JEE MAIN - Chemistry (2023 - 6th April Evening Shift - No. 18)
Consider the following data
Heat of combustion of $$\mathrm{H}_{2}(\mathrm{g})\quad\quad=-241.8 \mathrm{~kJ} \mathrm{~mol}^{-1}$$
Heat of combustion of $$\mathrm{C}(\mathrm{s})\quad\quad=-393.5 \mathrm{~kJ} \mathrm{~mol}^{-1}$$
Heat of combustion of $$\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(\mathrm{l})\quad=-1234.7 \mathrm{~kJ}~{\mathrm{mol}}^{-1}$$
The heat of formation of $$\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(\mathrm{l})$$ is $$(-)$$ ___________ $$\mathrm{kJ} ~\mathrm{mol}^{-1}$$ (Nearest integer).
Answer
278
Explanation
The reaction for the formation of $$\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(\mathrm{l})$$ is given by:
$$2\mathrm{C}(\mathrm{s}) + 6\mathrm{H}_{2}(\mathrm{g}) + \frac{1}{2}\mathrm{O}_{2}(\mathrm{g}) \rightarrow \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(\mathrm{l}) + 3\mathrm{H}_{2}\mathrm{O}(\mathrm{l})$$
The heat of combustion for $$\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(\mathrm{l})$$ is:
$$\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(\mathrm{l}) + 3\mathrm{O}_{2}(\mathrm{g}) \rightarrow 2\mathrm{CO}_{2}(\mathrm{g}) + 3\mathrm{H}_{2}\mathrm{O}(\mathrm{l})$$
By adding the formation reaction with the combustion reaction times -1, we get:
$$2\mathrm{C}(\mathrm{s}) + 3\mathrm{H}_{2}(\mathrm{g}) + \frac{3}{2}\mathrm{O}_{2}(\mathrm{g}) \rightarrow 2\mathrm{CO}_{2}(\mathrm{g}) + 3\mathrm{H}_{2}\mathrm{O}(\mathrm{l})$$
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:
$$-1234.7 \mathrm{~kJ/mol} + 2*(-393.5 \mathrm{~kJ/mol}) + 3*(-241.8 \mathrm{~kJ/mol})$$
Calculating this gives:
$$\Delta H_f(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(\mathrm{l})) = -277.7 \approx -278 \mathrm{~kJ/mol}$$ (rounded to the nearest integer)
$$2\mathrm{C}(\mathrm{s}) + 6\mathrm{H}_{2}(\mathrm{g}) + \frac{1}{2}\mathrm{O}_{2}(\mathrm{g}) \rightarrow \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(\mathrm{l}) + 3\mathrm{H}_{2}\mathrm{O}(\mathrm{l})$$
The heat of combustion for $$\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(\mathrm{l})$$ is:
$$\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(\mathrm{l}) + 3\mathrm{O}_{2}(\mathrm{g}) \rightarrow 2\mathrm{CO}_{2}(\mathrm{g}) + 3\mathrm{H}_{2}\mathrm{O}(\mathrm{l})$$
By adding the formation reaction with the combustion reaction times -1, we get:
$$2\mathrm{C}(\mathrm{s}) + 3\mathrm{H}_{2}(\mathrm{g}) + \frac{3}{2}\mathrm{O}_{2}(\mathrm{g}) \rightarrow 2\mathrm{CO}_{2}(\mathrm{g}) + 3\mathrm{H}_{2}\mathrm{O}(\mathrm{l})$$
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:
$$-1234.7 \mathrm{~kJ/mol} + 2*(-393.5 \mathrm{~kJ/mol}) + 3*(-241.8 \mathrm{~kJ/mol})$$
Calculating this gives:
$$\Delta H_f(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(\mathrm{l})) = -277.7 \approx -278 \mathrm{~kJ/mol}$$ (rounded to the nearest integer)
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