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Thermodynamics question

2014 · Q98
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Thermodynamics question

2014 · Q98

NEETChemistryThermodynamicsMCQ+4 / −1
For the reaction, X2O4(l)  →  2XO2(g){X_2}{O_{4\left( l \right)}}\,\, \to \,\,2X{O_{2(g)}}X2​O4(l)​→2XO2(g)​

Δ\DeltaΔU = 2.1 kcal, Δ\DeltaΔS = 20 cal K−-−1 at 300 K

Hence, G is
  1. A
    2.7 kcal
  2. B
    −-− 2.7 kcal
  3. C
    9.3 kcal
  4. D
    −-− 9.3 kcal
View written solutionFree

Correct answer: B

ΔH=ΔU+ΔngRT\Delta H = \Delta U + \Delta {n_g}RTΔH=ΔU+Δng​RT

Given ΔU=2.1 kcal, Δng=2\Delta U = 2.1\,kcal,\,\Delta {n_g} = 2ΔU=2.1kcal,Δng​=2

R=2×10−3kcal,T=300KR = 2 \times {10^{ - 3}}kcal,T = 300KR=2×10−3kcal,T=300K

∴\therefore∴ ΔH=2.1+2×2×10−3×300=3.3 kcal\Delta H = 2.1 + 2 \times 2 \times {10^{ - 3}} \times 300 = 3.3\,kcalΔH=2.1+2×2×10−3×300=3.3kcal

Again, ΔG=ΔH+TΔS\Delta G = \Delta H + T\Delta SΔG=ΔH+TΔS

Given Δ\Delta ΔS = 20 ×\times× 10-3 kcal K-1

On putting the values of Δ\Delta ΔH and Δ\Delta ΔS in the equation, we get

ΔG=3.3−300×20×10−3\Delta G = 3.3 - 300 \times 20 \times {10^{ - 3}}ΔG=3.3−300×20×10−3

⇒3.3−6×103×10−3=−2.7 kcal \Rightarrow 3.3 - 6 \times {10^3} \times {10^{ - 3}} = - 2.7\,kcal⇒3.3−6×103×10−3=−2.7kcal

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