Can Adding Heat Slow Down a Reaction?

[alex caps]

I had a dispute with a professor where he argued that adding heat to a reaction will shorten the time it takes to reach equilibrium. I feel as though adding heat will increase collisions and and therefore increase the rate which A+B goes to C. However, if you are looking at the reverse, where C goes to A+B, I think that adding heat will cause the A+B reaction to speed up, causing the process to take longer. Can anyone see a flaw in my analysis or if it is logical, point out any reactions where adding heat will actually slow the it down? Thanks.

 

[Pengwuino]

What do you mean causing the A+B reaction to speed up? It sounds like you're talking about a decomposition...

 

[alex caps]

Sorry, my question was unclear. What I mean is the reaction between A+B --> C will eventually reach an equilibrium. Adding heat to the forward reaction will cause this to happen sooner. Are there any reactions which will reach equilibrium slower with the addition of heat?

 

[Bystander]

Please make a distinction between "heat" and "temperature," and restate your question.

 

[kmarinas86]

I had a dispute with a professor where he argued that adding heat to a reaction will shorten the time it takes to reach equilibrium. I feel as though adding heat will increase collisions and and therefore increase the rate which A+B goes to C. However, if you are looking at the reverse, where C goes to A+B, I think that adding heat will cause the A+B reaction to speed up, causing the process to take longer. Can anyone see a flaw in my analysis or if it is logical, point out any reactions where adding heat will actually slow the it down? Thanks.

What kind of equlibrium?

http://en.wikipedia.org/wiki/Equilibrium

In the physical sciences

* Thermodynamic equilibrium, when internal processes of a system cause no overall change in temperature or pressure
* Dynamic equilibrium, when two reversible processes (in thermodynamics) occur at the same rate
* Chemical equilibrium, when a chemical reaction proceeds at the same rate as its reverse reaction, with no change in the amount of each compound
* Mechanical equilibrium, when the sum of the forces and moments on each particle of a system is zero
* Quasistatic equilibrium, when internal processes change gradually from one equilibrium state to the next

 

[kmarinas86]

Sorry, my question was unclear. What I mean is the reaction between A+B --> C will eventually reach an equilibrium. Adding heat to the forward reaction will cause this to happen sooner. Are there any reactions which will reach equilibrium slower with the addition of heat?

http://www.wpbschoolhouse.btinterne...we speed up or slow down chemical reactions?"

Unit 1 " How can we speed up or slow down chemical reactions?"

Being able to control the speed of chemical reactions is important both in everyday life (for example in cooking) and when making new materials on an industrial scale.

For FT and HT students: The speed (rate) of chemical reactions increases - if the temperature increases; if the concentration of dissolved reactants or the pressure of gases increases; if solid reactants are in smaller pieces (greater surface area); if a catalyst is used.

A catalyst increases the rate of a chemical reaction but it is not used up during the reaction. It is used over and over again to speed up the conversion of reactants to products. Different reactions need different catalysts.

Increasing the rates of chemical reactions is important in industry because it helps to reduce costs.

The rate of a chemical reaction can be followed by measuring the rate at which the products are formed or the rate at which the reactants are used up. This allows a comparison to be made of the changing rate of a chemical reaction under different conditions. You should be able to interpret graphs showing the amount of product formed (or reactant used up) with time in terms of the above principles.

Chemical reactions can only occur when reacting particles collide with each other and with sufficient energy. The minimum amount of energy particles must have to react is the activation energy. Increasing the temperature increases the speed of the reacting particles so that they collide more frequently and more energetically. This increases the rate of reaction.

Increasing the concentration of reactants in solutions and increasing the pressure of reacting gases also increases the frequency of collisions and so increases the rate of reaction.

 

[alex caps]

I realize I am still being unclear, let me explain the question asked which caused this dilemma:

3I- (aq) + S2O82- (aq) ---> I3- (aq) + 2 SO42- (aq)

(sorry for not being familiar with the proper way to write this on this forum, but with the S2O82-, the 2 and 8 and subscripts and the 2- is the charge on the oxygen, the I3- is I with a -3 charge, the 2SO42-, 4 is the subscript and the 2- is the charge on it).

During this redox reaction, we are assuming that the reaction goes to completion.

"In an experimentm equal volumes of 0.0120 M I- and 0.0040 M S2O82- are mixed at 25ºC. The conentration of I3- over the following 80 minutes is shown in the graph below."
- a graph showing the molar concentration of the I3- is on the y-axis andtime in minutes on the x-axis. As the time progresses, the concentration starts to increase fast then levels off at about 35 minutes, where it has reached equilibrium. The graph looks like a graph of y=square root of x, increasing in that fashion.

Now it is asked, if the temperature is raised to 35ºC, how would the graph change. The answer is it would be steeper at the beginning, reaching the same equilibrium level at an earlier time.

Now my question is: Is there ever a time when increasing temperature would cause the reaction (increasing [I3-] until equilibrium) to slow down. I am not speaking particular of this question, but in general, are there any instances when increasing temperature would cause the equliibrium point to be reached slower (later) ?

 

[alex caps]

Okay, let's try this. Looking at the NO2 in this graph, it will eventually flatten out when it is in chemical equilibrium with N2O4. Adding heat with move the curve to the left because it will cause the reaction to speed up. Are they ANY instances when adding heat will cause the reaction to move to the right?

http://www.cartage.org.lb/en/themes/Sciences/Chemistry/Miscellenous/Helpfile/Equilibrium/Equilibrium/eq.gif

 

[GCT]

adding "heat" energy to a reaction medium will help in surpassing the activation energy barrier for both forward and reverse reactions. The reaction will speed towards equilibrium, however, the equilibrium constant may change. I may add to this later, but this should make things a lot clearer; you don't want to conclude on such absolutes though, in fact, under the right context perhaps the time it takes to reach equilibrium may slow down in a sense.

 

[movies]

Adding heat to an exothermic reaction would cause a shift towards the reactants.

Here is a restatement of what GCT wrote, which is right on:

For a reaction to reach equillibrium you have to have enough energy around to get past the activation barriers. Adding heat gives more energy to the system, and therefore there is more energy to overcome the reaction barriers. How much heat is added can change where the equillibrium lies, but it will always speed up how fast equillibrium is reached because the barriers are more accessible.

 

[alex caps]

Adding heat to an exothermic reaction would cause a shift towards the reactants.

Here is a restatement of what GCT wrote, which is right on:

For a reaction to reach equillibrium you have to have enough energy around to get past the activation barriers. Adding heat gives more energy to the system, and therefore there is more energy to overcome the reaction barriers. How much heat is added can change where the equillibrium lies, but it will always speed up how fast equillibrium is reached because the barriers are more accessible.

Thank you, I do understand that completely. What I was wondering though, is whether or not there will ever be a time when adding heat will cause equilibrium to be reached slower? I was thinking that if A+B gives you C and adding heat will cause more collisions, then of course the equilibrium will speed up, but if you started with all of C and wanted to reach equ. breaking C into A and B, will adding heat slow down how long it takes because as C breaks into A and B, more heat will cause more collisions of A and B going back to C? Was that clear?

 

[GCT]

That doesn't make any sense. You're focusing too much on one aspect of the reaction,

the reaction between A and B is dependent upon the decomposition of C, that is every C that decomposes, produces A and B, and only from there A and B can combine to produce C Thus the decomposition of C is somewhat of a "limiting" rate.

That is you've got to establish an equilibrium rate, that is when both forward and reverse rates are the same. You start with a higher rate of decomposition of C, the A+B reaction doesn't have an initial rate. As more A+B is produced, its rate of reaction can increase until it reaches the rate of decomposition of C (which is constant, assuming a simple reaction).

 

[alex caps]

But as the formation of A+B is increasing, adding heat would increase collisions and force A+B to go back to C before equilibrium hits.. but eventually the rate of C ->A+B will equal A+B --> C (equilibrium). So is it possible that adding heat during the decomposition of C would make the time it takes to reach equilibrium slow down?

 

[GCT]

no, not "before equilibrium hits" but rather a new equilibrium will be established. Remember equilibrium constants change with temperature. The equilibrium is fixed for a particular temperature, if you add heat, it will change.

 

[movies]

I think the part that you are neglecting is that in your example adding more heat would speed up the decomposition of C into A + B. This decomposition could be unimolecular and therefore could be initiated by adding energy to the system and increasing bond rotations, etc. that might lead to decomposition to A + B. Also, the decomposition could be initiated by a collision of two molecules of C to give 2A + 2B or C + A + B; in either case increasing the temperature would increase the frequency of the collisions between two molecules of C.

Also, the rate of the C -> A + B reaction will decrease as the concentration of C decreases, so it's not entirely that the rate of A + B -> C increases toward equilibrium, but also that C -> A + B decreases toward equillibrium.

 

[lalbatros]

Google for "LE CHATELIER'S PRINCIPLE" ...

... and discover that adding heat (and therefore increasing temperature) displaces an equilibrium in the heat-consuming direction of the (equilibrium) reaction.

This maybe linked to your question since displacing an equilibrium also mean "activating" or "accelerating" a reaction. Therefore the word "activity" used sometimes in analysing Gibbs energy changes under a reaction.

This is only valid around an equilibrium state. For example, "activating" an explosive triggers a heat-producing reaction. This is because the initial state is not a thermodynamic equilibrium. Obviously, this is also what justifies the "LE CHATELIER'S PRINCIPLE", starting from the definition of an equilibrium. Tautology, isn't it?

Hope it heats you up !