Answer:
The standard enthalpy change for the reaction at [tex]25^{0}\textrm{C}[/tex] is -2043.999kJ
Explanation:
Standard enthalpy change ([tex]\Delta H_{rxn}^{0}[/tex]) for the given reaction is expressed as:
[tex]\Delta H_{rxn}^{0}=[3mol\times \Delta H_{f}^{0}(CO_{2})_{g}]+[4mol\times \Delta H_{f}^{0}(H_{2}O)_{g}]-[1mol\times \Delta H_{f}^{0}(C_{3}H_{8})_{g}]-[5mol\times \Delta H_{f}^{0}(O_{2})_{g}][/tex]
Where [tex]\Delta H_{f}^{0}[/tex] refers standard enthalpy of formation
Plug in all the given values from literature in the above equation:
[tex]\Delta H_{rxn}^{0}=[3mol\times (-393.509kJ/mol)]+[4mol\times (-241.818kJ/mol)]-[1mol\times (-103.8kJ/mol)]-[5mol\times (0kJ/mol)]=-2043.999kJ[/tex]
