Question

Calculate the standard enthalpy change for the reaction at 25 ∘ C. Standard enthalpy of formation values can be found in this list of thermodynamic properties. C3H8(g)+ 5O2(g) ⟶ 3CO2(g)+ 4H2O(g)

Answer 1

Answer:

The standard enthalpy change for the reaction at $25^{0}\textrm{C}$ is -2043.999kJ

Explanation:

Standard enthalpy change ($\Delta H_{rxn}^{0}$) for the given reaction is expressed as:

$\Delta H_{rxn}^{0}=[3mol\times \Delta H_{f}^{0}(CO_{2})_{g}]+[4mol\times \Delta H_{f}^{0}(H_{2}O)_{g}]-[1mol\times \Delta H_{f}^{0}(C_{3}H_{8})_{g}]-[5mol\times \Delta H_{f}^{0}(O_{2})_{g}]$

Where $\Delta H_{f}^{0}$ refers standard enthalpy of formation

Plug in all the given values from literature in the above equation:

$\Delta H_{rxn}^{0}=[3mol\times (-393.509kJ/mol)]+[4mol\times (-241.818kJ/mol)]-[1mol\times (-103.8kJ/mol)]-[5mol\times (0kJ/mol)]=-2043.999kJ$