Question

question 13 of 15
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).
\\( \delta h_{f}^{*} \\) is equal to zero \\( \delta h_{f}^{*} \\) is not equal to zero
answer bank

Explanation:

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

• $Hg(l)$: Mercury's standard state at STP is liquid. So, $\Delta H_f^{\circ}=0$.
• $Br_2(l)$: Bromine's standard state at STP is liquid. So, $\Delta H_f^{\circ}=0$.
• $I_2(s)$: 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, $\Delta H_f^{\circ}=0$.

• $Br_2(g)$: Bromine's standard state is liquid, not gas. So, $\Delta H_f^{\circ}

eq0$.

• $Br_2(s)$: Bromine's standard state is liquid, not solid. So, $\Delta H_f^{\circ}

eq0$.

• $I_2(g)$: Iodine's standard state is solid, not gas. So, $\Delta H_f^{\circ}

eq0$.

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

eq0$.

Answer:

• $\Delta H_f^{\circ}$ is equal to zero: $Hg(l)$, $Br_2(l)$, $I_2(s)$, $Cl_2(g)$
• $\Delta H_f^{\circ}$ is not equal to zero: $Br_2(g)$, $Br_2(s)$, $I_2(g)$, $Hg(s)$