Chemistry · Thermodynamics & Thermochemistry

JEE Main 2025 — 24 January, Evening Shift — Question 37

S(g)+32O2( g)→SO3( g)+2xkcal\mathrm{S}(\mathrm{g})+\frac{3}{2} \mathrm{O}_{2}(\mathrm{~g}) \rightarrow \mathrm{SO}_{3}(\mathrm{~g})+2 \mathrm{xkcal}

SO2( g)+12O2( g)→SO3( g)+ykcal\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 SO2( g)\mathrm{SO}_{2}(\mathrm{~g}) is given by :

  1. Option A:

    2xykcal\frac{2 x}{y} \mathrm{kcal}

  2. Option B:

    y−2xy-2 x kcal

    Correct
  3. Option C:

    2x+y2 x+y kcal

  4. Option D:

    x+yx+y kcal

Answer: B

Step-by-step solution

SO2(g)+12O2( g)⟶SO3( g)ΔH=−y\underset{(\mathrm{g})}{\mathrm{SO}_{2}}+\frac{1}{2} \underset{(\mathrm{~g})}{\mathrm{O}_{2}} \longrightarrow \underset{(\mathrm{~g})}{\mathrm{SO}_{3}} \quad \Delta \mathrm{H}=-\mathrm{y}

ΔHr=(ΔHf)SO3−(ΔHf)SO2\Delta \mathrm{H}_{\mathrm{r}}=\left(\Delta \mathrm{H}_{\mathrm{f}}\right)_{\mathrm{SO}_{3}}-\left(\Delta \mathrm{H}_{\mathrm{f}}\right)_{\mathrm{SO}_{2}}

−y=−2x−(ΔHf)SO2-\mathrm{y}=-2 \mathrm{x}-\left(\Delta \mathrm{H}_{\mathrm{f}}\right)_{\mathrm{SO}_{2}}

(ΔHf)SO2=y−2x\left(\Delta \mathrm{H}_{\mathrm{f}}\right)_{\mathrm{SO}_{2}}=\mathrm{y}-2 \mathrm{x}

Option (2)

Answer key and solution verified before publishing.

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Exam
JEE Main 2025
Subject
Chemistry
Chapter
Thermodynamics & Thermochemistry
Topic
Thermochemistry and Enthalpy Changes
S ( g )+3/2 O 2 ( g ) rightarrow SO 3 ( g )+2 xkcal SO 2 ( g )+1/2 O… | JEE Main 2025 PYQ with Solution · DhiX AI