Question

The bond dissociation enthalpy of \(\text{H}_2\), \(\text{Cl}_2\) and HCl are \(434\), \(242\) and \(431\) kJ \(\text{mol}^{-1}\) respectively. Calculate the enthalpy of formation of HCl.

• A. \(-93\) kJ \(\text{mol}^{-1}\)
• B. \(245\) kJ \(\text{mol}^{-1}\)
• C. \(93\) kJ \(\text{mol}^{-1}\)
• D. \(-245\) kJ \(\text{mol}^{-1}\)

Hint:

Break 1/2 mole H2 and 1/2 mole Cl2, then form 1 mole of H-Cl, and subtract energy released from energy absorbed.

Answer 1

The Correct Option is A

Step 1: Use the full reaction first:
Consider $\text{H}_2 + \text{Cl}_2 \rightarrow 2\text{HCl}$. Energy absorbed: $434 + 242 = 676$ kJ. Energy released: $2 \times 431 = 862$ kJ.

Step 2: Find the enthalpy for 2 moles:
$\Delta H = 676 - 862 = -186$ kJ for 2 moles of HCl.

Step 3: Divide for 1 mole:
Enthalpy of formation refers to 1 mole, so $\Delta_fH = -186/2 = -93$ kJ/mol.

Step 4: Check:
Heat is released, so the sign is negative, which matches option A.

Final Answer:
Halving the enthalpy of the 2HCl reaction gives -93 kJ per mole. \[ \boxed{\text{(A) }-93\ \text{kJ mol}^{-1}} \]