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Thermodynamics question

2019 · 10 Jan · Shift 1 · Q23
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Thermodynamics question

2019 · 10 Jan · Shift 1 · Q23

JEE MainChemistryThermodynamicsMCQ+4 / −1
A process has Δ\DeltaΔ H = 200 J mol–1 and Δ\DeltaΔ S = 40 JK–1 mol–1. Out of the values given below, choose the minimum temperature above which the process will be spontaneous :
  1. A
    4 K
  2. B
    20 K
  3. C
    5 K
  4. D
    12 K
View written solutionFree

Correct answer: C

  1. For spontaneity at constant temperature and pressure, the criterion is:

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

A process is spontaneous when:

ΔG<0\Delta G < 0ΔG<0

  1. Given:

ΔH=200 J mol−1\Delta H = 200\, \text{J mol}^{-1}ΔH=200J mol−1 ΔS=40 J K−1mol−1\Delta S = 40\, \text{J K}^{-1}\text{mol}^{-1}ΔS=40J K−1mol−1

So,

ΔG=200−40T\Delta G = 200 - 40TΔG=200−40T

  1. For the minimum temperature above which the process becomes spontaneous, set:

ΔG=0\Delta G = 0ΔG=0

Thus,

200−40T=0200 - 40T = 0200−40T=0

40T=20040T = 20040T=200

T=20040=5 KT = \frac{200}{40} = 5\, \text{K}T=40200​=5K

  1. Therefore, the process will be spontaneous for:

T>5 KT > 5\, \text{K}T>5K

So the threshold temperature is 5 K5\,\text{K}5K, and among the given options, the correct choice is:

Option C: 5 K5\,\text{K}5K.

  1. Checking options briefly:
  • A: 4 K4\,\text{K}4K — below threshold, not correct
  • B: 20 K20\,\text{K}20K — spontaneous above this too, but not the minimum
  • C: 5 K5\,\text{K}5K — correct threshold value
  • D: 12 K12\,\text{K}12K — not minimum
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