ThermochemistryHard

Question

Sublimation energy of Ca is 121 kJ/ mol. Dissociation energy of Cl2 is 242.8 kJ/ mol, the total ionization energy of Ca(g) → Ca2+ (g) is 2422 kJ/mol and electron affinity of Cl is −355 kJ/ mol. Lattice energy of CaCl2 is −2430.8 kJ/ mol. What is ΔH for the process Ca(s) + Cl2(g) → CaCl2(s)?

Options

A.−355 kJ mol–1
B.+3550 kJ mol–1
C.−35.5 kJ mol–1
D.−1720 kJ mol–1

Solution

$\Delta H = 121 + 242.8 + 2422 - 2 \times (355) - 2430.8 = - 355\text{ kJ}$

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