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QJEEM22Chef39: JEE Main 2022 Chemistry - Class 11 JEEM22Chef39 Solution: The enthalpy of combustion of propane, graphite and dihydrogen…

The enthalpy of combustion of propane, graphite and dihydrogen at 298 K298 \mathrm{~K} are −2220.0 kJ mol−1,−393.5 kJ mol−1-2220.0 \mathrm{~kJ} \mathrm{~mol}^{-1},-393.5 \mathrm{~kJ} \mathrm{~mol}^{-1} and −285.8 kJ mol−1-285.8 \mathrm{~kJ} \mathrm{~mol}^{-1} respectively. The magnitude of enthalpy of formation of propane (C3H8)\left(\mathrm{C}_{3} \mathrm{H}_{8}\right) is _______________ kJ mol−1\mathrm{kJ} \,\mathrm{mol}^{-1} . (Nearest integer)

Step-by-Step Verified Solution

Enthalpy of combustion of propane, graphite and H2\mathrm{H}_{2} at 298 K298 \mathrm{~K} are

C3H8( g)+5O2( g)→3CO2( g)+4H2O(I),ΔH1=−2220 kJ mol−1\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}

C(\mathrm{C}( graphite )+O2( g)→CO2( g),ΔH2=−393.5 kJ mol−1)+\mathrm{O}_{2}(\mathrm{~g}) \rightarrow \mathrm{CO}_{2}(\mathrm{~g}), \quad \Delta \mathrm{H}_{2}=-393.5 \mathrm{~kJ} \mathrm{~mol}^{-1}

H2( g)+12O2( g)→H2O(I),ΔH3=−285.8 kJ mol−1\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}

The desired reaction is

3C3 \mathrm{C} (graphite) + 4H2( g)→C3H8( g)4 \mathrm{H}_{2}(\mathrm{~g}) \rightarrow \mathrm{C}_{3} \mathrm{H}_{8}(\mathrm{~g})

ΔHf=3ΔH2+4ΔH3−ΔH1\Delta \mathrm{H}_{\mathrm{f}}=3 \Delta \mathrm{H}_{2}+4 \Delta \mathrm{H}_{3}-\Delta \mathrm{H}_{1}

amp;=3(−393.5)+4(−285.8)−(−2220)amp;=−103.7 kJ mol−1\begin{aligned}&=3(-393.5)+4(-285.8)-(-2220) \\&=-103.7 \mathrm{~kJ} \mathrm{~mol}^{-1}\end{aligned}

∣ΔHf∣≃104 kJ mol−1\left|\Delta \mathrm{H}_{\mathrm{f}}\right| \simeq 104 \mathrm{~kJ} \mathrm{~mol}^{-1}

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