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BITSAT 2022
Exam
BITSAT
Year
2022
Subject
Chemistry
Chapter
Thermodynamics (C)

BITSAT 2022 · Chemistry · Thermodynamics (C)

Standard entropy of X_2, Y_2 and X Y_2 are 60,40 and 50 JK^ -1 mol^...

Standard entropy of X2,Y2X_2, Y_2 and XY2X Y_2 are 60,40 and 50JK−1mol−150 \mathrm{JK}^{-1} \mathrm{ mol}^{-1}, respectively. For the reaction, 12X2+32Y2⟶XY3,ΔH=−30kJ\frac{1}{2} X_2+\frac{3}{2} Y_2 \longrightarrow X Y_3, \Delta H=-30 \mathrm{ kJ}, to be at equilibrium, the temperature will be

Options

  1. A1250K1250 \mathrm{ K}
  2. B500K500 \mathrm{ K}
  3. C750K750 \mathrm{ K}
  4. D1000K1000 \mathrm{ K}

Correct answer

C. 750K750 \mathrm{ K}

Step-by-step solution

ΔS=S(XY3)−12S(X2)−32S(Y2)=50−30−60=−40Jmol−1K−1ΔH=−30kJ=−3000JΔG=ΔH−TΔS\Delta S=S\left(X Y_3\right)-\frac{1}{2} S\left(X_2\right)-\frac{3}{2} S\left(Y_2\right)=50-30-60=-40 \mathrm{ J} \mathrm{ mol}^{-1} \mathrm{ K}^{-1}\Delta H=-30 \mathrm{ kJ}=-3000 \mathrm{ J}\Delta G=\Delta H-T \Delta S At equilibrium, ΔG=0T=ΔHΔS=−3000−40=750K\Delta G=0T=\frac{\Delta H}{\Delta S}=\frac{-3000}{-40}=750 \mathrm{ K}
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