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?

  • 1
    $\begingroup$ 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. $\endgroup$
    – Mithoron
    Commented Jan 8 at 0:14
  • $\begingroup$ See also en.wikipedia.org/wiki/Van_%27t_Hoff_equation $\endgroup$
    – Poutnik
    Commented Jan 8 at 4:18


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.