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

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

Step-by-step explanation:

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

`\Delta H_(rxn)^(0)=[3mol* \Delta H_(f)^(0)(CO_(2))_(g)]+[4mol* \Delta H_(f)^(0)(H_(2)O)_(g)]-[1mol* \Delta H_(f)^(0)(C_(3)H_(8))_(g)]-[5mol* \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* (-393.509kJ/mol)]+[4mol* (-241.818kJ/mol)]-[1mol* (-103.8kJ/mol)]-[5mol* (0kJ/mol)]=-2043.999kJ`