# Temperature dependence on the feasibility of an equilibrium

If we consider a reaction A + B ⟶ C + D, with $$\Delta$$S $$>0$$, then regardless of whether $$\Delta$$H is positive or negative, increasing the temperature will make $$\Delta$$H - T$$\Delta$$S more negative, and the reaction will be more feasible

If we now consider a reaction A + B ⇌ C + D, still with $$\Delta$$S $$>0$$ for the forward reaction, if we express $$ln(K)= -\frac{\Delta H}{RT}+\frac{\Delta S}{R}$$, now increasing temperature only makes $$ln(K)$$ larger if $$\Delta H >0$$ and is independent of $$\Delta S$$

Why has the dependence of the feasibility and equilibrium constant of the reaction switched from entropy to ethalpy. Am I confusing a relationship between the feasibility of a reaction itself and the equilibrium constant, i.e. a higher equilibrium constant does not mean the same thing as the forward reaction being more feasible?

• That's not true. Dividing by T doesn't really change anything. If T makes S component larger, or H one smaller, doesn't matter. Commented Jan 8 at 0:14
• Commented Jan 8 at 4:18