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NEET CHEMISTRYThermodynamicsEasy

Question

The values of ΔH\Delta H and ΔS\Delta S for the given reaction are 170 kJ170 \text{ kJ} and 170 J K1170 \text{ J K}^{-1}, respectively. C(graphite)+CO2(g)2CO(g)\text{C(graphite)} + \text{CO}_2\text{(g)} \rightarrow 2\text{CO(g)} This reaction will be spontaneous at:

A

710 K710 \text{ K}

B

910 K910 \text{ K}

C

1110 K1110 \text{ K}

D

510 K510 \text{ K}

Step-by-Step Solution

For a reaction to be spontaneous, the Gibbs free energy change (ΔG\Delta G) must be negative (ΔG<0\Delta G < 0). We know that, ΔG=ΔHTΔS\Delta G = \Delta H - T\Delta S So, ΔHTΔS<0    T>ΔHΔS\Delta H - T\Delta S < 0 \implies T > \frac{\Delta H}{\Delta S} Given: ΔH=170 kJ=170×103 J\Delta H = 170 \text{ kJ} = 170 \times 10^3 \text{ J} ΔS=170 J K1\Delta S = 170 \text{ J K}^{-1} Substituting the values: T>170×103 J170 J K1=1000 KT > \frac{170 \times 10^3 \text{ J}}{170 \text{ J K}^{-1}} = 1000 \text{ K} Thus, the reaction will be spontaneous at any temperature strictly greater than 1000 K1000 \text{ K}. Among the given options, 1110 K1110 \text{ K} is the only temperature greater than 1000 K1000 \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.

CHEMISTRYThermodynamicsvaluesreactionrespectivelytextcgraphitetextcotextg

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