JEE MAIN - Chemistry (2022 - 25th June Evening Shift - No. 2)
At 25$$^\circ$$C and 1 atm pressure, the enthalpy of combustion of benzene (I) and acetylene (g) are $$-$$ 3268 kJ mol$$-$$1 and $$-$$1300 kJ mol$$-$$1, respectively. The change in enthalpy for the reaction 3 C2H2(g) $$\to$$ C6H6 (I), is :
+324 kJ mol$$-$$1
+632 kJ mol$$-$$1
$$-$$ 632 kJ mol$$-$$1
$$-$$ 732 kJ mol$$-$$1
Explanation
I. $\mathrm{C}_{6} \mathrm{H}_{6}(\ell)+\frac{15}{2} \mathrm{O}_{2}(\mathrm{~g}) \rightarrow 6 \mathrm{CO}_{2}(\mathrm{~g})+3 \mathrm{H}_{2} \mathrm{O}(\mathrm{g})$
$$ \Delta \mathrm{H}_{1}=-3268 \mathrm{~kJ} / \mathrm{mol} $$
II. $\mathrm{C}_{2} \mathrm{H}_{2}(\mathrm{~g})+\frac{5}{2} \mathrm{O}_{2}(\mathrm{~g}) \rightarrow 2 \mathrm{CO}_{2}(\mathrm{~g})+\mathrm{H}_{2} \mathrm{O}(\mathrm{g})$
$$ \Delta \mathrm{H}_{2}=-1300 \mathrm{~kJ} / \mathrm{mol} $$
III. $3 \mathrm{C}_{2} \mathrm{H}_{2}(\mathrm{~g}) \rightarrow \mathrm{C}_{6} \mathrm{H}_{6}(\ell) \quad \Delta \mathrm{H}_{3}$
Applying Hess's law of constant heat summation
$$ \begin{aligned} \Delta \mathrm{H}_{3} &=3 \times \Delta \mathrm{H}_{2}-\Delta \mathrm{H}_{1} \\\\ &=3 \times(-1300)-(-3268) \\\\ &=-632 \mathrm{~kJ} / \mathrm{mol} \end{aligned} $$
$$ \Delta \mathrm{H}_{1}=-3268 \mathrm{~kJ} / \mathrm{mol} $$
II. $\mathrm{C}_{2} \mathrm{H}_{2}(\mathrm{~g})+\frac{5}{2} \mathrm{O}_{2}(\mathrm{~g}) \rightarrow 2 \mathrm{CO}_{2}(\mathrm{~g})+\mathrm{H}_{2} \mathrm{O}(\mathrm{g})$
$$ \Delta \mathrm{H}_{2}=-1300 \mathrm{~kJ} / \mathrm{mol} $$
III. $3 \mathrm{C}_{2} \mathrm{H}_{2}(\mathrm{~g}) \rightarrow \mathrm{C}_{6} \mathrm{H}_{6}(\ell) \quad \Delta \mathrm{H}_{3}$
Applying Hess's law of constant heat summation
$$ \begin{aligned} \Delta \mathrm{H}_{3} &=3 \times \Delta \mathrm{H}_{2}-\Delta \mathrm{H}_{1} \\\\ &=3 \times(-1300)-(-3268) \\\\ &=-632 \mathrm{~kJ} / \mathrm{mol} \end{aligned} $$
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