\(C_6^{}H_6^{} + \frac{{15}}{2}O_2^{}\xrightarrow{{{t^o}}}6CO_2^{} + 3H_2^{}O\)

\(\Delta H_{298}^o = 6.\Delta H_{298}^o(CO_2^{}) + 3\Delta H_{298}^o(H_2^{}O) - 1.\Delta H_{298}^o(C_6^{}H_6^{}) - \frac{{15}}{2}\Delta H_{298}^o(O_2^{})\)

\(\Delta H_{298}^o = 6.( - 393,5) + 3.( - 285,84) - 49 - \frac{{15}}{2}.0\)

\(\Delta H_{298}^o =  - 3267,52kJ\)

\(n_{C_6^{}H_6^{}}^{} = \frac{1}{{78}}\;{\text{mol}}\)

Cứ 1 mol \(C_6^{}H_6^{}\) thì tỏa ra 3267,52kJ

Vậy 1/78 mol \(C_6^{}H_6^{}\) thì tỏa ra \(\frac{1}{{78}}.(  3267,52).1 = 41,89kJ\)

\(C_3^{}H_8^{} + 5O_2^{}\xrightarrow{{{t^o}}}3CO_2^{} + 4H_2^{}O\)

\(\Delta H_{298}^o = 3.\Delta H_{298}^o(CO_2^{}) + 4\Delta H_{298}^o(H_2^{}O) - 1.\Delta H_{298}^o(C_3^{}H_8^{}) - 5\Delta H_{298}^o(O_2^{})\)

\(\Delta H_{298}^o = 3.( - 393,5) + 4.( - 285,84) - ( - 105) - 5.0\)

\(\Delta H_{298}^o =  - 2218,86kJ\)

\(n_{C_3^{}H_8^{}}^{} = \frac{1}{{44}}{\text{mol}}\)

Cứ 1 mol \(C_3^{}H_8^{}\) thì tỏa ra 2218,86kJ

Vậy 1/44 mol \(C_3^{}H_8^{}\) sẽ tỏa ra \(\frac{1}{{44}}.( - 2218,86).1 = 50,43kJ\)