Câu `2` 

`a)` $\rm CH_3-CH(CH_3)-CH(CH_3)-CH(CH_3)-CH_3+\dfrac{25}{2} O_2 \xrightarrow{to} 8CO_2 +9H_2O$ 

`b)` $\rm \Delta_rH_{298}(2,3,4-trimethylpentane)=8.\Delta_fH_{298}^0(CO_2)+9.\Delta_fH_{298}^0(H_2O) - \Delta_fH_{298}(2,3,4-trimethylpentane)$ 

`<=>` $\rm \Delta_fH_{298}(2,3,4-trimethylpentane)=-(-5064,9)+8.(-393,51)+9.(-241,82)=-295,56(kJ/mol)$