Question

If $S+O_2\to S{O}_2$ ; $\Delta H=-298.2$
$S{O}_2+\frac{1}{2}{O}_2\to S{O}_3$ ; $\Delta H=-98.7$
$S{O}_3+H_{2}O\to H_{2}S{O}_4$ ; $\Delta H=-130.2$
$H_2+\frac{1}{2}{O}_2\to H_{2}O$ ; $\Delta H=-287.3$
Then the enthalpy of formation of $H_{2}S{O}_4$ at 298 K is

• A. $-814.4kJ$
• B. $-650.3kJ$
• C. $-320.5kJ$
• D. $-433.5kJ$

Answer 1

The correct option is A $-814.4kJ$

The thermochemical reactions are as given below.
$S+O_2\to S{O}_2,\Delta H=-298.2$
$S{O}_2+\frac{1}{2}{O}_2\to S{O}_3,\Delta H=-98.7$
$S{O}_3+H_{2}O\to H_{2}S{O}_4,\Delta H=-130.2$
$H_2+\frac{1}{2}{O}_2\to H_{2}O,\Delta H=-287.3$
The four reactions are added to obtain the reaction $S+H_2+2O_2\to H_{2}S{O}_4$.

This is the reaction for the formation of $H_{2}S{O}_4$.
Hence, the enthalpy of formation of $H_{2}S{O}_4$ is $-298.2-98.7-130.2-287.3$
$=-814.4$kJ