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C3H8+3O2 = 3CO2+4H2O what is the enthalpy combustion

User Bland
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Answer:


\Delta H_(comb)=2043.85kJ/mol

Step-by-step 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:


\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)

Thus, given the values of the enthalpies of formation on the attached file, we obtain:
-\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

Best regards!

C3H8+3O2 = 3CO2+4H2O what is the enthalpy combustion-example-1
User Bodzio
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