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View Full Version : how to guess the reaction will be cool or hot?


aster
July 17th, 2003, 10:56 AM
since we may mix many things together, i am sometimes wonder, when i am reacting mercury with nitric acid it is cool, but when i am reacting methylamine with nitric acid there are a lot of heat released, how to predict a reaction will be cool or hot, in the another word will be exothermic or endothermic, right? sorry because i am not chemist, so i think there must be a guidance to determine whether a reaction will be, or we just know this by trial and learn from previous try? thanks alot :D

knowledgehungry
July 17th, 2003, 11:01 AM
You can look at the Delta H of the reaction if it is negative energy is released, if positive than energy iss required for the reaction.

aster
July 17th, 2003, 11:29 AM
where i can get learn about "delta H" in simple word how to calculate them?:confused:

knowledgehungry
July 17th, 2003, 11:58 AM
Well you can look at the heat of formation of a chemical, look for it online maybe chemfinder.com has? and add the heat of formation of the reactents and then subtract the sum of the heats of formation of the products from it and that lets you know the free energy, if positive than requires enrgy/heat, if negative than gives off.

Example:

H2SO4@-450kj mole+2KNO3@-215kj mole-->2HNO3@-200kj mole+ K2SO4@-150kj mole

-450+2(-215)=-880Kj
2(-200)+-150=-550KJ
-880Kj-(-550KJ)=-330kj so you have energy being released
*heat of formation numbers pulled out of my ass for demonstrtion purposes.

aster
July 17th, 2003, 01:05 PM
thanks, it's helpfull

Arthis
July 17th, 2003, 01:39 PM
The merck index will give you the "delta H", which is btw called enthalpy of formation of the compound.

Then you sum the different enthalpies, pondered by the coeff. before them in the reaction. Note that a product will be counted as negative.

A + p*B --> m*C

p, m are numbers, A,B,C compounds.

Enthalpy of reaction : DeltaH(A) + p*DeltaH(B) - m*DeltaH(C)

inFinie
July 18th, 2003, 06:36 AM
There is a search form on NIST
http://webbook.nist.gov/chemistry/form-ser.html
Results seems complicated, maybe it'll help to someone

Tuatara
July 18th, 2003, 07:11 PM
Or if you can't find exactly what you are looking for, go back to first principles and add up the bond energies of the reactants, then add up the bond energies of the products. If you have more energy on the reactant side then your reaction will be exothermic. Simple.

TheMaddHatter
August 7th, 2003, 09:02 PM
It may also be worth your time to look up Gibb's Free Energy Equation....it can be use to predict whether a reaction will take place based on entropy and enthalpy and tempature. Though it has many forms the basic is G = H- TS or Free Energy Constant = enthalpy - tempature(entropy). Where if G is negative than the reaction will likely take place. It can be interesting as it can really bring a lot together. Anyway just a further application for delta H.

The/\/\addhatter

P.S. i think i got the uses and spelling of enthalpy and entropy right but they could be switched....main thing to remember is that H is (in a nutshell) exothermic or endothermic and S is the "messiness" of the reaction and T is of coarse tempature.

vulture
August 8th, 2003, 04:08 AM
Messiness or chaos are bad examples to explain entropy.

A better explanation would be that if the entropy of a compound increased, it has lost some of it's ability to perform work.

For example, you have 2l of water in an thermically isolated vessel, separated by a wall. 1l is at 300C, the other at 350C. If you pull the wall away the water will mix, forming 2l at 325C. While there has not been any energy exchange with the environment, the entropy did increase.

You can easily see, that if you want to reverse the process, you'll need to invest energy, even if your not changing the total internal energy of the water. That's the energy that got "lost" with the entropy increase.

zeocrash
August 8th, 2003, 05:58 AM
you also have to remember that just because a reaction doesn't feel hot doesn't mean it's not exothermic. mixing sodium bicarbonate with vinegar doesn't feel like it's givving off any heat, but it's very slightly exothermic.

vulture
August 8th, 2003, 07:16 AM
True zeocrash and very important to consider when you are thinking about upscaling.
Slight exothermic reactions will be slow in small quantities because they produce just enough heat to activate the reaction.

In larger batches (+1kg) you can easily get a chain reaction and your slow and easy reaction will derail into vigorous boiling and splattering.

metafractal
August 8th, 2003, 09:13 AM
Knowledgehungry, those calculations are correct for the amount of engergy released in a reaction. The energy given off by a reaction can be expressed in a number of ways, another being light, for example. While energy is often expressed as heat, this is not a reliable way of calculating the precise enthalpy of a reaction.
...or is knowledehungry no longer with us?

knowledgehungry
August 8th, 2003, 09:33 AM
I'm still here, I was merely giving him a general idea of a simple way to tell whether the reaction will most likely be hot or cold, I know energy is not only heat, but in most cases the majority of energy is given off as heat. I'm just taking a break from making anything, im into theory only for a while at least.