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Thermodynamics question

2019 · 10 Apr · Shift 1 · Q11
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Thermodynamics question

2019 · 10 Apr · Shift 1 · Q11

JEE MainChemistryThermodynamicsMCQ+4 / −1
A process will be spontaneous at all temperatures if :
  1. A
    Δ\DeltaΔ H < 0 and Δ\DeltaΔ S > 0
  2. B
    Δ\DeltaΔ H < 0 and Δ\DeltaΔ S < 0
  3. C
    Δ\DeltaΔ H > 0 and Δ\DeltaΔ S < 0
  4. D
    Δ\DeltaΔ H > 0 and Δ\DeltaΔ S > 0
View written solutionFree

Correct answer: A

  1. For spontaneity, the Gibbs free energy change must be negative:

ΔG=ΔH−TΔS\Delta G = \Delta H - T\Delta SΔG=ΔH−TΔS

A process is spontaneous when:

ΔG<0\Delta G < 0ΔG<0

  1. The question asks: spontaneous at all temperatures. So ΔG\Delta GΔG must remain negative for every value of TTT.

  2. Analyze the sign combinations:

  • If ΔH<0\Delta H < 0ΔH<0 and ΔS>0\Delta S > 0ΔS>0: ΔG=ΔH−TΔS\Delta G = \Delta H - T\Delta SΔG=ΔH−TΔS Here, ΔH\Delta HΔH is negative and −TΔS-T\Delta S−TΔS is also negative. So, ΔG<0 for all T\Delta G < 0 \text{ for all } TΔG<0 for all T

  • If ΔH<0\Delta H < 0ΔH<0 and ΔS<0\Delta S < 0ΔS<0: Then −TΔS-T\Delta S−TΔS becomes positive, so spontaneity only at low temperature.

  • If ΔH>0\Delta H > 0ΔH>0 and ΔS>0\Delta S > 0ΔS>0: Spontaneity only at high temperature.

  • If ΔH>0\Delta H > 0ΔH>0 and ΔS<0\Delta S < 0ΔS<0: ΔG>0\Delta G > 0ΔG>0 for all temperatures, so never spontaneous.

  1. Therefore, the required condition is:

ΔH<0 and ΔS>0\boxed{\Delta H < 0 \text{ and } \Delta S > 0}ΔH<0 and ΔS>0​

  1. Compare with options:

Option A is intended to represent:

ΔH<0, ΔS>0\Delta H < 0,\ \Delta S > 0ΔH<0, ΔS>0

Hence, Option A is correct.

  1. Comparison with stored answer:

Stored correct answer = A, which matches the derived answer.

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