continuation:
"=-782771\\frac{J}{mol}\\approx-783\\frac{kJ}{mol}"
Then ΔHf="534.9\\frac{kJ}{mol}\u200b+U_{L}=534.9\\frac{kJ}{mol}\u200b-783\\frac{kJ}{mol}\u200b\\approx-248\\frac{kJ}{mol}" < 0 - an exothermic process
Thus we show that CaCl is an exothermic compound.
for the reaction 2 CaCl(s) → Ca(s) + CaCl2(s).
"\\Delta H=\\Delta H_{f,CaCl2}-2\\cdot\n\\Delta H_{f,CaCl}=-785.8\\frac{kJ}{mol}-2\\cdot(-248\\frac{kJ}{mol})\\approx-289.8\\frac{kJ}{mol}" - exotermic. Thus nonexistent CaCl would decompose spontaneously to CaCl2 and Ca.
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