Medium INTEGER +4 / -1 PYQ · JEE Mains 2022

The enthalpy of combustion of propane, graphite and dihydrogen at $298 \mathrm{~K}$ are $-2220.0 \mathrm{~kJ} \mathrm{~mol}^{-1},-393.5 \mathrm{~kJ} \mathrm{~mol}^{-1}$ and $-285.8 \mathrm{~kJ} \mathrm{~mol}^{-1}$ respectively. The magnitude of enthalpy of formation of propane $\left(\mathrm{C}_{3} \mathrm{H}_{8}\right)$ is _______________ $\mathrm{kJ} \,\mathrm{mol}^{-1}$. (Nearest integer)

Answer (integer) 104

Solution

Enthalpy of combustion of propane, graphite and $\mathrm{H}_{2}$ at $298 \mathrm{~K}$ are <br/><br/> $$\mathrm{C}_{3} \mathrm{H}_{8}(\mathrm{~g})+5 \mathrm{O}_{2}(\mathrm{~g}) \rightarrow 3 \mathrm{CO}_{2}(\mathrm{~g})+4 \mathrm{H}_{2} \mathrm{O}(\mathrm{I}), \Delta \mathrm{H}_{1}=-2220 \mathrm{~kJ} \mathrm{~mol}^{-1}$$<br/><br/> $\mathrm{C}($ graphite $$)+\mathrm{O}_{2}(\mathrm{~g}) \rightarrow \mathrm{CO}_{2}(\mathrm{~g}), \quad \Delta \mathrm{H}_{2}=-393.5 \mathrm{~kJ} \mathrm{~mol}^{-1}$$ <br/><br/>$$\mathrm{H}_{2}(\mathrm{~g})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{~g}) \rightarrow \mathrm{H}_{2} \mathrm{O}(\mathrm{I}), \quad \Delta \mathrm{H}_{3}=-285.8 \mathrm{~kJ} \mathrm{~mol}^{-1}$$<br/><br/> The desired reaction is <br/><br/> $3 \mathrm{C}$ (graphite) + $$4 \mathrm{H}_{2}(\mathrm{~g}) \rightarrow \mathrm{C}_{3} \mathrm{H}_{8}(\mathrm{~g})$$ <br/><br/> $$\Delta \mathrm{H}_{\mathrm{f}}=3 \Delta \mathrm{H}_{2}+4 \Delta \mathrm{H}_{3}-\Delta \mathrm{H}_{1}$$ <br/><br/> $$ \begin{aligned} &=3(-393.5)+4(-285.8)-(-2220) \\ &=-103.7 \mathrm{~kJ} \mathrm{~mol}^{-1} \end{aligned} $$ <br/><br/> $\left|\Delta \mathrm{H}_{\mathrm{f}}\right| \simeq 104 \mathrm{~kJ} \mathrm{~mol}^{-1}$

About this question

Subject: Chemistry · Chapter: Thermodynamics · Topic: Zeroth and First Law

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