Question

The bond dissociation energies of gaseous H₂, Cl₂, and HCl are 104, 58, and 103 kcal, respectively. The enthalpy of formation of HCl gas would be:

• A. -44 kcal
• B. -88 kcal
• C. -22 kcal
• D. -11 kcal

Hint:

The enthalpy of formation of a compound from its elements can be calculated using bond dissociation energies. Remember to apply the stoichiometric coefficients to the BDEs of the reactants and products, and subtract the energy of bonds formed from the energy of bonds broken. For standard enthalpy of formation, ensure the equation forms one mole of the compound from elements in their standard states.

Answer 1

The Correct Option is C

Step 1: Write the balanced chemical equation for HCl gas formation.
The enthalpy of formation (ΔH_f^°) is the enthalpy change when one mole of a compound forms from its elements in standard states. For HCl gas, the elements are hydrogen (H₂) and chlorine (Cl₂), both diatomic gases in their standard states.
The reaction for one mole of HCl gas formation is:
$\frac{1}{2}$H₂(g) + $\frac{1}{2}$Cl₂(g) → HCl(g)

Step 2: Relate enthalpy of reaction to bond dissociation energies.
The enthalpy change of a reaction (ΔH_reaction) can be estimated from bond dissociation energies (BDEs) using:
ΔH_reaction = ∑ (BDE of bonds broken) − ∑ (BDE of bonds formed)

Step 3: Identify broken/formed bonds and their energies.
In the reaction $\frac{1}{2}$H₂(g) + $\frac{1}{2}$Cl₂(g) → HCl(g):
Bonds broken:
$\frac{1}{2}$ mole of H−H bonds. H₂ BDE = 104 kcal.
Energy to break $\frac{1}{2}$ H₂ bonds = $\frac{1}{2}$ × 104 kcal = 52 kcal.
$\frac{1}{2}$ mole of Cl−Cl bonds. Cl₂ BDE = 58 kcal.
Energy to break $\frac{1}{2}$ Cl₂ bonds = $\frac{1}{2}$ × 58 kcal = 29 kcal.
Total energy to break bonds = 52 kcal + 29 kcal = 81 kcal.

Bonds formed:
1 mole of H−Cl bonds. HCl BDE = 103 kcal.
Energy released forming 1 mole of HCl bonds = 1 × 103 kcal = 103 kcal.

Step 4: Calculate HCl gas enthalpy of formation.
ΔH_f^°(HCl) = (Energy of bonds broken) − (Energy of bonds formed)
ΔH_f^°(HCl) = 81 kcal − 103 kcal
ΔH_f^°(HCl) = −22 kcal

Step 5: Evaluate the options.

• (1) -44 kcal: Incorrect.
• (2) -88 kcal: Incorrect.
• (3) -22 kcal: This matches the calculated value.
• (4) -11 kcal: Incorrect.

Step 6: Conclusion.
The enthalpy of formation of HCl gas is −22 kcal.

\[ \n\boxed{\text{-22 kcal}} \n\]