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NEET CHEMISTRYThermodynamicsMedium

Question

Standard entropies of X2X_2, Y2Y_2 and XY3XY_3 are 6060, 4040 and 50 J K1mol150 \text{ J K}^{-1} \text{mol}^{-1} respectively. For the reaction 12X2+32Y2XY3\frac{1}{2}X_2 + \frac{3}{2}Y_2 \rightleftharpoons XY_3; ΔH=30 kJ\Delta H = -30 \text{ kJ} to be at equilibrium, the temperature should be:

A

750 K750 \text{ K}

B

1000 K1000 \text{ K}

C

1250 K1250 \text{ K}

D

500 K500 \text{ K}

Step-by-Step Solution

For the reaction: 12X2+32Y2XY3\frac{1}{2}X_2 + \frac{3}{2}Y_2 \rightleftharpoons XY_3 The entropy change of the reaction (ΔS\Delta S) is: ΔS=S(XY3)[12S(X2)+32S(Y2)]\Delta S = S(XY_3) - \left[ \frac{1}{2}S(X_2) + \frac{3}{2}S(Y_2) \right] ΔS=50[12(60)+32(40)]\Delta S = 50 - \left[ \frac{1}{2}(60) + \frac{3}{2}(40) \right] ΔS=50[30+60]=5090=40 J K1mol1\Delta S = 50 - [30 + 60] = 50 - 90 = -40 \text{ J K}^{-1} \text{mol}^{-1} At equilibrium, the Gibbs free energy change ΔG=0\Delta G = 0. Since ΔG=ΔHTΔS\Delta G = \Delta H - T\Delta S, we have: ΔH=TΔS    T=ΔHΔS\Delta H = T\Delta S \implies T = \frac{\Delta H}{\Delta S} Given ΔH=30 kJ=30000 J\Delta H = -30 \text{ kJ} = -30000 \text{ J} T=30000 J40 J K1=750 KT = \frac{-30000 \text{ J}}{-40 \text{ J K}^{-1}} = 750 \text{ K}

Exam Context & Concepts Covered

This question aligns with the NEET CHEMISTRY syllabus, specifically targeting concepts from Thermodynamics. Mastering this topic is crucial for scoring well in the upcoming medical entrance examinations. Solving conceptually related problems will help you understand the nuances of these concepts and improve your problem-solving speed.

CHEMISTRYThermodynamicsstandardentropiestextmolrespectivelyreaction

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