The enthalpy change for which reaction depicted below is equivalent to the standard enthalpy of formation for sodium chloride? A. Na+ (aq) + Cl- (aq) → NaCl (s) B. Na+ (aq) + Cl2 (aq) → NaCl (aq) C. 2Na (s) + Cl2 (g) → 2NaCl (s) D. Na (s) + ½ Cl2 (g) → NaCl (s)
The Correct Answer and Explanation is:
Correct Answer: D. Na (s) + ½ Cl₂ (g) → NaCl (s)
Explanation:
The standard enthalpy of formation (ΔH°f) of a compound is defined as the enthalpy change when one mole of a compound is formed from its elements in their standard states under standard conditions (298 K and 1 atm pressure). This process involves the formation of one mole of a compound from its constituent elements in their most stable forms at standard state.
In the case of sodium chloride (NaCl), the relevant elements involved are:
- Sodium (Na) in its standard state: a solid metal, denoted as Na (s).
- Chlorine (Cl₂) in its standard state: a diatomic gas, denoted as Cl₂ (g).
Since the goal is to form one mole of NaCl (s) from the elements in their standard states, the reaction must use:
- 1 mole of sodium (Na) solid.
- ½ mole of chlorine gas (Cl₂), because chlorine exists as Cl₂ molecules and each NaCl requires only one Cl atom.
Therefore, the correct reaction that corresponds to the standard enthalpy of formation is:
Na (s) + ½ Cl₂ (g) → NaCl (s)
This reaction satisfies all criteria:
- Reactants are in standard elemental forms.
- Exactly one mole of product (NaCl) is formed.
- No other substances are present.
Option C is similar but forms 2 moles of NaCl, which does not represent the formation of one mole and therefore does not match the definition of standard enthalpy of formation.
Option A and B involve ions or aqueous states, not elemental forms, and thus do not reflect formation directly from standard-state elements.
Hence, option D accurately represents the enthalpy of formation for sodium chloride.
