question_answer

At a certain temperature 'T' the endothermic reaction \[A\xrightarrow{\,}B\] proceeds virtually to the end. Determine (i) sign of \[\Delta S\] for this reaction (ii) sign of \[\Delta G\] for the reaction \[B\xrightarrow{\,}A\] at the temperature T, and (iii) the possibility of reaction \[B\xrightarrow{\,}A\] proceeding at a low temperature.

Answer:

                (i) Energy factor opposes. So enthalpy factor must favour, i.e., \[\Delta S\] must be positive. (ii) For \[A\xrightarrow{\,}B,\,\Delta G\]is -ve, therefore, for \[B\xrightarrow{\,}A,\,\Delta G\] will be +ve. (iii) For \[B\xrightarrow{\,}\,A,\,\Delta H=-ve\] and \[\Delta S\] is also -ve, i.e., \[\Delta S\] opposes the process but at low temp., \[T\Delta S\] may be so low that \[\Delta H\]  is greater in magnitude than \[T\Delta S\] and the process will be spontaneous.