# Why enthalpy change of reaction does not affect the rate of reaction?

For example, an exothermic reaction will release energy to provide energy to overcome the activation energy. So, why the enthalpy change doesnt affect the rate of reaction?

• If the reaction heats up the environment the rate will change. – DSVA Aug 14 '17 at 16:53
• Your usage of the word "affect" is a misnomer. You can't take a reaction, change its enthalpy and see what would happen to the rate. – Ivan Neretin Aug 14 '17 at 22:41
• @Ivan Neretin I believe they mean that the heat produced by a reaction (the enthalpy change) would affect the surrounding temperature, which would affect the reaction rate. – Tyberius Aug 15 '17 at 3:02
• The definition of the enthalpy change (heat of reaction) is based on the temperature being the same in the initial and final states thermodynamic equilibrium states. – Chet Miller Aug 15 '17 at 12:19

The rate of a chemical reaction is directly dependant to the activation energy of that particular reaction. These two values are inversely proportional. The higher the activation energy, the lower the rate. As it has been mentioned though, the rate can be increased, in such cases, by increasing the temperature. All of the above is described as kinetics, which shows how fast the reaction will reach its end phase.

Enthalpy is linked with thermodynamics, which shows what kind of entities will be present at the end of the reaction. At what degree the reactants will turn to products.

I don't know your chemical background, so I tried to make this as simple as possible.

• " the rate can be increased, in such cases, by increasing the temperature. " Not necessarily since the energy of activation is also temperature dependent. – DSVA Aug 14 '17 at 22:22
• Isn't activation energy the same at all times? For a reaction to happen you need a certain amount of energy to overcome the transition phase. I thought that this amount is fixed for a given reaction and by changing the temperature you change the amount of molecules that have this appropriate energy and thus increasing the rate of it. – Αντώνιος Κελεσίδης Aug 14 '17 at 22:30
• The gibbs free energy of activation, the one that counts if we are talking about transition state theory, is temperature dependend. $\Delta G^{\ddagger}= \Delta H^{\ddagger}-T \Delta S^{\ddagger}$. For reactions with negative $\Delta S^{\ddagger}$ the free energy of activation rises with temperature, however as you said also the average energy of the molecule rises. In most cases this leads to faster reactions. – DSVA Aug 14 '17 at 22:36
• I agree with what you've mentioned but I think that activation energy is fixed. It's like a energy barrier for the reactants to overcome in order to become products. I've just made a small google search and they've mentioned the same thing. – Αντώνιος Κελεσίδης Aug 14 '17 at 22:42
• Ok first we need to be clear if we are talking about Arrhenius (which I don't like since it's an empirical relationship) or transition state theory. However, in both cases the activation energy (Arrhenius) or free energy of activation(transition state theory) is temperature dependend. I've explained the transition state theory case, here's an answer explaining the temperature dependency of the activation energy and why we usually get away with taking it as a constant: chemistry.stackexchange.com/a/75680/37313 – DSVA Aug 14 '17 at 22:52

The enthalpy change of a reaction is just the difference in the bond ( potential energy) stored within the products compared with that within reactants. However, the activation energy is simply the amount of energy you need to break the bonds and start the reaction. To clarify the specific case you suggested: the only reason that an exothermic reaction would proceed faster than a hypothetical endothermic reaction of the same activation energy would be that the exothermic reaction is releasing energy that raises the temperature. However, if you could keep the temperature constant that would not happen and they would proceed at the same rate.

• Your characterization of the activation energy is not quite correct. For an S$_N$2 reaction bonds form and break simultaneously, but the transition state is reached before any bond is "broken". Similarly for a Diels-Alder reaction. – Deathbreath Mar 2 '18 at 15:44