Endothermic and exotermic reactions

[kylegg4]

I'm a little unsure on endothermic and exotermic reactions. an endothermic reaction absorbs heat (the product will become hot) and an exothermic reaction gives off heat (the product will become cold) But if i react two products together will exo and endo occur at the same time? Because, for example in a replacement reaction one product gains some more energy and the other loses energy.

Edited February 29, 2012 by kylegg4

[JohnStu]

Yes endothermic reaction means the stuff has "eaten" some heat, becoming more hot, and exotermir reaction means the stuff has given off heat, becoming more cold.

 

These reactions talk about the result, not the process. In some complex reactions it goes endothermic, then exotermic, then edothermic and you get a net of exothermic reaction as result due to the magnitudes.

 

 

[mississippichem]

I'm a little unsure on endothermic and exotermic reactions. an endothermic reaction absorbs heat (the product will become hot) and an exothermic reaction gives off heat (the product will become cold) But if i react two products together will exo and endo occur at the same time? Because, for example in a replacement reaction one product gains some more energy and the other loses energy.

 

Good question.

 

Remember that the total enthalpy of a reaction is the sum of the enthalpies of the elementary steps. Many reactions will involve elementary steps that are both endothermic and exothermic (good example: bond breaking is endothermic, bond forming is exothermic. Both bond making and breaking often happen in the same reaction). The total enthalpy of a reaction will always be net endothermic, net exothermic, or net zero.

 

Look at wikipedia's image of a Born-Haber cycle for the formation of LiF.

 

 

Also, a small side note. Your wording is a bit crunchy with the way you have defined endothermic. An endothermic reaction may not necessarily absorb heat from the surroundings. What we can say is that for all endothermic reactions [math] \Delta H [/math] is less than zero. The whole "getting hot/getting cold" way of thinking about this can lead to problems later in your understanding as temperature and heat are not the same thing.

 

I'm being a bit picky here but usually picky leads itself to good understanding with respect to science.

 

Yes endothermic reaction means the stuff has "eaten" some heat, becoming more hot, and exotermir reaction means the stuff has given off heat, becoming more cold.

 

These reactions talk about the result, not the process. In some complex reactions it goes endothermic, then exotermic, then edothermic and you get a net of exothermic reaction as result due to the magnitudes.

 

Be careful here though. The reaction probably never "goes exothermic then endothermic" as the kinetics of the reaction proceed. The energy of any collection of molecules in the system will display statistical behavior. Some molecules may already be in the product state while others are only just beginning the first elementary step.

 

The take home message here is that enthalpy is a state-function and is not dependent on path. Some molecules may even take different kinetic paths to the products!

 

EDIT: for wording

Edited February 29, 2012 by mississippichem

[kylegg4]

"These reactions talk about the result, not the process." Yes ok so say i do a reaction like A+BC--->AC+B then one product is losing energy and one is gaining, so what do i say the end result is? A gain more energy and B losing energy but C also gains???? Has this any relevance as to whether it's exothermic or endothermic?

[mississippichem]

"These reactions talk about the result, not the process." Yes ok so say i do a reaction like A+BC--->AC+B then one product is losing energy and one is gaining, so what do i say the end result is? A gain more energy and B losing energy but C also gains???? Has this any relevance as to whether it's exothermic or endothermic?

 

Enthalpy measurements are made on the system. You could consider the internal energy, U, of a given product or reactant but all this involves fairly complicated physical chemistry/statistical mechanics and usually doesn't yield useful information (It's more an exercise in thermodynamic book keeping).

[kylegg4]

"bond breaking is endothermic, bond forming is exothermic." if endothermic reactions are [math] \Delta H [/math] less than zero then does that mean energy of that system is decreasing so bonds contain energy you break bonds energy is lost(endo) you create bonds with energy (exo) then [math] \Delta H [/math] is larger than zero?

 

Ok, so you ignore single properties of the whole system you're just talking about the system as a whole...much like entropy.

Edited February 29, 2012 by kylegg4

[mississippichem]

"bond breaking is endothermic, bond forming is exothermic." if endothermic reactions are [math] \Delta H [/math] less than zero then does that mean energy of that system is decreasing so bonds contain energy you break bonds energy is lost(endo) you create bonds with energy (exo) then [math] \Delta H [/math] is larger than zero?

 

You've got it backwards. Endothermic reaction have positive enthalpies, exothermic reactions have negative enthalpies. Of course this is all by convention but the convention makes sense as it is in line with the generally accepted physics sig convention for work being done on/by a system.

 

Also, you don't really create bonds with energy. A bond is a lower lying quantum state than the orbitals that comprised either of the orbitals in the atoms that will participate in the bonding. You can think of a bond as a low potential energy state of electrons in a molecule. When the energy of a molecule is lowered by the formation of a bond, the excess heat has to go somewhere, so we observe exothermy. It takes energy for the electrons to climb out of this potential so bond breaking events are endothermic.

[kylegg4]

"Endothermic reaction have positive enthalpies" is this one of those negative is positive and positive is negative. because you said earlier that endothermic reactions can be described as [math] \Delta H [/math]<0 which means a decrease in enthalpy? or am i just way off!?

Edited February 29, 2012 by kylegg4

[mississippichem]

I love this way to remeber:( [math] \Delta H [/math]>0 endothermic)( [math] \Delta H [/math]<0 exothermic) if i've understood it right then thats a good way for me to remeber:) thanks.

 

No problem. Helping you helps me to be a better communicator of scientific knowledge.

 

But yeah, I'm not usually a fan of telling people to memorize (mathematical and or conceptual understanding is FAR superior in every way), but unfortunately there are things like sign conventions that are just best memorized. So as it pains me to say it , just memorize it in any way you see fit.

 

Welsome to SFN by the way.

Edited February 29, 2012 by mississippichem

[kylegg4]

Ok so i havnt actually got the whole exo and endo round the wrong way i've just got the wrong idea about bonds, if bonds are broken enthalpy decreases? if bonds are formed enthalpy increases? so enthalpy increases because heat is released when the electrons go into what you called "low energy states? but when these bonds are broken the enthalpy decreases because energy is used to break the bonds so energy is "used up" and can nolonger be measure as part of the enthalpy of the system?

Edited February 29, 2012 by kylegg4

[mississippichem]

Ok so i havnt actually got the whole exo and endo round the wrong way i've just got the wrong idea about bonds, if bonds are broken enthalpy decreases? if bonds are formed enthalpy increases? so enthalpy increases because heat is released when the electrons go into what you called "low energy states? but when these bonds are broken the enthalpy decreases because energy is used to break the bonds so energy is "used up" and can nolonger be measure as part of the enthalpy of the system?

 

No. Bond forming events are exothermic, [math] \Delta H < 0 [/math]. Bond breaking events are endothermic, [math] \Delta H > 0 [/math]. Try to think of enthalpy in terms of how it changes without respect to time. Instead of saying that the enthalpy increases when a bond is broken, say that bond breaking is an endothermic process. I know it is seems trivial but it isn't. Read into the concept of state function.

 

Also, the energy is not "used up". It is transferred to a molecule from somewhere and can be used to break a bond. Energy is conserved if you count everything in the system and the surroundings (remember energy can either exit or enter the system if our experiment is set up that way, which it usually is in solution chemistry settings).

[kylegg4]

if it is a completely closed system then enthalpy does not change?

[mississippichem]

if it is a completely closed system then enthalpy does not change?

 

If the process is adiabadic (no heat transferred across the system boundary), and there is no work done on or by the system. The change in enthalpy is zero. Yes.

 

[math] dH=dU+Vdp+pdV [/math]

Edited March 1, 2012 by mississippichem

[dinzie]

URGENT! NEED HELP!

Hi, I am doing a project and would like to ask if it is possible to create gel insoles that are able to produce both endothermic and exothermic reaction based on the current temperature of your feet? If it is, what is the equation and chemicals needed? How long can the chemicals last/continue reacting?

 

 

 

 

Good question.

 

Remember that the total enthalpy of a reaction is the sum of the enthalpies of the elementary steps. Many reactions will involve elementary steps that are both endothermic and exothermic (good example: bond breaking is endothermic, bond forming is exothermic. Both bond making and breaking often happen in the same reaction). The total enthalpy of a reaction will always be net endothermic, net exothermic, or net zero.

 

Look at wikipedia's image of a Born-Haber cycle for the formation of LiF.

 

 

Also, a small side note. Your wording is a bit crunchy with the way you have defined endothermic. An endothermic reaction may not necessarily absorb heat from the surroundings. What we can say is that for all endothermic reactions [math] \Delta H [/math] is less than zero. The whole "getting hot/getting cold" way of thinking about this can lead to problems later in your understanding as temperature and heat are not the same thing.

 

I'm being a bit picky here but usually picky leads itself to good understanding with respect to science.

 

 

 

Be careful here though. The reaction probably never "goes exothermic then endothermic" as the kinetics of the reaction proceed. The energy of any collection of molecules in the system will display statistical behavior. Some molecules may already be in the product state while others are only just beginning the first elementary step.

 

The take home message here is that enthalpy is a state-function and is not dependent on path. Some molecules may even take different kinetic paths to the products!

 

EDIT: for wording