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 the formation of HCl gas.
The enthalpy of formation (ΔH_f^°) refers to the enthalpy change when one mole of a compound is formed from its constituent elements in their standard states. For HCl gas, the constituent elements are hydrogen (H₂) and chlorine (Cl₂), both of which are diatomic gases in their standard states.
The reaction for the formation of one mole of HCl gas 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 the formula:
ΔH_reaction = ∑ (BDE of bonds broken) − ∑ (BDE of bonds formed)
Step 3: Identify bonds broken and bonds formed and their respective 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. Bond dissociation energy of H₂ = 104 kcal.
Energy required to break $\frac{1}{2}$ H₂ bonds = $\frac{1}{2}$ × 104 kcal = 52 kcal.
$\frac{1}{2}$ mole of Cl−Cl bonds. Bond dissociation energy of Cl₂ = 58 kcal.
Energy required to break $\frac{1}{2}$ Cl₂ bonds = $\frac{1}{2}$ × 58 kcal = 29 kcal.
Total energy required to break bonds = 52 kcal + 29 kcal = 81 kcal.
Bonds formed:
1 mole of H−Cl bonds. Bond dissociation energy of HCl = 103 kcal.
Energy released by forming 1 mole of HCl bonds = 1 × 103 kcal = 103 kcal.
Step 4: Calculate the enthalpy of formation of HCl gas.
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.
\[ \boxed{\text{-22 kcal}} \]