Question

Comment Both propane and benzene are hydrocarbons. As a rule,

the energy obtained from the combustion of a gram of hydrocar-
bon is between 40 and 50 kJ.
Practice Exercise 1
Calculate the enthalpy change for the reaction
2 H2O2(l)-→ 2 H2O(l) + O2(g)
using enthalpies of formation:
Change in enthalpy(f)H2O2(l)= -187.8 kJ/mol
Change in enthalpy (f) [H2O(l)= -285.8 kJ/mol

Answer 1

The enthalpy change : -196.2 kJ/mol

Further explanation

The change in enthalpy in the formation of 1 mole of the elements is called enthalpy of formation

The enthalpy of formation measured in standard conditions (25 ° C, 1 atm) is called the standard enthalpy of formation (ΔHf °)

(ΔH) can be positive (endothermic = requires heat) or negative (exothermic = releasing heat)

The value of ° H ° can be calculated from the change in enthalpy of standard formation:

∆H ° rxn = ∑n ∆Hf ° (product) - ∑n ∆Hf ° (reactants)

Reaction

2 H₂O₂(l)-→ 2 H₂O(l) + O₂(g)

∆H ° rxn = 2. ∆Hf ° H₂O - 2. ∆Hf °H₂O₂

`\tt \Delta H_(rxn)=2.(-285.8)-2.(-187.8)\\\\\Delta H_(rxn)=-571.6+375.4=-196.2~kJ/mol\rightarrow \Delta Hf~O_2=0`