Q.Predict the feasibility of a reaction when i) both $\Delta H$ and $\Delta S$ positive ii) both $\Delta H$ and $\Delta S$ negative iii) $\Delta H$ decreases but $\Delta S$ increases

Step 1. Spontaneity is governed by $\Delta G=\Delta H-T\Delta S$; a process is spontaneous when $\Delta G<0$.

Step 2. (i) Both $\Delta H$ and $\Delta S$ positive: $\Delta G=(+)-T(+)$ -- at low $T$ the positive $\Delta H$ term dominates ($\Delta G>0$, non-spontaneous), but at sufficiently high $T$ the $-T\Delta S$ term dominates and $\Delta G$ turns negative. So this combination is spontaneous only ABOVE some crossover temperature (matching Table 7.5's 'melting of a solid' row).

Step 3. (ii) Both $\Delta H$ and $\Delta S$ negative: $\Delta G=(-)-T(-)=(-)+T|\Delta S|$ -- at low $T$ the negative $\Delta H$ dominates ($\Delta G<0$, spontaneous), but at high $T$ the $+T|\Delta S|$ term grows and eventually makes $\Delta G>0$. So this is spontaneous only BELOW some crossover temperature (Table 7.5's 'adsorption of gases' row). …

Substitute each $\Delta H,\Delta S$ sign combination into $\Delta G=\Delta H-T\Delta S$ and see how the balance …

• Assuming a positive $\Delta S$ alone guarantees spontaneity without checking the competing $\Delta H$ term -- spontaneity always needs the COMBINED sign of …