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QJEEM23Chafec: JEE Main 2023 Chemistry - Class 11 JEEM23Chafec Solution: Consider the following data Heat of combustion of Heat of…

Consider the following data

Heat of combustion of H2(g)=−241.8 kJ mol−1\mathrm{H}_{2}(\mathrm{g})\quad\quad=-241.8 \mathrm{~kJ} \mathrm{~mol}^{-1}

Heat of combustion of C(s)=−393.5 kJ mol−1\mathrm{C}(\mathrm{s})\quad\quad=-393.5 \mathrm{~kJ} \mathrm{~mol}^{-1}

Heat of combustion of C2H5OH(l)=−1234.7 kJ mol−1\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(\mathrm{l})\quad=-1234.7 \mathrm{~kJ}~{\mathrm{mol}}^{-1}

The heat of formation of C2H5OH(l)\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(\mathrm{l}) is (−)(-) ___________ kJ mol−1\mathrm{kJ} ~\mathrm{mol}^{-1} (Nearest integer).

Step-by-Step Verified Solution

The reaction for the formation of C2H5OH(l)\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(\mathrm{l}) is given by:

2C(s)+6H2(g)+12O2(g)→C2H5OH(l)+3H2O(l)2\mathrm{C}(\mathrm{s}) + 6\mathrm{H}_{2}(\mathrm{g}) + \frac{1}{2}\mathrm{O}_{2}(\mathrm{g}) \rightarrow \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(\mathrm{l}) + 3\mathrm{H}_{2}\mathrm{O}(\mathrm{l})

The heat of combustion for C2H5OH(l)\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(\mathrm{l}) is:

C2H5OH(l)+3O2(g)→2CO2(g)+3H2O(l)\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(\mathrm{l}) + 3\mathrm{O}_{2}(\mathrm{g}) \rightarrow 2\mathrm{CO}_{2}(\mathrm{g}) + 3\mathrm{H}_{2}\mathrm{O}(\mathrm{l})

By adding the formation reaction with the combustion reaction times -1, we get:

2C(s)+3H2(g)+32O2(g)→2CO2(g)+3H2O(l)2\mathrm{C}(\mathrm{s}) + 3\mathrm{H}_{2}(\mathrm{g}) + \frac{3}{2}\mathrm{O}_{2}(\mathrm{g}) \rightarrow 2\mathrm{CO}_{2}(\mathrm{g}) + 3\mathrm{H}_{2}\mathrm{O}(\mathrm{l})

This reaction has a heat equal to the heat of formation of ethanol times -1.

Therefore, the heat of formation of ethanol is equal to:

−1234.7 kJ/mol+2∗(−393.5 kJ/mol)+3∗(−241.8 kJ/mol)-1234.7 \mathrm{~kJ/mol} + 2*(-393.5 \mathrm{~kJ/mol}) + 3*(-241.8 \mathrm{~kJ/mol})

Calculating this gives:

ΔHf(C2H5OH(l))=−277.7≈−278 kJ/mol\Delta H_f(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(\mathrm{l})) = -277.7 \approx -278 \mathrm{~kJ/mol} (rounded to the nearest integer)

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