Question

calculate the electron affinity of Cl from the following data for NaCl.
Bond energy of Cl2= 239 kJ/mol; ΔHsub(Na)=109kJ/mol; IE(Na)=496kJ/mol; lattice enthalpy(Nacl)=-789kJ/mol and ΔH_f, NaCl=-413kJ/mol.

Answer 1

Answer 2

---

Given Data:

1. Bond energy of ${\text{Cl}}_2$ ($D$) = $239\text{kJ/mol}$

2. Enthalpy of sublimation of Na ($\mathrm{\Delta }{H}_{\text{sub}}(\text{Na})$) = $109\text{kJ/mol}$

3. Ionization energy of Na ($\text{IE}(\text{Na})$) = $496\text{kJ/mol}$

4. Lattice enthalpy of NaCl ($\mathrm{\Delta }{H}_{\text{lattice}}$) = $-789\text{kJ/mol}$

5. Enthalpy of formation of NaCl ($\mathrm{\Delta }{H}_f$) = $-413\text{kJ/mol}$

---

Born-Haber Cycle Steps:

The enthalpy of formation ($\mathrm{\Delta }{H}_f$) of NaCl can be expressed as:

$$\mathrm{\Delta }{H}_f=\mathrm{\Delta }{H}_{\text{sub}}(\text{Na})+\text{IE}(\text{Na})+\frac{1}{2}D({\text{Cl}}_2)+E_{\text{ea}}(\text{Cl})+\mathrm{\Delta }{H}_{\text{lattice}}$$

Rearranging to solve for $E_{\text{ea}}(\text{Cl})$:

$$E_{\text{ea}}(\text{Cl})=\mathrm{\Delta }{H}_f-\mathrm{\Delta }{H}_{\text{sub}}(\text{Na})-\text{IE}(\text{Na})-\frac{1}{2}D({\text{Cl}}_2)-\mathrm{\Delta }{H}_{\text{lattice}}$$

---

Plugging in the Values:

$$E_{\text{ea}}(\text{Cl})=-413\text{kJ/mol}-109\text{kJ/mol}-496\text{kJ/mol}-\frac{1}{2}\times 239\text{kJ/mol}-(-789\text{kJ/mol})$$$$=-413-109-496-119.5+789$$$$=(-413-109-496-119.5)+789$$$$=-1137.5+789$$$$=-348.5\text{kJ/mol}$$

---

Interpretation:

The negative sign indicates that energy is released when an electron is added to a gaseous Cl atom. Thus, the electron affinity of Cl is:

$$E_{\text{ea}}(\text{Cl})=\boxed{{\displaystyle -348.5\text{kJ/mol}}}$$

---

Key Notes:

• The electron affinity is typically reported as a positive value (magnitude) in many contexts, but the thermodynamic sign convention here shows energy release (exothermic process).

• 
  

• If the question expects a positive value, the answer is . Clarify the convention if needed.