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JEE Main 202621 JanuaryMorning Shift
Exam
JEE Main
Year
2026
Date
21 January
Shift
Morning Shift
Subject
Chemistry
Chapter
Thermodynamics (C)

JEE Main 2026 · 21 January · Morning Shift · Chemistry · Thermodynamics (C)

Use the following data : ( array |c|c|c| Substance & Δ_f H^ (500...

Use the following data :
(Substance ΔfH⊖(500 K)kJmol−1 S⊖(500 K)JK−1 mol−1AB( g)32222 A2(g)6146 B2(g)x280\begin{array}{|c|c|c|} \hline \text{Substance } & \frac{\Delta_f \mathrm{H}^{\ominus}(500 \mathrm{~K})}{\mathrm{kJmol}^{-1}} & \frac{\mathrm{~S}^{\ominus}(500 \mathrm{~K})}{\mathrm{JK}^{-1} \mathrm{~mol}^{-1}} \\ \hline \mathrm{AB}(\mathrm{~g}) & 32 & 222 \\ \hline \mathrm{~A}_2(\mathrm{g}) & 6 & 146 \\ \hline \mathrm{~B}_2(\mathrm{g}) & x & 280 \\ \hline \end{array})
One mole each of A2( g)\mathrm{A}_{2}(\mathrm{~g}) and B2( g)\mathrm{B}_{2}(\mathrm{~g}) are taken in a 1 L closed flask and allowed to establish the equilibrium at 500 K. A2( g)+B2( g)⇌2AB( g)\mathrm{A}_{2}(\mathrm{~g})+\mathrm{B}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{AB}(\mathrm{~g}) The value of x(inkJmol−1)x\left(\mathrm{in} \mathrm{kJ} \mathrm{mol}^{-1}\right) is ____\_\_\_\_. (Nearest integer)
(Given : log⁡K=2.2R=8.3 J K−1 mol−1\log \mathrm{K}=2.2 \quad \mathrm{R}=8.3 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1})

Correct answer

0

Step-by-step solution

Reaction: A2(g)+B2(g)⇌2AB(g)A_2(g) + B_2(g) \rightleftharpoons 2AB(g)
ΔG°=−2.303RTlog⁡K=−2.303×8.3×500×2.2=−21026\Delta G° = -2.303RT \log K = -2.303 \times 8.3 \times 500 \times 2.2 = -21026 J/mol = −21.03-21.03 kJ/mol
ΔS°=2(222)−[146+280]=444−426=18\Delta S° = 2(222) - [146 + 280] = 444 - 426 = 18 J K⁻¹ mol⁻¹
ΔH°=2(32)−[6+x]=(58−x)\Delta H° = 2(32) - [6 + x] = (58 - x) kJ/mol
Using ΔG°=ΔH°−TΔS°\Delta G° = \Delta H° - T\Delta S°:
−21.03=(58−x)−(500×0.018)-21.03 = (58 - x) - (500 \times 0.018)
−21.03=58−x−9=49−x-21.03 = 58 - x - 9 = 49 - x
x=49+21.03=70.03≈70x = 49 + 21.03 = 70.03 \approx 70 kJ/mol
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