Born-haber Cycle
Using the Born-Haber cycle for the formation of MgCl(_2)(s), given the following data — sublimation enthalpy of Mg: +148 kJ/mol, first ionisation energy of Mg: +738 kJ/mol, second ionisation energy of Mg: +1451 kJ/mol, bond dissociation enthalpy of Cl(_2): +242 kJ/mol, electron affinity of Cl: -349 kJ/mol, lattice enthalpy of MgCl(_2): -2523 kJ/mol — calculate the standard enthalpy of formation of MgCl(_2)(s).
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Solution
−642kJ/mol
The Born-Haber cycle applies Hess's Law to ionic compound formation by constructing a thermodynamic cycle through elementary steps. The standard enthalpy of formation of MgCl(2) is found by summing all the steps in the cycle: (\Delta_f H^\circ = \Delta{sub}H(Mg) + IE_1(Mg) + IE_2(Mg) + \Delta_{bond}H(Cl_2) + 2 \times EA(Cl) + \Delta_{lattice}H). Note that we need 2 Cl atoms, so we use one full bond dissociation ((+242) kJ/mol, producing 2 Cl atoms) and apply electron affinity twice ((2 \times (-349) = -698) kJ/mol). Substituting: (\Delta_f H^\circ = +148 + 738 + 1451 + 242 + 2(-349) + (-2523)). Summing the positive terms: (148 + 738 + 1451 + 242 = 2579) kJ. Summing the negative terms: (-698 + (-2523) = -3221) kJ. Therefore (\Delta_f H^\circ = 2579 - 3221 = -642) kJ/mol. Option -493 kJ/mol omits the second ionisation energy. Option -754 kJ/mol uses an incorrect lattice energy. Option -521 kJ/mol incorrectly applies electron affinity only once. This is a JEE Advanced standard Born-Haber cycle calculation requiring careful bookkeeping of all cycle steps. Plausibility check: MgCl(_2) is a stable ionic compound, so a large negative formation enthalpy is expected; -642 kJ/mol is physically consistent with known thermochemical data.
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About This Question
- Subject
- chemistry
- Chapter
- chemical thermodynamics
- Topic
- born-haber cycle
- Difficulty
- Hard
- Year
- 2025
Solution
Correct Answer:
−642kJ/mol
The Born-Haber cycle applies Hess's Law to ionic compound formation by constructing a thermodynamic cycle through elementary steps. The standard enthalpy of formation of MgCl(2) is found by summing all the steps in the cycle: (\Delta_f H^\circ = \Delta{sub}H(Mg) + IE_1(Mg) + IE_2(Mg) + \Delta_{bond}H(Cl_2) + 2 \times EA(Cl) + \Delta_{lattice}H). Note that we need 2 Cl atoms, so we use one full bond dissociation ((+242) kJ/mol, producing 2 Cl atoms) and apply electron affinity twice ((2 \times (-349) = -698) kJ/mol). Substituting: (\Delta_f H^\circ = +148 + 738 + 1451 + 242 + 2(-349) + (-2523)). Summing the positive terms: (148 + 738 + 1451 + 242 = 2579) kJ. Summing the negative terms: (-698 + (-2523) = -3221) kJ. Therefore (\Delta_f H^\circ = 2579 - 3221 = -642) kJ/mol. Option -493 kJ/mol omits the second ionisation energy. Option -754 kJ/mol uses an incorrect lattice energy. Option -521 kJ/mol incorrectly applies electron affinity only once. This is a JEE Advanced standard Born-Haber cycle calculation requiring careful bookkeeping of all cycle steps. Plausibility check: MgCl(_2) is a stable ionic compound, so a large negative formation enthalpy is expected; -642 kJ/mol is physically consistent with known thermochemical data.
This hard difficulty chemistry question is from the chapter chemical thermodynamics, covering the topic of born-haber cycle. It appeared in the 2025 exam.
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