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प्रश्न
Answer the following question.
Obtain the relation between ΔG and `triangle "S"_"total"`. Comment on the spontaneity of the reaction.
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उत्तर
1) The total entropy change that accompanies a process is given by,
`triangle "S"_"total" = triangle "S"_"sys" + triangle "S"_"surr"`
OR
`triangle "S"_"total" = triangle "S" + triangle "S"_"surr"` ....(1)
2) According to second law of thermodynamics for a process to be spontaneous, `triangle "S"_"total"` > 0.
3) If ∆H is the enthalpy change accompanying a reaction (system), the enthalpy change of the surroundings is –∆H. Thus,
`triangle "S"_"surr" = - (triangle "H")/"T"` ....(2)
4) Substituting equation (2) in equation (1), we get,
`triangle "S"_"total" = triangle "S" - (triangle "H")/"T"`
Rearranging above expression, we get,
`"T" triangle "S"_"total" = "T" triangle "S" - triangle "H"`
or
`- "T" triangle "S"_"total" = triangle "H" - "T" triangle "S"` .....(3)
5) The change in Gibbs energy at constant temperature and constant pressure is given by,
∆G = ∆H - T ∆S …(4)
6) Substituting equation (3) in equation (4), we get,
∆G = `- "T" triangle "S"_"total"`
7) For a spontaneous reaction, `"S"_"total"` > 0 and hence, ∆G < 0. At constant temperature and pressure Gibbs energy of the system decreases in a spontaneous process.
8) The second law leads to the conditions of spontaneity as follows:
i) `triangle "S"_"total"` > 0 and ∆G < 0, the process is spontaneous.
ii) `triangle "S"_"total"` < 0 and ∆G > 0, the process is nonspontaneous.
iii) `triangle "S"_"total"` = 0 and ∆G = 0, the process is at equilibrium.
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