Q44 / 180 · COMEDK 2024 (Evening Shift) · Chemistry
Given :
$
\begin{gathered}
\Delta \mathrm{H}^0 \mathrm{f}_{\text {of }} \mathrm{CO}_2(\mathrm{~g})=-393.5 \mathrm{~kJ} / \mathrm{mol} \\
\Delta \mathrm{H}^0{ }_{\mathrm{f}} \text { of } \mathrm{H}_2 \mathrm{O}(\mathrm{l})=-286 \mathrm{~kJ} / \mathrm{mol} \\
\Delta \mathrm{H}^0{ }_{\mathrm{f}} \text { of } \mathrm{C}_3 \mathrm{H}_6(\mathrm{~g})=+20.6 \mathrm{~kJ} / \mathrm{mol}
\end{gathered}
$
$\Delta \mathrm{H}^0$ isomerisation of Cyclopropane to Propene $=-33 \mathrm{~kJ} / \mathrm{mol}$
What is the standard enthalpy of combustion of Cyclopropane?
- A$-1985$ kJ/mol
- B$+2384$ kJ/mol
- Correct answerC$-2092$ kJ/mol
- D$-2051$ kJ/mol
Solution
Combustion: $C_3H_6(g) + \dfrac{9}{2}O_2(g) \to 3CO_2(g) + 3H_2O(l)$
$\Delta H_f^0$ of cyclopropane using isomerisation data:
Cyclopropane $\to$ Propene, $\Delta H_{iso} = -33$ kJ/mol
$\Delta H_f^0$(Cyclopropane) $= \Delta H_f^0$(Propene) $- \Delta H_{iso} = 20.6 - (-33) = 53.6$ kJ/mol
$\Delta H_c^0 = [3 \times (-393.5) + 3 \times (-286)] - [53.6]$
$= [-1180.5 - 858] - 53.6$
$= -2038.5 - 53.6$
$= -2092.1 \approx -2092$ kJ/mol
$\Delta H_f^0$ of cyclopropane using isomerisation data:
Cyclopropane $\to$ Propene, $\Delta H_{iso} = -33$ kJ/mol
$\Delta H_f^0$(Cyclopropane) $= \Delta H_f^0$(Propene) $- \Delta H_{iso} = 20.6 - (-33) = 53.6$ kJ/mol
$\Delta H_c^0 = [3 \times (-393.5) + 3 \times (-286)] - [53.6]$
$= [-1180.5 - 858] - 53.6$
$= -2038.5 - 53.6$
$= -2092.1 \approx -2092$ kJ/mol
Hence, the correct answer is option: 'C'.