# Why would a reaction be nonspontaneous at higher temperatures?

Typically we think of a higher temperature speeding up the reaction rate and/or supplying the activation energy of a reaction. So why is it the case that some reactions are only spontaneous at lower temperatures?

Using the gibbs free energy equation $$\Delta G = \Delta H - T \Delta S$$, If I have a reaction where $$\Delta H$$ is negative (exothermic?) and $$\Delta S$$ is negative it makes $$\Delta G$$ positive at higher temperatures which means the reaction is nonspontaneous at higher temperatures. Why would this be the case?

• LeChatelier's principle says a reaction that creates heat get's slower (and ultimately goes backwards) at higher temperature. And btw. it's totally arbitrary which one is the "forward" direction of any reaction.
– Karl
Sep 23, 2022 at 20:49
• @Karl Not slower, the net reaction ultimately is backwards. LeChatelier talks about equilibrium thermodynamics, not kinetics.
– Karsten
Sep 23, 2022 at 21:46
• @Karsten I agree, but Karl isn't completely wrong. If the equilibrium constant shifts, then the forward and backward rate constants do too (as $K = k_f/k_r$). A smaller equilibrium constant does indicate a slower forward reaction (and/or faster backward reaction). Sep 24, 2022 at 0:38
• You are mixing the thermodynamics and kinetics of a reaction, Thermodynamics just predicts if the reaction will happen or not, Kinetics predicts how fast will it happen, If the reaction is feasible then we look at its kinetics Oct 25, 2022 at 5:04

Thermodynamics studies systems at equilibrium so changing temperature moves the equilibrium from one position to another. For a given change we wish to know if the change of state will occur spontaneously or whether some amount of work is needed to make the change occur. In the gas phase this work usually called 'PV' work such as a change in volume. From the second law there must be overall entropy production so that the total change in entropy of the 'system' i.e. the reaction, plus that of the surroundings is always positive but which can be zero if the conditions are reversible. It can be shown using the First and Second Laws that the Gibbs free energy must decrease if the total entropy production is to be positive, i.e. the Gibbs free energy change must be negative for a spontaneous reaction.

When the temperature changes we can use the Van't Hoff Iscohore to predict what happens;

$$\frac{d\ln(K_p)}{dT}=\frac{\Delta H^{\text{o}}}{RT^2}$$

which when integrated (assuming that $$\Delta H^{\text{o}}$$ is independent of temperature over a small temp range which is often a good approximation) produces a gradient of $$-\Delta H^{\text{o}}/R$$ when $$\ln(K)$$ is plotted vs $$1/T$$.

(An aside. Notice that the entropy is not involved, the species involved are the same but in different proportions and there is always entropy of mixing which compensates. You can see this also from $$G=H-TS$$ where $$\displaystyle \left( \frac{\partial G}{\partial T} \right)_p = -S$$ is a constant as temperature changes.)

Thus for an exothermic reaction such as the formation of ammonia from hydrogen and nitrogen since $$\Delta H<0$$ the equilibrium constant $$K_p$$ must decrease as the temperature increases and less ammonia is in the equilibrium mixture as the temperature is increased, as is experimentally verified. The opposite is true for an endothermic reaction, i.e. dissociation of $$\ce{N_2O_4}$$.

If you have measured the forward and reverse rate constants, defining the equilibrium constant as the ratio of rate constants and using the Arrhenius equation produces

$$K_e= \frac{k_f}{k_b}=\frac{k_f^0}{k_b^0}e^{-(E_f-E_b)/RT}$$

where $$k_f, k_b$$ are the rate constants and $$E_f, E_b$$ the activation energies. An exothermic reaction has $$E_b > E_f$$ and the equilibrium constant decreases on increasing the temperature just as predicted by the thermodynamic argument.

Finally the Gibbs energy is related to the equilibrium constant as

$$\displaystyle \Delta G^{\text{o}}=-RT\ln(K_p)$$

so the sign of $$\Delta G^{\text{o}}$$, positive or negative, depends only on whether the equilibrium constant is less than or greater than one.

You expect both the forward and the reverse reaction to proceed faster at higher temperature. If the reverse reaction speeds up by a higher factor, this will affect the equilibrium.

"Spontaneous" is a technical term that does not reflect our day-to-day use of spontaneous. If you substitute the longer "goes forward or would have to go forward, starting from standard state, to attain equilibrium", it might make more sense.

If I start at standard state at a given temperature, either the forward reaction will be faster than the reverse reaction, or vice versa. If I repeat the experiment at higher temperature, I will see an increased forward rate and an increased reverse rate. These increases are typically not of the same magnitude, however.

So the direction that the reaction takes to reach equilibrium, starting from standard conditions, can be different depending on temperature. With respect to the direction of the reaction as written, it can change from spontaneous to non-spontaneous, or from non-spontaneous to spontaneous. If you write the reaction in the other direction, as Karl mentions, you come to the opposite conclusion.

In conclusion, there is always one net direction of the reaction that becomes "more favored", and the other net direction that becomes "less favored" as you increase the temperature and individual rates go up.

• Thank you for your answer. So basically what it means is that its opposite reaction would be favorable at the high temperature instead. Sep 24, 2022 at 17:11
• Yes, that is always the case. If a reaction is non-spontaneous, the reverse reaction is spontaneous. It just says that equilibrium is either reached in the forward or the reverse direction (unless we are already at equilibrium, in which case we don't see any net reaction).
– Karsten
Sep 24, 2022 at 17:38