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: Understanding the Concept:
The enthalpy of formation of HCl is the enthalpy change for forming one mole of HCl(g) from its elements in their standard states. Bonds are broken (energy absorbed) and new bonds are formed (energy released).

Step 2: Write the reaction:
\[ \tfrac{1}{2}\text{H}_2(g) + \tfrac{1}{2}\text{Cl}_2(g) \rightarrow \text{HCl}(g) \]

Step 3: Apply bond enthalpies:
\[ \Delta_fH = \sum(\text{bonds broken}) - \sum(\text{bonds formed}) \]
\[ \Delta_fH = \tfrac{1}{2}(434) + \tfrac{1}{2}(242) - 431 = 217 + 121 - 431 \]
\[ \Delta_fH = 338 - 431 = -93 \text{ kJ/mol} \]

Step 4: Why the other options are wrong.
+93 kJ/mol has the wrong sign. The values 245 and -245 come from using full moles of \(\text{H}_2\) and \(\text{Cl}_2\), but the equation for one mole of HCl needs only half a mole of each.

Final Answer:
The enthalpy of formation of HCl is -93 kJ per mole. \[ \boxed{\text{(A) }-93\ \text{kJ mol}^{-1}} \]