Question:

In the Born-Haber cycle, the heat of formation of CuCl is ______ kJ/mol.
[Given: Heat of atomization of Cu = +338 kJ/mol, 
Ionization energy of Cu = +746 kJ/mol, 
Heat of atomization of Cl2 = +121 kJ/mol, 
Electron affinity of Cl = −349 kJ/mol, and 
Lattice energy of CuCl = −973 kJ/mol] 
(round off to the nearest integer)

Updated On: Nov 17, 2025
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Correct Answer: -117

Solution and Explanation

Heat of Formation of CuCl Using the Born-Haber Cycle 

To find the heat of formation of \( \text{CuCl} \) using the Born-Haber cycle, we must consider the sequential steps involved in forming \( \text{CuCl} \) from its constituent elements, copper (\( \text{Cu} \)) and chlorine (\( \text{Cl}_2 \)). The relevant steps are:

  1. Atomization of Cu: \[ \text{Cu(s)} \rightarrow \text{Cu(g)}, \Delta H = +338 \, \text{kJ/mol}. \]
  2. Ionization of Cu: \[ \text{Cu(g)} \rightarrow \text{Cu}^+(g) + e^-, \Delta H = +746 \, \text{kJ/mol}. \]
  3. Atomization of \( \text{Cl}_2 \): \[ \frac{1}{2} \, \text{Cl}_2(g) \rightarrow \text{Cl(g)}, \Delta H = +121 \, \text{kJ/mol}. \] (Since half a mole of \( \text{Cl}_2 \) is used, the atomization energy is already adjusted accordingly.)
  4. Electron affinity of Cl: \[ \text{Cl(g)} + e^- \rightarrow \text{Cl}^-(g), \Delta H = -349 \, \text{kJ/mol}. \]
  5. Formation of \( \text{CuCl(s)} \) from gaseous ions: \[ \text{Cu}^+(g) + \text{Cl}^-(g) \rightarrow \text{CuCl(s)}, \Delta H = -973 \, \text{kJ/mol}. \]

Applying Hess's Law:

The heat of formation of \( \text{CuCl} \) (\( \Delta H_f \)) can be calculated as the sum of these enthalpy changes:

\[ \Delta H_f = (+338 \, \text{kJ/mol}) + (+746 \, \text{kJ/mol}) + (+121 \, \text{kJ/mol}) + (-349 \, \text{kJ/mol}) + (-973 \, \text{kJ/mol}). \]

Final Calculation:

Adding these values:

\[ \Delta H_f = 338 + 746 + 121 - 349 - 973 = -117 \, \text{kJ/mol}. \]

Conclusion:

The computed heat of formation of \( \text{CuCl} \) is \( -117 \, \text{kJ/mol} \), which fits within the expected range of \( -117 \, \text{kJ/mol} \) to \( -117 \, \text{kJ/mol} \), confirming the solution's accuracy.

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