\(CH_4^{} + 2O_2^{} \to CO_2^{} + 2H_2^{}O\)

\(\Delta _r^{}H_{298}^o = E_b^{}(CH_4^{}) + 2.E_b^{}(O_2^{}) - E_b^{}(CO_2^{}) - 2.E_b^{}(H_2^{}O)\)

\(\Delta _r^{}H_{298}^o = E_b^{}(C - H) + 2.E_b^{}(O = O) - 2.E_b^{}(C = O) - 2.2E_b^{}(H - O)\)

\(\Delta _r^{}H_{298}^o = 4.414 + 2.498 - 2.736 - 2.2.464\)

\(\Delta _r^{}H_{298}^o =  - 676kJ\) ( lưu ý đây là nhiệt lượng khi đốt cháy 1 mol)

Chỉ cần tính số mol của 1 gam ra rồi tính là được

\(n_{CH_4^{}}^{} = \frac{1}{{16}} \to \Delta _r^{}H_{298}^o(1gam) = 676.\frac{1}{{16}} = 42,25kJ\)

\(CH_3^{}Cl + 3O_2^{} \to CO_2^{} + 2H_2^{}O + 2HCl\)

\(\Delta _r^{}H_{298}^o = E_b^{}(CH_3^{}Cl) + 3.E_b^{}(O_2^{}) - E_b^{}(CO_2^{}) - 2.E_b^{}(H_2^{}O) - 2.E_b^{}(HCl)\)

\(\Delta _r^{}H_{298}^o = 2.(E_b^{}(C - Cl) + 3E_b^{}(C - H)) + 3.E_b^{}(O = O) - \;2.2.E_b^{}(C = O{\text{)}}--{\text{ }}2.2E_b^{}(O - H)2 - 2.E_b^{}(H - Cl)\)

\(\Delta _r^{}H_{298}^o = 2.(339 + 3.414) + 3.498 - 2.2.736 - 2.2.464 - 2.431\)

\(\Delta _r^{}H_{298}^o =  - 1006kJ\)

\(n_{CH_3^{}Cl}^{} = \frac{1}{{50,5}}\,{\text{mol}} \to \Delta _r^{}H_{298}^o(1Gam) = \frac{1}{{50,5}}.\frac{{1006}}{2} = 9,96kJ\) 

\(2CH_2^{}Cl_2^{} + 3O_2^{} \to 2CO_2^{} + 2H_2^{}O + 2Cl_2^{}\)

\(\Delta _r^{}H_{298}^o = 2E_b^{}(CH_2^{}Cl_2^{}) + 3.E_b^{}(O_2^{}) - 2E_b^{}(CO_2^{}) - 2.E_b^{}(H_2^{}O) - 2.E_b^{}(Cl_2^{})\)

\(\Delta _r^{}H_{298}^o = 2.(2.E_b^{}(C - Cl) + 2E_b^{}(C - H)) + 3.E_b^{}(O = O) - \;2.2.E_b^{}(C = O{\text{)}}--{\text{ }}2.2E_b^{}(O - H)2 - 2.E_b^{}(Cl - Cl)\)

\(\Delta _r^{}H_{298}^o = 2.(2.339 + 2.414) + 3.498 - 2.2.736 - 2.2.464 - 2.243\)

\(\Delta _r^{}H_{298}^o =  - 780kJ\)

\(n_{CH_2^{}Cl_2^{}}^{} = \frac{1}{{85}}\,{\text{mol}} \to \Delta _r^{}H_{298}^o(1Gam) = \frac{1}{{85}}.\frac{{780}}{2} = 4,59kJ\) Ở đây chia 2 vì hệ số của CH2Cl2

\(4CHCl_3^{} + 5O_2^{} \to 4CO_2^{} + 2H_2^{}O + 6Cl_2^{}\)

\(\Delta _r^{}H_{298}^o = 4E_b^{}(CHCl_3^{}) + 5.E_b^{}(O_2^{}) - 4E_b^{}(CO_2^{}) - 2.E_b^{}(H_2^{}O) - 6.E_b^{}(Cl_2^{})\)

\(\Delta _r^{}H_{298}^o = 4.(3.E_b^{}(C - Cl) + E_b^{}(C - H)) + 5.E_b^{}(O = O) - \;2.4.E_b^{}(C = O{\text{)}}--{\text{ }}2.2E_b^{}(O - H)2 - 6.E_b^{}(Cl - Cl)\)

\(\Delta _r^{}H_{298}^o = 4.(3.339 + 414) + 5.498 - 4.2.736 - 2.2.464 - 6.243\)

\(\Delta _r^{}H_{298}^o =  - 988kJ\)

\(n_{CHCl_3^{}}^{} = \frac{1}{{119,5}}\,{\text{mol}} \to \Delta _r^{}H_{298}^o(1Gam) = \frac{1}{{119,5}}.\frac{{988}}{4} = 2,07kJ\) 

\(CCl_4^{} + O_2^{} \to CO_2^{} + 2Cl_2^{}\)

\(\Delta _r^{}H_{298}^o = E_b^{}(CCl_4^{}) + E_b^{}(O_2^{}) - E_b^{}(CO_2^{}) - 2E_b^{}(Cl_2^{})\)

\(\Delta _r^{}H_{298}^o = 4E_b^{}(C - Cl) + E_b^{}(O = O) - 2.E_b^{}(C = O) - 2.E_b^{}(Cl - Cl)\)

\(\Delta _r^{}H_{298}^o = 4.339 + 498 - 2.736 - 2.243 =  - 104kJ\)

\(n_{CCl_4^{}}^{} = \frac{1}{{119,5}}\,{\text{mol}} \to \Delta _r^{}H_{298}^o(1Gam) = \frac{1}{{154}}.104 = 0,68kJ\)

 2.

Delta H càng âm phản ứng cháy càng mãnh liệt

\(CCl_4^{} < CHCl_3^{} < CH_2^{}Cl_2^{} < CH_3^{}Cl < CH_4^{}\)