If, \(\Delta H>0\) and \(\Delta S>0\), the reaction can proceed spontaneously at
If, \(\Delta H>0\) and \(\Delta S>0\), the reaction can proceed spontaneously at
low temperature
high temperature
all temperature
will never be spontaneous
Solution
When both \(\Delta H\) and \(\Delta S\) are positive then the process will be an endothermic which involves an increase in system entropy. In this case, \(\Delta G\) will be negative, if the magnitude of the \(T \Delta S\) term is greater than \(\Delta H\). If the \(T \Delta S\) term is less than \(\Delta H\), the free energy change will be positive. Such a process is spontaneous at high temperatures and non-spontaneous at low temperatures. Hence, the correct option is (b).