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If $\mathrm{S}+\mathrm{O}_2 \rightarrow \mathrm{SO}_2 ;(\Delta H=-298.2)$
$\mathrm{SO}_2+\frac{1}{2} \mathrm{O}_2 \rightarrow \mathrm{SO}_3 ;(\Delta H=-98.2)$
$\mathrm{SO}_3+\mathrm{H}_2 \mathrm{O} \rightarrow \mathrm{H}_2 \mathrm{SO}_4 ;(\Delta H=-130.2)$
$\mathrm{H}_2+\frac{1}{2} \mathrm{O}_2 \rightarrow \mathrm{H}_2 \mathrm{O} ;(\Delta \mathrm{H}=-287.3)$
then the enthalpy of formation of $\mathrm{H}_2 \mathrm{SO}_4$ at $298 \mathrm{~K}$ will be
Options:
$\mathrm{SO}_2+\frac{1}{2} \mathrm{O}_2 \rightarrow \mathrm{SO}_3 ;(\Delta H=-98.2)$
$\mathrm{SO}_3+\mathrm{H}_2 \mathrm{O} \rightarrow \mathrm{H}_2 \mathrm{SO}_4 ;(\Delta H=-130.2)$
$\mathrm{H}_2+\frac{1}{2} \mathrm{O}_2 \rightarrow \mathrm{H}_2 \mathrm{O} ;(\Delta \mathrm{H}=-287.3)$
then the enthalpy of formation of $\mathrm{H}_2 \mathrm{SO}_4$ at $298 \mathrm{~K}$ will be
Solution:
2499 Upvotes
Verified Answer
The correct answer is:
$-814.4 \mathrm{~kJ/mole}$
$\mathrm{S}+\mathrm{O}_2 \rightarrow \mathrm{SO}_2 ; \Delta \mathrm{H}=-298.2 \mathrm{~kJ} / \mathrm{mole}$
$\mathrm{SO}_2+12 \mathrm{O}_2 \rightarrow \mathrm{SO}_3 \quad \& \mathrm{H}=-98.7 \mathrm{~kJ} / \mathrm{mole}$
$\mathrm{SO}_3+\mathrm{H}_2 \mathrm{O} \rightarrow \mathrm{H}_2 \mathrm{SO}_4 ; \Delta \mathrm{H}=-130.2 \mathrm{~kJ} / \mathrm{mole}$
$\mathrm{H}_2+12 \mathrm{O}_2 \rightarrow \mathrm{H}_2 \mathrm{O} \underset{\sim}{ } \Delta \mathrm{H}=-287.3 \mathrm{~kJ} / \mathrm{mole}$
The formation of $\mathrm{H}_2 \mathrm{SO}_4$
$\mathrm{H}_2+\mathrm{S}+2 \mathrm{O}_2 \rightarrow \mathrm{H}_2 \mathrm{SO}_4$ can be calculated by adding the above four equations
$\Delta \mathrm{H}=-(298.2+98.7+130.2+287.3)=-814.4 \mathrm{~kJ} / \mathrm{mole}$
$\mathrm{SO}_2+12 \mathrm{O}_2 \rightarrow \mathrm{SO}_3 \quad \& \mathrm{H}=-98.7 \mathrm{~kJ} / \mathrm{mole}$
$\mathrm{SO}_3+\mathrm{H}_2 \mathrm{O} \rightarrow \mathrm{H}_2 \mathrm{SO}_4 ; \Delta \mathrm{H}=-130.2 \mathrm{~kJ} / \mathrm{mole}$
$\mathrm{H}_2+12 \mathrm{O}_2 \rightarrow \mathrm{H}_2 \mathrm{O} \underset{\sim}{ } \Delta \mathrm{H}=-287.3 \mathrm{~kJ} / \mathrm{mole}$
The formation of $\mathrm{H}_2 \mathrm{SO}_4$
$\mathrm{H}_2+\mathrm{S}+2 \mathrm{O}_2 \rightarrow \mathrm{H}_2 \mathrm{SO}_4$ can be calculated by adding the above four equations
$\Delta \mathrm{H}=-(298.2+98.7+130.2+287.3)=-814.4 \mathrm{~kJ} / \mathrm{mole}$
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