Final answer:
The correct formation equation for ethanol with a standard enthalpy of formation of 484.3 kJ/mol is the combination of elemental carbon in graphite form, hydrogen gas, and oxygen gas to form one mole of ethanol in the liquid state, which is represented as: 2 C(s, graphite) + 3 H₂(g) + ½ O₂(g) → CH₃COOH(l).
Step-by-step explanation:
The student is asking about the standard enthalpy of formation for ethanol (CH₃COOH(l)), which is given as 484.3 kJ/mol. The correct formation equation corresponding to this standard enthalpy change (ΔH°f) is the one that represents the formation of one mole of ethanol from its elements in their standard states. Therefore, the balanced chemical equation is:
2 C(s, graphite) + 3 H₂(g) + ½ O₂(g) → CH₃COOH(l)
This equation represents the formation of ethanol from elemental carbon (graphite), hydrogen gas, and oxygen gas. In this formation reaction, carbon is in its standard state as graphite, hydrogen is in gas form, and oxygen is also in its gaseous state as molecular oxygen (O₂). The combination of these elements in the specified stoichiometry to form ethanol is an example of a standard enthalpy of formation reaction for a chemical compound.