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How does free energy change explain the criterion of spontaneity? |
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Answer» Solution :The total entropy CHANGE for a system and its surroundings ACCOMPANYING a process is given by, `DeltaS_("total") = DeltaS_("system") +DeltaS_("surr")` By second law, for a spontaneous process, `DeltaS_("total") gt 0`. If `+DeltaH` is the enthalpy change (or enthalpy increase) for the process, or a reaction at CONSTANT temperature (T) and pressure, then enthalpy change (or enthalpy decrease) for the surroundings will be `-DeltaH` . `thereforeDeltaS_(Surr) = (-DeltaH)/(T)` `becauseDeltaS_("total")= DeltaS_("system") + DeltaS_(surr)` `thereforeDeltaS_("total") =DeltaS_("system") -(DeltaT)/(T)` `thereforeT DeltaS_("total") = T DeltaS_("system") - DeltaH` `therefore - T Delta S_("total") = - T Delta S_("System") -DeltaH` `therefore -T Delta S_("total") = DeltaH - T DeltaS_("system")` By Gibbs equation, `DeltaG = DeltaH -T DeltaS` By comparing above two equations, `therefore DeltaG = -T DeltaS_("total")` As `DeltaS_("total")` increase, `DeltaG` decrease. For a spontaneous process, `DeltaS_("total") gt 0` which is according to second law of thermodynamics. `thereforeDeltaG lt 0` Hence in a spontaneous process, Gibbs free energy decreases `(DeltaG lt 0)` while entrophyincrease `(DeltaS gt 0)` . Therefore for a non-spontaneous process Gibbs free energy INCREASES `(Delta G gt 0)`. It can be concluded that for a process at EQUILIBRIUM,`Delta G = 0` Hence (i) For the spontaneous process, `DeltaG lt0`. (ii) For the non-spontaneous process, `DeltaG gt 0`. (iii) For the process at equilibrium, `DeltaG =0`. |
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