Question

The enthalpies of combustion of $S _{(s)}, SO _{2(g)}$ and $H _{2(g)}$ are -298.2,-98.7 and $-287.3\, kJ\, mol ^{-1}$. If the enthalpy of reaction $SO _{3(g)}+ H _{2} O _{(l)} \rightarrow H _{2} SO _{4(l)}$ is $-130.2 \,kJ\, mol ^{-1},$ the enthalpy of formation of $H _{2} SO _{4(l)}$ would be

• A. $-814.4 \,kJ\, mol^{-1}$
• B. $-650.3 \,kJ\, mol^{-1}$
• C. $-554.2 \,kJ\, mol^{-1}$
• D. $-435.5 \,kJ\, mol^{-1}$

Answer 1

Answer: (A)

$S + O _{2} \rightarrow SO _{2} ; \Delta H=-298.2 \,kJ\, mol^{-1}$

$SO _{2}+\frac{1}{2} O _{2} \rightarrow SO _{3} ; \Delta H=-98.7 kJ\, mol ^{-1}$

$H _{2}+\frac{1}{2} O _{2} \rightarrow H _{2} O ; \Delta H=-287.3 \,kJ\, mol ^{-1}$

and $SO _{3}+ H _{2} O \rightarrow H _{2} SO _{4} ; \Delta H=-130.2\, kJ\, mol^{-1}$.

Formation of $H _{2} SO _{4}$ implies $H _{2(g)}+ S _{(s)}+2 O _{2(g)} \rightarrow H _{2} SO _{4(l)}$ and this reaction can be

obtained by adding the given four reactions.

$\therefore \Delta H=-298.2+(-98.7)+(-287.3)+(-130.2)$

$=-814.4\, kJ\, mol ^{-1}$