Which of the following equations represents BOTH the standard enthalpy of formation of zinc oxide and the standard enthalpy of combustion of zinc?

To represent both the standard enthalpy of formation of zinc oxide and the standard enthalpy of combustion of zinc, the equation must satisfy two conditions:

• Standard enthalpy of formation ($\Delta H_f^\circ$): One mole of the compound (ZnO) must be formed from its constituent elements (Zn and O) in their standard states. The standard state of zinc is solid (Zn(s)), and the standard state of oxygen is diatomic gas (O$_2$(g)).
• Standard enthalpy of combustion ($\Delta H_c^\circ$): One mole of the substance (Zn) must react completely with oxygen (O$_2$(g)) to form its most stable oxide (ZnO(s)).

Let's examine the given options:

• A) Zn(s) + O$_2$(g) $\longrightarrow$ 2ZnO(s) This equation forms two moles of ZnO, so it does not represent the standard enthalpy of formation (which is defined for one mole).
• B) Zn(s) + $\frac{1}{2}$O$_2$(g) $\longrightarrow$ ZnO(s) This equation forms one mole of ZnO from its elements in their standard states (Zn(s) and O$_2$(g)). Thus, it represents the standard enthalpy of formation of ZnO. It also shows one mole of zinc (Zn(s)) reacting with oxygen to form zinc oxide, which is the combustion of zinc. Thus, it represents the standard enthalpy of combustion of zinc.
• C) Zn(s) + O(g) $\longrightarrow$ ZnO(s) Oxygen is shown as O(g), which is not its standard state (O$_2$(g)). Therefore, this does not represent the standard enthalpy of formation or combustion.
• D) Zn(g) + $\frac{1}{2}$O$_2$(g) $\longrightarrow$ ZnO(s) Zinc is shown as Zn(g), which is not its standard state (Zn(s)). Therefore, this does not represent the standard enthalpy of formation or combustion.

Based on the analysis, option B correctly represents both definitions.

The final answer is $\boxed{\text{B}}$.