To determine the standard enthalpy of combustion of Cyclopropane, we need to follow a systematic thermodynamic approach. First, we need the balanced chemical equation for the combustion of Cyclopropane:
C3H6(g)+O2(g)⟶3CO2(g)+3H2O(l)The combustion reaction involves the transformation of Cyclopropane into carbon dioxide and water. We need to calculate the change in enthalpy for this reaction.
Given data:
∆Hf0(CO2(g))=−393.5kJ∕mol ∆Hf0(H2O(l))=−286kJ∕mol ∆Hf0(C3H6(g))=+20.6kJ∕molAnd the given isomerization enthalpy:
∆H0( isomerization of Cyclopropane to Propene
)=−33kJ∕molNow, let's determine the enthalpy of formation of Cyclopropane
(∆Hf0(C3H6(g))Cyclopropane ). From the given isomerization enthalpy of Cyclopropane to Propene, we can write:
∆Hf0( Cyclopropane )=∆Hf0( Propene )+33kJ∕mol
Substituting the values:
∆Hf0( Cyclopropane )=20.6kJ∕mol+33kJ∕mol=53.6kJ∕mol
Next, the enthalpy change of combustion reaction
(∆Hcombustion ) is found using the formula:
∆Hcombustion 0=∑∆Hf0 (products) −∑∆Hf0 (reactants) For the combustion reaction of Cyclopropane:
Note that the standard enthalpy of formation for
O2(g) is zero:
∆Hf0(O2(g))=0Therefore:
∆Hcombustion 0=[3(−393.5kJ∕mol)+3(−286kJ∕mol)]−[53.6kJ∕mol+0]∆Hcombustion 0=[3(−393.5)+3(−286)]−53.6
=[−1180.5−858]−53.6
=−2038.5−53.6=−2092.1kJ∕molThus, the standard enthalpy of combustion of Cyclopropane is closest to:
Option A:
−2092kJ∕mol