Verified Solution • 2025–26 Session moderate

QJEEM20Ch8545: JEE Main 2020 Chemistry - Class 11 JEEM20Ch8545 Solution: For the reaction : A( ) 2B(g) at 300 K Hence G in kcal is :

For the reaction :

A( ll ) →\to 2B(g)

ΔU=2.1 kcal, ΔS=20 cal K−1\Delta U = 2.1\,kcal,\,\Delta S = 20\,cal\,{K^{ - 1}} at 300 K

Hence Δ\Delta G in kcal is :

Step-by-Step Verified Solution

A( ll ) →\to 2B(g)

We know, Δ\Delta H = Δ\Delta U + Δ\Delta ngRT

and Δ\Delta G = Δ\Delta H - T Δ\Delta S

∴\therefore Δ\Delta G = Δ\Delta U + Δ\Delta ngRT - T Δ\Delta S

= 2.1 + 2×2×3001000{{2 \times 2 \times 300} \over {1000}} - 300×201000{{300 \times 20} \over {1000}}

= 2.1 + 1.2 - 6 = – 2.70 Kcal/mol

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