# can a thermodynamically unfavorable reaction be made favorable by a catalyst?

The question i am asking about is pertaining simply to chemical use of a catalyst. It drives a reaction along a different path without taking part chemically. I know this fact. Is it possible so like this?

• No. That would be against the energy conservation law. – Ivan Neretin Feb 6 '18 at 15:27
• Catalysts don't change thermodynamics. – Zhe Feb 6 '18 at 16:02
• The catalyst increases the rates of both the forward and reverse reactions equally, so the equilibrium point does not change. – Chet Miller Feb 6 '18 at 21:13
• Also a catalyst most definitely takes part in the reaction chemically. – bon Feb 18 '18 at 9:57

Chemically unfavourable means non-spontaneous which means Gibbs' free energy change ($\Delta G$) is positive, i.e. $\Delta G > 0$.
But we know that $\Delta G = \Delta H - T \Delta S$. Now $\Delta H$ and $\Delta S$ (enthalpy and entropy change, respectively) are state functions. So $\Delta G$ is also a state function.
Catalysts can only change the pathway of a reaction (in most of the cases, by lowering the activation energy of the reaction ($E_\pu{a}$)) but after using the catalysts also, the energy states of the reactants and products don't alter. So $\Delta H$ and $\Delta S$ remain the same as uncatalysed reaction. So $\Delta G$ also doesn't change at all. So if the reaction is non-spontaneous, it remains like that.
But changing the temperature to a very high value can make $\Delta G < 0$, so the reaction can be made favourable by abruptly increasing temperature.