continuation:
=−782771molJ≈−783molkJ
Then ΔHf=534.9molkJ+UL=534.9molkJ−783molkJ≈−248molkJ < 0 - an exothermic process
Thus we show that CaCl is an exothermic compound.
for the reaction 2 CaCl(s) → Ca(s) + CaCl2(s).
ΔH=ΔHf,CaCl2−2⋅ΔHf,CaCl=−785.8molkJ−2⋅(−248molkJ)≈−289.8molkJ - exotermic. Thus nonexistent CaCl would decompose spontaneously to CaCl2 and Ca.
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