition reaction is the reverse of the formation reaction, so:

For a formation reaction: $\text{Elements}
ightarrow \text{Compound}$, $\Delta G_f^\circ$ (formation) is given.
For decomposition: $\text{Compound}
ightarrow \text{Elements}$, $\Delta G_{\text{decomp}}^\circ = -\Delta G_f^\circ$ (formation).

Step 1: Recall the relationship

A reaction is spontaneous when $\Delta G < 0$. For decomposition, $\Delta G_{\text{decomp}}^\circ = -\Delta G_f^\circ$ (formation). Thus, the less negative (or more positive) $\Delta G_f^\circ$ (formation) is, the more negative $\Delta G_{\text{decomp}}^\circ$ becomes (making decomposition more spontaneous).

Step 2: Compare $\Delta G_f^\circ$ values

We analyze each compound:

  • $\text{NaBr}(s)$: $\Delta G_f^\circ = -349.0\ \text{kJ/mol}$
  • $\text{HBr}(g)$: $\Delta G_f^\circ = -53.4\ \text{kJ/mol}$ (least negative)
  • $\text{CdF}_2(s)$: $\Delta G_f^\circ = -647.7\ \text{kJ/mol}$
  • $\text{BaCO}_3(s)$: $\Delta G_f^\circ = -1134.4\ \text{kJ/mol}$ (most negative)
Step 3: Determine decomposition spontaneity

For decomposition, $\Delta G_{\text{decomp}}^\circ = -\Delta G_f^\circ$. The least negative $\Delta G_f^\circ$ (formation) gives the most negative $\Delta G_{\text{decomp}}^\circ$ (most spontaneous decomposition).

Among the options, $\text{HBr}(g)$ has the least negative $\Delta G_f^\circ$ (formation) ($-53.4\ \text{kJ/mol}$). Thus, its decomposition ($\text{HBr}(g)
ightarrow \text{H}_2(g) + \text{Br}_2(l)$) will have $\Delta G_{\text{decomp}}^\circ = -(-53.4) = +53.4\ \text{kJ/mol}$? Wait, no—wait, the formation reaction for $\text{HBr}(g)$ is $\frac{1}{2}\text{H}_2(g) + \frac{1}{2}\text{Br}_2(l)
ightarrow \text{HBr}(g)$, so decomposition is $\text{HBr}(g)
ightarrow \frac{1}{2}\text{H}_2(g) + \frac{1}{2}\text{Br}_2(l)$, and $\Delta G_{\text{decomp}}^\circ = -\Delta G_f^\circ = -(-53.4) = +53.4\ \text{kJ/mol}$? Wait, no—wait, spontaneity: if $\Delta G_f^\circ$ (formation) is negative, the compound is stable (formation is spontaneous). For decomposition, we need $\Delta G_{\text{decomp}} < 0$. So $\Delta G_{\text{decomp}} = -\Delta G_f^\circ$ (formation). Thus, if $\Delta G_f^\circ$ (formation) is *less negative* (closer to zero), then $-\Delta G_f^\circ$ (decomposition) is more positive? Wait, no—let’s correct:

Wait, $\Delta G_f^\circ$ (formation) is the free energy change to form 1 mol of the compound from its elements. For decomposition, the reaction is the reverse: compound → elements. So $\Delta G_{\text{decomp}} = -\Delta G_f^\circ$ (formation).

A reaction is spontaneous when $\Delta G < 0$. So for decomposition to be spontaneous, $\Delta G_{\text{decomp}} < 0$ → $-\Delta G_f^\circ < 0$ → $\Delta G_f^\circ > 0$? No, wait:

Wait, if $\Delta G_f^\circ$ (formation) is negative, the formation reaction is spontaneous (elements form compound). For decomposition (compound → elements) to be spontaneous, $\Delta G_{\text{decomp}} < 0$ → $-\Delta G_f^\circ < 0$ → $\Delta G_f^\circ > 0$? No, that can’t be. Wait, no—let’s take an example:

If $\Delta G_f^\circ$ (formation) is negative (e.g., $\text{H}_2\text{O}(l)$: $\Delta G_f^\circ = -237\ \text{kJ/mol}$), formation is spontaneous (elements → water). Decomposition (water → elements) has $\Delta G_{\text{decomp}} = +237\ \text{kJ/mol}$ (non -…

Answer:

$\boldsymbol{\text{HBr}(g)}$ (with $\Delta G_f^\circ = -53.4\ \text{kJ/mol}$)