Sublimation energy of Ca is 121 kJ/mol. Dissociation energy of is 242.8 kJ/mol, the total ionization — Thermodynamics and Thermochemistry Chemistry Question
Question
Sublimation energy of Ca is 121 kJ/mol. Dissociation energy of $\text{Cl}_2$ is 242.8 kJ/mol, the total ionization energy of $\text{Ca(g)} \to \text{Ca}^{2+}\text{(g)}$ is 2422 kJ/mol and electron affinity of Cl is –355 kJ/mol. Lattice energy of $\text{CaCl}_2$ is –2430.8 kJ/mol. What is $\Delta H$ for the process $\text{Ca(s)} + \text{Cl}_2\text{(g)} \to \text{CaCl}_2\text{(s)}$?
Answer: A
💡 Solution & Explanation
Using the Born-Haber cycle: $\Delta H_f = \Delta H_{\text{sub}}(\text{Ca}) + \text{IE}_{1+2}(\text{Ca}) + \text{BE}(\text{Cl}_2) + 2\text{EA}(\text{Cl}) + \text{LE}(\text{CaCl}_2)$. $\Delta H_f = 121 + 2422 + 242.8 + 2(-355) + (-2430.8) = 2785.8 - 710 - 2430.8 = -355\text{ kJ/mol}$.
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