JEE Main 2021 · 24 February · Evening Shift · Chemistry · Thermodynamics (C)
3HC≡CH(g)⇌C6H6(l)3HC ≡ {CH}_{( g )} ⇌ {C}_{6}{H}_{6 ( l )}3HC≡CH(g)⇌C6H6(l)
[Given: ΔfG°(HC≡CH)=−2.04×105Jmol−1;ΔfG°(C6H6)=−1.24×105Jmol−1;R=8.314JK−1mol−1{\Delta}_{f}G ° \left(HC ≡ CH\right) = - 2 . 04 \times {10}^{5}J{mol}^{- 1}; {\Delta}_{f}G ° \left({C}_{6}{H}_{6}\right) = - 1 . 24 \times {10}^{5}J{mol}^{- 1}; R = 8 . 314J{K}^{- 1}{mol}^{- 1}ΔfG°(HC≡CH)=−2.04×105Jmol−1;ΔfG°(C6H6)=−1.24×105Jmol−1;R=8.314JK−1mol−1]
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3HC≡CH(g)→C6H6(ℓ):ΔG0=−RTlnK3HC ≡ {CH}_{\left(g\right)} \to {C}_{6}{H}_{6}\left(ℓ\right): {\Delta G}^{0} = - RTlnK3HC≡CH(g)→C6H6(ℓ):ΔG0=−RTlnK
ΔGf0−2.04×105Jmol−1.24×105J/mol{\Delta G}_{f}^{0} - 2 . 04 \times {10}^{5}\dfrac{J}{mol} - 1 . 24 \times {10}^{5}J / molΔGf0−2.04×105molJ−1.24×105J/mol
⇒ΔG0=Σ(ΔGf0)P−Σ(ΔGf0)R⇒ {\Delta G}^{0} = Σ{\left({\Delta G}_{f}^{0}\right)}_{P} - Σ{\left({\Delta G}_{f}^{0}\right)}_{R}⇒ΔG0=Σ(ΔGf0)P−Σ(ΔGf0)R
⇒−RTℓnK=1×(−1.24×105)−(−3×2.04×105)⇒ - RTℓnK = 1 \times \left(- 1 . 24 \times {10}^{5}\right) - \left(- 3 \times 2 . 04 \times {10}^{5}\right)⇒−RTℓnK=1×(−1.24×105)−(−3×2.04×105)
⇒−2.303×R×TlogK=4.88×105⇒ - 2 . 303 \times R \times TlogK = 4 . 88 \times {10}^{5}⇒−2.303×R×TlogK=4.88×105
⇒logK=−4.88×1052.303×R×T=−4880005705.848=−85.52⇒ logK = - \dfrac{4 . 88 \times {10}^{5}}{2 . 303 \times R \times T} = - \dfrac{488000}{5705 . 848} = - 85 . 52⇒logK=−2.303×R×T4.88×105=−5705.848488000=−85.52
=855×10−1= 855 \times {10}^{- 1}=855×10−1
⇒x=855⇒ x = 855⇒x=855
