4.6 kJ heat is liberated on burning 0.5 g of sulphur. The enthalpy of formation of SO2 is: [molecular weight of S = 32, O = 16]

Question

4.6 kJ heat is liberated on burning 0.5 g of sulphur. The enthalpy of formation of SO2 is: [molecular weight of S = 32, O = 16] (A) + 294.4 kJ (B) - 294.4 kJ (C) + 462.4 kJ (D) - 462.4 kJ

Tardigrade Admin · Accepted Answer

Sulphur burns in oxygen according to the following equation

32 g S + O_{2} ⟶ S O_{2} + Δ H_{f}

0.5 g

S liberate

= 4.6 k J

heat

1 g S

liberate

= \frac{4.6}{0.5} k J

32 g S

will liberate

= \frac{4.6}{0.5} \times 32

= 294.4 k J

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

S O_{2} = - 294.4 k J

Q. 4.6kJ heat is liberated on burning 0.5g of sulphur. The enthalpy of formation of SO2​ is: [molecular weight of S=32,O=16]

+ 294.4 kJ

- 294.4 kJ

+ 462.4 kJ

- 462.4 kJ

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