Post by: Over9000 on July 02, 2009, 04:04:42 pm
Could anyone verify this or have any thoughts about it?
Post by: TrueTears on July 02, 2009, 04:06:05 pm
Post by: Over9000 on July 02, 2009, 04:07:48 pm
What is the reaction that affects skin growth?Not sure, I just assume that the skin would undergo reactions with the elements in the air such as C02, maybe im wrong tho?
Post by: TrueTears on July 02, 2009, 04:16:29 pm
Post by: Over9000 on July 02, 2009, 04:23:22 pm
Increasing the temperature increases the proportion of particles that has average kinetic energy equal or larger than the activation energy for the reaction. Assuming the activation energy for the reaction that causes skin growth is very high, there would be a negligible difference between the proportion of particles which have this amount of energy between the 2 temperatures (1 higher, 1 lower). Hence skin growth should be approximately the same.That seems logical, kk thanks TT
Post by: NE2000 on July 02, 2009, 04:26:08 pm
If there turns out to be substance to your theory it's the sort of thing you could write a thesis on :P
Post by: Over9000 on July 02, 2009, 04:34:43 pm
Increasing the temperature increases the proportion of particles that has average kinetic energy equal or larger than the activation energy for the reaction. Assuming the activation energy for the reaction that causes skin growth is very high, there would be a negligible difference between the proportion of particles which have this amount of energy between the 2 temperatures (1 higher, 1 lower). Hence skin growth should be approximately the same.That seems logical, kk thanks TT
Problem is this is bio as well as chem, there's a lot more to it than just reactions with components of the air. Firstly what exactly do you mean by skin ageing? Do you mean physical changes in the skin that you can see because I'm pretty sure that is more to do with the endocrine system rather than chemical reactions. In either case I also agree with TT, if there is a reaction between components of the air and skin that causes this (skin isn't really a reactant obviously but yeah just some chemicals on the outer surface of the skin) then temperature differences like that would have a negligible effect. Also there is some influence from blood vessels which are at a higher temperature; although in colder temperatures blood vessels tend to constrict.Yeh, I was just thinking lightly that reactions should be slower (even if extremely small) in Russia because theyre skin does seem to be in better condition than us, but now that I think about it, its probably mostly to do with sun exposure.
If there turns out to be substance to your theory it's the sort of thing you could write a thesis on :P
Post by: Over9000 on July 02, 2009, 05:28:18 pm
Say with the equation
If the coefficents were halved, would the activation energy be halved along with the "heat of reaction" (delta H) or would only the "heat of reaction be halved"
Post by: TrueTears on July 02, 2009, 05:28:50 pm
Post by: Over9000 on July 02, 2009, 05:29:29 pm
Thx alot TT :)would half.
Post by: Over9000 on July 02, 2009, 05:46:10 pm
Oh, page 248 of heineman text book said heat of reaction is given the symbolis not the heat of reaction, it is the difference in chemical energy between the reactants and products. Called the "heat content" or "enthalpy"
EDIT: Wait, I think I got confused, it just said heat of reaction is equal to the difference in enthalpy, its a confusing paragraph.
Post by: TrueTears on July 02, 2009, 05:49:37 pm
Yet on page 248, it says the energy released IS
Post by: Over9000 on July 02, 2009, 05:51:24 pm
No, wait, I thought the energy released (look at page 250) is theYeh same here, im thinking theyve got the wording mixed up on pge 248 coz its pretty confusing.. Because in a question I did a long time ago it said something like "4000 kJ of energy was RELEASED" what was the
to get the
Yet on page 248, it says the energy released IS
Post by: TrueTears on July 02, 2009, 05:53:51 pm
Post by: Over9000 on July 02, 2009, 05:59:51 pm
Okay I found it, on pg 259 of heinmann check out Q 11 b). It says calculate the heat of reaction ,Yeh, I did that question and found the answer fine, I think the wording on page 248 is what should be in question here, it just doesnt sound right, I sense some contradiction in it as well.
Post by: Over9000 on July 02, 2009, 06:05:07 pm
Post by: TrueTears on July 02, 2009, 06:06:30 pm
The wording on pg 248 is perfectly correct just depends on how you interpret it.
Post by: TrueTears on July 02, 2009, 06:09:41 pm
Another question, why is it that when you light a match, you start a combustion reaction (it provides the activation energy), yet even when the match burns out, the reaction continues?The reaction continues until it reaches equilibrium.
It's like saying if you light a fire, the flame you start off to light the twigs provides the activational energy, does the fire just stop when you take away the flame? No, the reaction proceeds.
Post by: Over9000 on July 02, 2009, 06:11:59 pm
So the match provides the activation energy to break the bonds at the start, but where does the energy come from to break the bonds that need to be broken after?Another question, why is it that when you light a match, you start a combustion reaction (it provides the activation energy), yet even when the match burns out, the reaction continues?The reaction continues until it reaches equilibrium.
It's like saying if you light a fire, the flame you start off to light the twigs provides the activational energy, does the fire just stop when you take away the flame? No, the reaction proceeds.
Post by: TrueTears on July 02, 2009, 06:13:03 pm
You already broke the bonds, what more bonds do you want to break?So the match provides the activation energy to break the bonds at the start, but where does the energy come from to break the bonds that need to be broken after?Another question, why is it that when you light a match, you start a combustion reaction (it provides the activation energy), yet even when the match burns out, the reaction continues?The reaction continues until it reaches equilibrium.
It's like saying if you light a fire, the flame you start off to light the twigs provides the activational energy, does the fire just stop when you take away the flame? No, the reaction proceeds.
Post by: Over9000 on July 02, 2009, 06:18:25 pm
Oh, I seeYou already broke the bonds, what more bonds do you want to break?So the match provides the activation energy to break the bonds at the start, but where does the energy come from to break the bonds that need to be broken after?Another question, why is it that when you light a match, you start a combustion reaction (it provides the activation energy), yet even when the match burns out, the reaction continues?The reaction continues until it reaches equilibrium.
It's like saying if you light a fire, the flame you start off to light the twigs provides the activational energy, does the fire just stop when you take away the flame? No, the reaction proceeds.
Post by: chem-nerd on July 03, 2009, 12:41:46 pm
You initially have to provide the activation energy - lighting the match to start the combustion reaction. The heat released from this combustion reaction is greater than the activation energy and so allows the bunsen burner to continue burning fuel without having to continuously apply external energy.