i) S+O2→SO2 ΔH=-298.2 kJii) SO2+1/2O2→SO3 ΔH=-98.7 kJiii) SO3+H2O→H2SO4 ΔH=-130.2 kJiii) H2+1/2O2→H2O ΔH=-287.3 kJThe enthalpy of formation of H2SO4 at 298 K is:

Enthalpy of formation of H2SO4 is represented by the reaction as follows:H2+S+2O2→H2SO4From the given reactions,S+O2→SO2ΔH1=-298.2 kJSO2+1/2O2→SO3ΔH2=-98.7 kJSO3+H2O→H2SO4ΔH3=-130.2 kJH2+1/2O2→H2OΔH4=-287.3 kJOn (i) +( ii )+( iii )+( iv) we get,H2+S+2O2→H2SO4 ΔH=ΔH1+ΔH2+ΔH3+ΔH4⇒ ΔH=-298.2-98.7-130.2-287.3⇒ ΔH=-814.4Hence, option (A) is correct.

i) S+O2→SO2 ΔH=-298.2 kJii) SO2+1/2O2→SO3 ΔH=-98.7 kJiii) SO3+H2O→H2SO4 ΔH=-130.2 kJiii) H2+1/2O2→H2O ΔH=-287.3 kJThe enthalpy of formation of H2SO4 at 298 K is:

Enthalpy of formation of H2SO4 is represented by the reaction as follows:H2+S+2O2→H2SO4From the given reactions,S+O2→SO2ΔH1=-298.2 kJSO2+1/2O2→SO3ΔH2=-98.7 kJSO3+H2O→H2SO4ΔH3=-130.2 kJH2+1/2O2→H2OΔH4=-287.3 kJOn (i) +( ii )+( iii )+( iv) we get,H2+S+2O2→H2SO4 ΔH=ΔH1+ΔH2+ΔH3+ΔH4⇒ ΔH=-298.2-98.7-130.2-287.3⇒ ΔH=-814.4Hence, option (A) is correct.

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i) $S + O_{2} \to {SO}_{2} ΔH = - 298.2 kJ$
ii) $S O_{2} + 1 / 2 O_{2} \to {SO}_{3} ΔH = - 98.7 kJ$
iii) ${SO}_{3} + H_{2} O \to H_{2} {SO}_{4} ΔH = - 130.2 kJ$
iii) $H_{2} + 1 / 2 O_{2} \to H_{2} O ΔH = - 287.3 kJ$
The enthalpy of formation of $H_{2} {SO}_{4}$ at $298 K$ is:

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$- 814.4 k J$

$- 650.3 k J$

$320.5 k J$

$- 433.5 k J$

answer is A.

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Detailed Solution

Enthalpy of formation of $H_{2} {SO}_{4}$ is represented by the reaction as follows:

$H_{2} + S + 2 O_{2} \to H_{2} {SO}_{4}$

From the given reactions,

$\begin{array}{l} S + O_{2} \to {SO}_{2} {ΔH}_{1} = - 298.2 kJ \\ {SO}_{2} + 1 / 2 O_{2} \to {SO}_{3} {ΔH}_{2} = - 98.7 kJ \\ {SO}_{3} + H_{2} O \to H_{2} {SO}_{4} {ΔH}_{3} = - 130.2 kJ \\ H_{2} + 1 / 2 O_{2} \to H_{2} {OΔH}_{4} = - 287.3 kJ \end{array}$

On (i) $+ ($ ii $) + ($ iii $) + ($ iv) we get,

$\begin{array}{cl} H_{2} + S + 2 O_{2} \to H_{2} {SO}_{4} \\ ΔH = {ΔH}_{1} + {ΔH}_{2} + {ΔH}_{3} + {ΔH}_{4} \\ \Rightarrow ΔH = - 298.2 - 98.7 - 130.2 - 287.3 \\ \Rightarrow ΔH = - 814.4 \end{array}$