Question

classify the molecules based on whether the molecule has a standard enthalpy of formation, \\( delta h_{f}^{*} \\), equal to 0. assume all conditions are at standard pressure and temperature (stp).

Explanation:

The standard enthalpy of formation ($\Delta H_{f}^{\circ}$) of an element in its standard state (most stable form at standard pressure and temperature - STP) is zero.

• $I_{2}(s)$: Iodine's standard state at STP is solid. So, $\Delta H_{f}^{\circ}=0$.
• $Cl_{2}(g)$: Chlorine's standard state at STP is gas. So, $\Delta H_{f}^{\circ}=0$.
• $Br_{2}(l)$: Bromine's standard state at STP is liquid. So, $\Delta H_{f}^{\circ}=0$.
• $Hg(l)$: Mercury's standard state at STP is liquid. So, $\Delta H_{f}^{\circ}=0$.
• $Br_{2}(g)$: Bromine's standard state is liquid ($Br_{2}(l)$), not gas. So, $\Delta H_{f}^{\circ}

eq0$.

• $I_{2}(g)$: Iodine's standard state is solid ($I_{2}(s)$), not gas. So, $\Delta H_{f}^{\circ}

eq0$.

• $Br_{2}(s)$: Bromine's standard state is liquid ($Br_{2}(l)$), not solid. So, $\Delta H_{f}^{\circ}

eq0$.

• $Hg(s)$: Mercury's standard state is liquid ($Hg(l)$), not solid. So, $\Delta H_{f}^{\circ}

eq0$.

Answer:

• $\Delta H_{f}^{\circ}$ is equal to zero: $I_{2}(s)$, $Cl_{2}(g)$, $Br_{2}(l)$, $Hg(l)$
• $\Delta H_{f}^{\circ}$ is not equal to zero: $Br_{2}(g)$, $I_{2}(g)$, $Br_{2}(s)$, $Hg(s)$