Answer:
(i) Yes, the given statement is true. This is because \[\Delta G=\Delta H-T\Delta S\]. For exothermic reaction, \[\Delta H\] is -ve. If \[T\Delta S\] is +ve (i.e., entropy factor opposes the process) and\[T\Delta S>\Delta H\] in magnitude, \[\Delta G\] will be +ve and process will not be spontaneous.
(ii) Yes, the given statement is true. This is because at the same temperature, gaseous state is more disordered than the liquid state.
(iii)\[-\Delta {{G}^{o}}\,=RT\,\ln \,K\]. Thus, if \[\Delta {{G}^{o}}\] is less than zero, i.e., it is -ve, then \[\,\ln \,K\]will be +ve and hence K will be greater than 1.
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