# Affect of temperature on exothermic reactions

1. ### NewtonsFellow

434
Does heat affect the rate of an exothermic reaction?? In my point of view it shouldn't affect it since exothermic reactions are spontaneous so it already works as fast as it can.

2. ### Yanick

401
Thermodynamics has nothing to say about kinetics. You can have a thermodynamically favorable reaction which is kinetically very slow: See diamond/graphite, nucleic acid hydrolysis etc.

Temperature however, can affect the free energy of a reaction to make it less favorable or more favorable. It is Gibbs' Free Energy which dictates reaction spontaneity which has a term encompassing temperature. If you search the forums you will find several of my own posts talking about this.

3. ### NewtonsFellow

434
I really appreciate your help, but I'm a high schooler so I don't really understand what do you mean, I just Need a brief clear answer for my question

4. ### sludger13

83
The primary cause of occurrence of chemical reactions is (very much) more probable reality.

The combinations of instant energy microstates of matter in (and around) chemical reactions characterize exactly the same state of it. When the matter increases it's entropy, that means it can occupy more combinations of energy microstates, so that state is more probable. Maximum entropy means the most probable state of matter, and that is towards what a closed system is always heading into (chemical reactions are heading into highest entropy, nuclear reactions and expanding gases as well...).

The heat (that changes the state of matter characterized by the temperature) changes the number of possible energy microstates, so it can deffinitely affect the chemical reaction.

This link nicely clarifies the whole thing:

Last edited: Mar 23, 2014
5. ### Yanick

401
Okay very briefly and basically we can say that thermodynamics teaches us how to calculate the energies, and other properties, of a system undergoing some process under specific conditions. You have probably come upon sentences such as, a reaction took place under constant pressure conditions and some amount of heat was released, or a gas was compressed adiabatically etc.

One of the key concepts in thermodynamics is that of a state function. It is basically a function which doesn't care about the path that a system took going from state 1 to state 2. This is very useful in some ways but a bit limited in other ways. It is useful because, for example, many of the proofs of thermodynamics have a condition in them which has the word "reversible." The concept of a reversible process is purely imaginary, it relies on an infinitesimal change away from equilibrium (dq or dP etc). This condition cannot actually be realized in real life, only approximated. But because we have this concept of state functions, we don't need to worry too much how the transition from state 1 to state 2 occurred in real life, we only need to know some things about state 1 and state 2 (such as the pressure, temperature, volume, number of moles etc). Then we can imagine this transition occurring in such a way that we can use the relations of thermodynamics to get useful information, such as enthalpy changes and such.

The limitations of this approach however is that we don't really get any information about time from thermodynamics. Aside from entropy being "the arrow of time" we cannot judge how LONG a process will last when transitioning from state 1 to state 2, simply because thermodynamics doesn't really care how long. That kind of question is in the realm of kinetics which aims to describe how much time a system will take to get to equilibrium from some non-equilibrium position.

What I meant about the diamond/graphite example is that these two forms of carbon have a difference in energy. It just so happens that graphite is lower energy than diamond. So thermodynamics predicts that all of the diamond in the universe will become graphite. We don't need to worry about our wives' jewelry becoming worthless because the transition takes a very long time. Same story with nucleic acids, and most macromolecules in biology. Having a long string of DNA or protein is mostly thermodynamically disfavored but we are not falling apart because the conversion is very slow. In the biology case we also have mechanisms to fix stuff and kind of "cheat" but I'm simplifying the discussion. For a bit more information you can google metastability or unstable equilibrium.

What I meant re: the temperature affecting the energetics is that the master energy parameter is not enthalpy or entropy but gibbs free energy which is defined as ΔG = ΔH - TΔS. Negative gibbs free energy → spontaneous and vice versa. The two terms can be imagined to be in conflict with each other. For any given process you can have ΔH </=/> 0 and ΔS </=/> 0. The simplest case is where ΔH < 0 and ΔS > 0 because ΔG will always be negative regardless of the enthalpy and entropy change numerical values. However, we can have the terms opposing. For example we can have ΔH > 0 and ΔS > 0. For this case we really need to know what the numbers are because if ΔH > 0 but very small and ΔS > 0 but very large, the ΔG will still be < 0 and we are good to go. On the other hand if the numerical values of ΔH ~ ΔS we look at the temperature. We can have ΔH ~ ΔS but at 1000K, TΔS >> ΔH and G < 0. There are many more situations to consider but I will leave that to you to work out for yourself.

Hope this helps a bit.

6. ### sludger13

83
What happens if you divide the equation by temperature?

... and the entropy has no dimension of energy.

7. ### Yanick

401
Then you have trivially rearranged the equation, the behavior remains the same.

I never claimed it does, but TΔS has units of energy.

8. ### sludger13

83
Maybe just my wrong translation (. I just wanted to emphasize that decreasing Gibbs energy is equivalent to inceasing entropy.

9. ### Yanick

401
Decreasing entropy of what? In Gibbs equation the ΔS term is describing the system, not the universe. Maybe I'm misunderstanding what you mean, and I can't claim to be a thermodynamics expert by any stretch.

Also keep in mind I was tailoring the response to a high school student who may never have considered the difference between kinetics and thermodynamics. I hope I got that point across without muddying the water too much. I'm sure there are many details we can nitpick but I'm not about to write a textbook chapter on a forum.

10. ### sludger13

83
0 ≤ ΔS(total) = ΔS(system) – ΔS(surrounding) = ΔS(system) – ΔH/T(surrounding) / . (-T)

The enthalpy is obviously universal, it's only opposite in another direction.
As temperature changes with a heat, this formula might be only for infinitesimal segment of heat.

Last edited: Mar 23, 2014

### Staff: Mentor

The rates of all chemical reactions increase with temperature. So if a reaction is exothermic and the heat isn't removed, the reaction speeds up as the temperature increases. However, you also have to be aware of what Yanick says. All reactions run forward and in reverse simultaneously. When the temperature rises, both rates increase, but not necessarily by the same amount. So by changing the temperature, you can also change the point at which the forward and reverse reactions reach equilibrium. Once you reach the equilibrium point, the net rate of the reaction is zero. So, in short, not only do you have to consider rates of the forward and reverse reactions, but also where their equilibrium point is located.

Chet

12. ### NewtonsFellow

434
So the raise in temperature will speed up the rate of both the endo and exothermic reaction, but since the change in the temperature of a reaction will disturb the equilibrium position this indicates that the rate constants of both the forward and the backward reaction aren't equally affected, it's known that if the forward reaction is exothermic then the backward one would be endothermic and vice versa, so which rate constant is more affected by the change of the temperature, the endothermic one or the exothermic's ????

### Staff: Mentor

The rate constant for the reverse reaction is equal to the rate constant for the forward reaction divided by the equilibrium constant. The temperature dependence of the equilibrium constant is determined by the heat of the reaction. The temperature dependence of the forward rate is determined by its activation energy. So, if you combine this information, you should be able to determine the activation energy for the reverse reaction in terms of the activation energy for the forward rate constant and the heat of the reaction.

Chet

14. ### NewtonsFellow

434

Yeah sure, isn't there a fixed rule for the effect of heat on both constants, which one does it affect more in case of exo and endo reactions ???

### Staff: Mentor

I'm suggesting that you make use of what I said in my previous post, and work it out for us. If you don't feel like doing this, let me know and I will provide the analysis myself. But it's pretty simple algebra.

Chet

16. ### NewtonsFellow

434

I would love to do it but the problem is I'm not currently taking a thermochemistry class I don't really remember its algebra, I'm studying chemical equilibrium and how temperature affects it, all what my book says about that topic is that an increase of temperature in exothermic reactions that have reached an equilibrium state makes it move backwards without mentioning reasons I know that Kc = Rate constant of the forward reaction / rate constant of the backward reaction so I deduced from what the book mentions that the increase in the temperature made the reaction move backwards because the rate constantof the backward was more affected. That's it, I came here on the forum to find a better reason, however I don't like using algebra to understand science, building intuition is a lot better

Last edited: Mar 24, 2014

### Staff: Mentor

OK. I'll do the math. Here goes.

If, for example, the reaction is A + B → C + D, and, if you can assume mass action kinetics, then the rates of the forward and reverse reactions are:

$$R_f=k_f[A]$$
$$R_r=k_r[C][D]$$
so, at equilibrium,
$$k_f[A]=k_r[C][D]$$
or, equivalently,
$$\frac{[C][D]}{[A]}=\frac{k_f}{k_r}=K_E$$
where KE is the equilibrium constant for the reaction. In terms of the activation energies for the forward and reverse reaction rate constants,
$$k_f=k_{f0}\exp\left(\frac{-E_f}{RT}\right)$$
$$k_r=k_{r0}\exp\left(\frac{-E_r}{RT}\right)$$
So,
$$K_E=\frac{k_{f0}}{k_{r0}}\exp\left(\frac{-(E_f-E_r)}{RT}\right)$$
But, from thermochemistry, we also know that:
$$\frac{d\ln K_E}{d(1/T)}=\frac{-ΔH}{R}$$
where ΔH is the heat of reaction at temperature T. This equation tells us that, if the reaction is exothermic (ΔH<0), increasing the temperature causes KE to decrease, and shift the equilibrium from the products toward the reactants. This is what you said in your quote.

If we combine the previous equations, we obtain:
$$E_f-E_r=ΔH$$
So, $$E_r=E_f-ΔH$$
According to this equation, if the reaction is exothermic (ΔH<0), the reverse rate constant increases more than the forward rate constant when you increase the temperature. This also agrees with what you said in your quote.

Chet

18. ### MathewsMD

386
I realize for most rxns, this is the overall trend. But before we go on, isn't this not entirely correct? I don't know of any specific examples off the top of my head, but my workbook only says "reactions tend to proceed faster at higher temperatures." I don't believe this is an absolute statement and feel like there are exceptions...

### Staff: Mentor

For reversible reactions, the rates of the forward and the reverse reactions always increase with temperature (for fixed values of the concentrations of the reactants and products). But, as we said, if the reaction is exothermic, the reverse reaction rate increases faster with temperature than the forward rate. So, for certain combinations of reactant and product concentrations, the net rate of the reaction slows down with temperature (because the equilibrium shifts more toward the reactants and away from the products).

Chet

20. ### Yanick

401
It should be noted that Chestermiller's analysis is valid for 1 step reactions or elementary steps in a reaction mechanism. It goes back to what I was saying regarding the innate differences between kinetics and thermodynamics. Thermodynamics doesn't care about the path (atleast the state functions don't) while kinetics is all about the path. In a multi-step mechanism the rate law, describing the concentrations/time of reactants or products, can have no relation to the balanced chemical equation. The stoichiometric coefficients need not be equal to the order of the reactants and the rate constant need not have any relation to the equilibrium constant. Kinetic studies are by and large driven by, sort of, guess and check methods of analysis. We don't really sit down and derive stuff from first principles, we try to fit various models to whatever data we have at hand to make inferences about mechanisms and the system in general. Kinetics is also not the most exact of sciences because the data is not always uniquely determined. That is, more than 1 model may explain the data, so there are always arguments about mechanisms going between groups.

Otherwise what people have been saying is pretty much correct, keeping in mind the limitations. Looking from a classical physics perspective, we imagine that the kinetic energy of the individual molecules/atoms is a reflection of the bulk property we call temperature. As we increase the temperature, the kinetic energy of the molecules increases accordingly. In addition, reaction rates depend on collisions between molecules. Not every collision will yield a successful reaction so we are in the realm of probability. More kinetic energy means that there is more collisions per unit time which means there should be more successful collisions per unit as well. This works symmetrically for the forward and reverse reaction rate constants (see the Arrhenius Equation that Chestermiller used in his analysis), however you are shifting the equilibrium constant with temperature so the system will "come to rest" at a different point than it would at some other temperature.