BotanyFor a particular process when ΔH=60kJ/mole and ΔS=200JKmole−1 then the process is non-spontaneous at

For a particular process when ΔH=60kJ/mole and ΔS=200JKmole1 then the process is non-spontaneous at

  1. A

    350K

  2. B

    400K

  3. C

    200K

  4. D

    800K

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    Solution:

    At equilibrium, ΔG=0

    ΔHTΔS=0

    ΔH=TΔS

    Teq=60×103200 =300 K

    ΔG=+veT+ve

    Process is spontaneous when ‘T’ is greater than Teq (300K)

    Process is non-spontaneous when ‘T’ is lesser than Teq (300K)

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