$\mathrm{S}(\mathrm{g})+\frac{3}{2} \mathrm{O}_{2}(\mathrm{~g}) \rightarrow \mathrm{SO}_{3}(\mathrm{~g})+2 \mathrm{x} \mathrm{kcal}$ $\mathrm{SO}_{2}(\mathrm{~g})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{~g}) \rightarrow \mathrm{SO}_{3}(\mathrm{~g})+\mathrm{y} \mathrm{kcal}$ The heat of formation of $\mathrm{SO}_{2}(\mathrm{~g})$ is given by :

(A) $\frac{2 x}{y} \mathrm{kcal}$

JEE Main · 2025 · 24 Jan 2025 · Shift 2 · Chemical Thermodynamics

S (g) + \frac{3}{2} O_{2} (g) \to SO_{3} (g) + 2 x kcal

SO_{2} (g) + \frac{1}{2} O_{2} (g) \to SO_{3} (g) + y kcal

The heat of formation of

SO_{2} (g)

is given by :

Answer: (B)

By Hess's law, the enthalpy of formation of

S O_{2} (g)

is obtained by combining the two given equations:

S (g) + \frac{3}{2} O_{2} (g) \to S O_{3} (g), Δ H_{1} = - 2 x kcal

S O_{2} (g) + \frac{1}{2} O_{2} (g) \to S O_{3} (g), Δ H_{2} = - y kcal

Reversing the second equation and adding it to the first cancels

S O_{3}

and

\frac{1}{2} O_{2}

, leaving:

S (g) + O_{2} (g) \to S O_{2} (g), Δ H_{f} = Δ H_{1} - Δ H_{2} = - 2 x - (- y) = y - 2 x kcal

The heat of formation of

S O_{2} (g)

y - 2 x kcal

JEE Main · 2025 · 24 Jan 2025 · Shift 2 · Single correct option