Re: If a reaction needs a match to happen, is it spontaneous or not?

The basis for the answers to your specific questions is rather lengthy. The answers themselves are at the end of this answer. I don’t know any short way to really answer this so here goes.

Any mixture of chemicals will react spontaneously. The tendency to produce one species at the expense of another (the direction of the reaction) depends on their concentrations relative to those at equilibrium. Please note: From a molecular point of view the reaction doesn’t stop running at equilibrium. It just runs as fast one way as it does the other. All chemical equilibria are dynamic not static. Using a match (raising the temperature) to “start” a reaction can involve the interplay of several things, so first the players and then the interplay.

The players: 1. The sign of the change in Gibbs Free Energy (delta G) determines the net direction a mix of chemicals at some concentrations and conditions (temperature, pressure, etc.) will run to try and reach equilibrium. Negative indicates it will run in the direction written and positive the reverse. It is zero at equilibrium. This free energy change does not tell us how fast the reaction will run. It also, by itself, does not tell us whether the reaction is exothermic or endothermic. 2. The change in Enthalpy (Delta H does that. 3. The activation energy can be thought of as a speed bump along the reaction direction. 4. Entropy, which has to do with how randomly the energy available is distributed among the various places the molecules can store it. 5. The absolute temperature, which is proportional to the random kinetic energy present.

The interplay: Gibbs Free Energy depends on Enthalpy, Temperature, and Entropy. Delta G = Delta H - T(Delta S) Please note this has nothing to do with activation energy. This is because the change terms in this equation are net changes from beginning to end for the reaction involved. This means that activation energy has nothing to do with whether the reaction is spontaneous or not for the direction written. Raising the temperature however can cause a variety of things to change.

1. Even if changing T does not change Delta H significantly (though it sometimes can, especially for higher temperatures) it will cause the T(Delta S) term to change. If the change in entropy (Delta S) is positive (this is the usual case, entropy increases) then the whole term becomes even more negative. If the reaction is already exothermic (Delta H is negative) then it is spontaneous at any temperature. If the reaction is endothermic (Delta H is positive) then there will be some temperature above which the reaction becomes spontaneous in the direction written. Below this temperature it runs spontaneously the other way.

2. Raising the temperature increases the number of molecules present whose kinetic energy exceeds any activation energy that may exist and so increases the rate at which the reaction reaches equilibrium.

3. If the reaction releases energy as it runs and this energy is not removed quickly enough it can begin to raise its own temperature which in turn raises the reaction rate and so on in a sort of thermal runaway.

4. It is also possible that raising the temperature enough can cause a cascade of reactions. In other words at some temperature one or more of the components of the mixture react to produce some chemical that wasn’t there before but which once produced is extremely reactive with the rest of the mixture. This situation gives the appearance of a sort of spontaneity requiring an energy input but is really two or more sequential reactions.

References: You can find all of this in principle in any college level Physical Chemistry text but…. It may be spread over several chapters and far from clear. I hope this helps tie it together a bit for you. Thanks for a really good question.

(Sorry about all the "Delta" stuff, I couldn't get the Greek symbol to display.)