Termochimica 2 – Soluzioni

1. NO NO SI NO NO
2. 1429 kJ/mol

DU=DH-PDV

AH=-2DH₁ +DH₂ + DH₃

PDV = DnRT = -0.5 x 0.0083144 x 298

3. 1875 kJ/mol

DH =-DH₁ -DH₂ -DH₃ -2DH₄ +DH₅ -2DH₆ -DH₇

4. -114.4kJ/mol; -111.9 kJ/mol; -92.31kJ/mole

DH =-2DH₁ +DH₂

5. 32.2°C

DH(solvatazione HCl) = -DH₁ +DH₂ +DH₃

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