Đáp án:

$\Delta_rH^{0}_{298}=-1238(kJ.mol^{-1})$

Giải thích các bước giải:

Xét phản ứng:

$CH_3COCH_3(g)+4O_2(g) \to 3CO_2(g)+3H_2O(g)$

$\Delta_rH^{0}_{298}=\Delta_rH^{0}_{298,CH_3COCH_3(g)}+4.\Delta_rH^{0}_{298,O_2(g)}-3.\Delta_rH^{0}_{298,CO_2(g)}-3.\Delta_rH^{0}_{298,H_2O(g)}\\\to\Delta_rH^{0}_{298}=[E_{b,C=O}+2.E_{b,C-C}+6.E_{b,C-H}]+4.E_{b,O=O}-3.[2.E_{b,C=O}]-3.[2.E_{b,O-H}]\\\to\Delta_rH^{0}_{298}=(732+2.346+6.418)+4.494-3.(2.732)-3.(2.459)-=-1238(kJ.mol^{-1})$