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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.
Yanick said: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.
What happens if you divide the equation by temperature?Yanick said: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.
sludger13 said:What happens if you divide the equation by temperature?
sludger13 said:... and the entropy has no dimension of energy.
sludger13 said:Maybe just my wrong translation (. I just wanted to emphasize that decreasing Gibbs energy is equivalent to inceasing entropy.
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.ElmorshedyDr said: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.
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.ElmorshedyDr said: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 ?
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 ?Chestermiller said: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
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.ElmorshedyDr said: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 ?
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 betterChestermiller said: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
OK. I'll do the math. Here goes.ElmorshedyDr said: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
Chestermiller said: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
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).MathewsMD said: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...
An exothermic reaction is a chemical reaction that releases energy in the form of heat. This means that the overall energy of the reactants is higher than the energy of the products, resulting in a release of heat.
Increasing the temperature of an exothermic reaction will typically increase the rate of the reaction. This is because higher temperatures provide more energy to the reactant molecules, allowing them to overcome the activation energy barrier more easily and resulting in more frequent and successful collisions.
The ideal temperature for an exothermic reaction depends on the specific reaction and its reactants. However, in general, a higher temperature will result in a faster reaction rate. It is important to carefully control the temperature in order to achieve the desired reaction rate and avoid potential hazards.
Yes, in some cases, extremely high temperatures can negatively affect exothermic reactions. This can occur if the temperature is too high and causes the reactant molecules to break down or change in a way that prevents them from reacting. Additionally, if the temperature is too low, the reaction may not proceed at a fast enough rate to be useful.
As the name suggests, exothermic reactions release energy in the form of heat. Therefore, increasing the temperature of the reaction will result in a larger overall energy change, as more energy is being released. On the other hand, decreasing the temperature will result in a smaller overall energy change as less heat is being released.