Answer:
[tex]\Delta H_{comb}=2043.85kJ/mol[/tex]
Explanation:
Hello there!
In this case, according to the given chemical reaction, it possible for us to set up the expression for the calculation of the enthalpy change as shown below:
[tex]\Delta H_r=-\Delta H_{comb}=3\Delta _fH_{CO_2}+4\Delta _fH_{H_2O}-\Delta _fH_{C_3H_8}-3\Delta _fH_{O_2}[/tex]
Thus, given the values of the enthalpies of formation on the attached file, we obtain:[tex]-\Delta H_{comb}=3(-393.5kJ/mol)+4(-241.8kJ/mol)-(-103.85kJ/mol)-3(0kJ/mol)\\\\-\Delta H_{comb}=-2043.85kJ/mol\\\\\Delta H_{comb}=2043.85kJ/mol[/tex]
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