Question

$4.6 \,kJ$ heat is liberated on burning $0.5 \,g$ of sulphur. The enthalpy of formation of $SO_{2}$ is: [molecular weight of $S = 32, O = 16]$

• A. #ERROR!
• B. - 294.4 kJ
• C. #ERROR!
• D. - 462.4 kJ

Answer 1

Answer: (B)

- 294.4 kJ

Answer 2

Sulphur burns in oxygen according to the following equation $\underset{32 g}{ S }+ O _{2} \longrightarrow SO _{2}+\Delta H_{f}$ $0.5\, g$ S liberate $=4.6 \,kJ$ heat $1\, g \,S$ liberate $=\frac{4.6}{0.5} kJ$ $32\, g \,S$ will liberate $=\frac{4.6}{0.5} \times 32$ $=294.4 \,kJ$ Heat of formation is the amount of heat liberated or absorbed when one mole of compound is formed from its constituent elements hence, Heat of formation of $SO _{2}=-294.4\, kJ$