COMEDK 2024 Shift 2 Paper

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)+‌

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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∕mol
And the given isomerization enthalpy:
∆H0( isomerization of Cyclopropane to Propene )=−33‌kJ∕mol
Now, 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:
Substituting the values:
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))=0
Therefore:

‌∆H‌combustion ‌0‌‌=[3(−393.5‌kJ∕mol)+3(−286‌kJ∕mol)]−[53.6‌kJ∕mol+0]
‌=−2038.5−53.6
‌=−2092.1‌kJ∕mol
Thus, the standard enthalpy of combustion of Cyclopropane is closest to:
Option A: −2092‌kJ∕mol