If
$S_{(s)} + {O_{2}}_{(g)} \overset{}{\to} S O_{2 (g)}; Δ H = - 398.2 K J;$
$S O_{2 (g)} + \frac{1}{2} O_{2 (g)} \to S O_{3 (g)}; Δ H = - 98.7 K J$
${SO}_{3 (g)} + H_{2} O_{(l)} \overset{}{\to} H_{2} {SO}_{4 (l)};$ $ΔH = - 130.2 KJ;$
$H_{2 (g)} + \frac{1}{2} O_{2 (g)} \overset{}{\to} H_{2} O_{(l)};$ $ΔH = - 227.3 KJ$
The enthalpy of formation of sulfuric acid at 298 K will be:

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-854.4kJ

-650.3kJ

-754.4kJ

-433.7kJ

answer is A.

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Detailed Solution

The image describes the enthalpy changes for various chemical reactions and calculates the enthalpy of formation for $\text{H}_2\text{SO}_4$ $H$ $2$ $$ $SO$ $4$ $$ (sulfuric acid) using the following reactions and data:

Reactions and Enthalpy Changes:

$S + O_2 \rightarrow SO_2$ $S$ $+$ $O$ $2$ $$ $\to$ $SO$ $2$ $$ ; $\Delta H = -398.2 \, \text{kJ}$ $Δ$ $H$ $=$ $-398.2$ $kJ$
$SO_2 + \frac{1}{2} O_2 \rightarrow SO_3$ $SO$ $2$ $$ $+$ $21$ $$ $O$ $2$ $$ $\to$ $SO$ $3$ $$ ; $\Delta H = -98.7 \, \text{kJ}$ $Δ$ $H$ $=$ $-98.7$ $kJ$
$SO_3 + H_2O \rightarrow H_2SO_4$ $SO$ $3$ $$ $+$ $H$ $2$ $$ $O$ $\to$ $H$ $2$ $$ $SO$ $4$ $$ ; $\Delta H = -130.2 \, \text{kJ}$ $Δ$ $H$ $=$ $-130.2$ $kJ$
$H_2 + \frac{1}{2} O_2 \rightarrow H_2O$ $H$ $2$ $$ $+$ $21$ $$ $O$ $2$ $$ $\to$ $H$ $2$ $$ $O$ ; $\Delta H = -227.3 \, \text{kJ}$ $Δ$ $H$ $=$ $-227.3$ $kJ$

To Find:

The enthalpy of formation of $\text{H}_2\text{SO}_4$ $H$ $2$ $$ $SO$ $4$ $$ for the reaction: $H_2 + S + 2O_2 \rightarrow H_2SO_4$ $H$ $2$ $$ $+$ $S$ $+$ $2$ $O$ $2$ $$ $\to$ $H$ $2$ $$ $SO$ $4$ $$

The overall reaction is derived by combining the given reactions:

4 + 1 + 2 + 3

$4$ $+$ $1$ $+$ $2$ $+$ $3$

Combining Enthalpy Changes:

\Delta H_{\text{formation}} = -398.2 \, \text{kJ} + (-98.7 \, \text{kJ}) + (-130.2 \, \text{kJ}) + (-227.3 \, \text{kJ})

$Δ$ $H$ $formation$ $$ $=$ $-398.2$ $kJ$ $+$ $($ $-98.7$ $kJ$ $)$ $+$ $($ $-130.2$ $kJ$ $)$ $+$ $($ $-227.3$ $kJ$ $)$ $\Delta H_{\text{formation}} = -854.4 \, \text{kJ}$ $Δ$ $H$ $formation$ $$ $=$ $-854.4$ $kJ$