question_answer 1)

  \[\Delta G\]is net energy available to do useful work and is thus a measure of "free energy". Show mathematically that \[\Delta G\] isa measure of free energy. Find the unit of\[\Delta G\]. If a reaction has positive enthalpy change and positive entropy change, under what condition will the reaction be spontaneous?

Answer:

  According to Gibbs Helmholtz equation: \[\Delta G=\Delta H-T\Delta S\] \[\Delta G=\Delta U+P\Delta V-T\Delta S\,\,\,\,\,\,\,\,\,\,.....(i)\] From second law of thermodynamics: \[\Delta S=\frac{q}{T}\] (Reversible process)                    ... (ii) From first law of thermodynamics:                                 \[\Delta U=q-{{w}_{rev}}\] \[{{w}_{rev}}\] = Reversible work done by the system From (ii) and (iii) \[\Delta S=\frac{\Delta U+{{w}_{rev}}}{T}\] or            \[T\Delta S=\Delta U+{{w}_{rev}}\] Putting the value of \[T\Delta S\] in (i) we get \[\Delta G=\Delta U+P\Delta V-[\Delta U+{{w}_{rev}}]\] \[=P\Delta V-{{w}_{rev}}\] \[\mathbf{-\Delta G=}{{\mathbf{w}}_{\mathbf{rev}}}\mathbf{-P\Delta V}\] Thus, decrease in Gibbs free energy is measure of reversible work done by the system exclusive of work of volume expansion. From (i). If\[\Delta H=+ve\], and \[\Delta S=+ve,\] then process will be spontaneous when \[T\Delta S>\Delta H\] ; in this case \[\Delta G\] will be negative.