Question

At 298 K and 1 atm, the molar enthalpies of combustion of cyclopropane and propene are −2091 kJ mol$^{-1}$ and −2058 kJ mol$^{-1}$, respectively. The enthalpy change (in kJ mol$^{-1}$) for the conversion of one mole of propene to one mole of cyclopropane is ..............

Hint:

Use combustion enthalpies to compare stability: less negative → more stable.

Answer 1

Correct Answer: 33

Step 1: Use combustion enthalpies to find relative stability.
More negative ΔH$_c$ means the compound is less stable.
Cyclopropane (−2091 kJ/mol) is less stable than propene (−2058 kJ/mol).
Step 2: Apply Hess’s law.
ΔH (propene → cyclopropane) = ΔH$_c$(propene) − ΔH$_c$(cyclopropane).
Step 3: Substitute values.
ΔH = (−2058) − (−2091)
= −2058 + 2091
= +33 kJ mol$^{-1}$.
But conversion of propene to cyclopropane requires energy, hence enthalpy change is **−33 kJ mol$^{-1}$** when written as combustion difference.
Step 4: Conclusion.
Thus, the conversion enthalpy is −33 kJ mol$^{-1}$.