Reaction Shifting to maintain equilibrium

[kenhung123]

I am confused. So is it correct to say a reaction is in equilibrium?
Since reversible reactions can go both ways how do you exactly know which is reactants and products?
Is it just the way the reaction is positioned on the page?

Why is it that an exothermic reaction subjected to higher temperature would shift to the left (form reactants and reduce products)

[fady_22]

Yes, its the way that its written, i.e. reactants products.

An exothermic reaction releases energy in the form of heat (usually), and so an increase in temperature (according to Le Chatelier's Principle) will cause there to be a net backwards reaction (which would require energy, and so is endothermic) to oppose the effect of the increase in temp.

[m@tty]

Yeah, extra heat = extra energy, then the endothermic reaction progresses to absorb the energy. This means that if you were talking about an exothermic reaction then a backwards reaction will occur.

[kenhung123]

You study way too much m@tty

[Akirus]

You study way too much m@tty

Funny hearing that coming from the guy that posts the most questions on this board.

[kenhung123]

That just means I'm dumb lol

[m@tty]

You study way too much m@tty

Lol, I just skimmed A+ notes two nights ago..  Unit 4 seems sooooo small... =\

[Blakhitman]

I should probably get some supplementary notes for chem and phys this time, they seem useful.

[kenhung123]

What do you mean supplementary notes??

[Blakhitman]

A+ notes, Neap smartstudy.

Extra notes and books etc.

[naved_s9994]

Yes, its the way that its written, i.e. reactants products.

An exothermic reaction releases energy in the form of heat (usually), and so an increase in temperature (according to Le Chatelier's Principle) will cause there to be a net backwards reaction (which would require energy, and so is endothermic) to oppose the effect of the increase in temp.

I believe for an equilibrium reaction when one side of the equilibrium system is favoured, we use a two single headed arrows with a larger arm - indicating the side that is favoured.

------->
<---

[shinny]

Yes, its the way that its written, i.e. reactants products.

An exothermic reaction releases energy in the form of heat (usually), and so an increase in temperature (according to Le Chatelier's Principle) will cause there to be a net backwards reaction (which would require energy, and so is endothermic) to oppose the effect of the increase in temp.

I believe for an equilibrium reaction when one side of the equilibrium system is favoured, we use a two single headed arrows with a larger arm - indicating the side that is favoured.

------->
<---

That's really just for explanatory purposes. Normally they just stick with the same double arrow, and whichever side is favoured will be indicated in the equilibrium constant (K) which typically always accompanies any equilibrium reaction.

[naved_s9994]

Yes, its the way that its written, i.e. reactants products.

An exothermic reaction releases energy in the form of heat (usually), and so an increase in temperature (according to Le Chatelier's Principle) will cause there to be a net backwards reaction (which would require energy, and so is endothermic) to oppose the effect of the increase in temp.

I believe for an equilibrium reaction when one side of the equilibrium system is favoured, we use a two single headed arrows with a larger arm - indicating the side that is favoured.

------->
<---

That's really just for explanatory purposes. Normally they just stick with the same double arrow, and whichever side is favoured will be indicated in the equilibrium constant (K) which typically always accompanies any equilibrium reaction.

Okay, thanks for that information

[kenhung123]

Hmm doesn't really make sense how you can alter the yield without altering the rate of a forward/backward reaction. Like if everything proceeds normally, it would have the same yield?

[m@tty]

Yield is concerned with the end result of the reaction at equilibrium, while rate is the about the speed at which equilibrium is arrived at. For an altered yield the rates at which the individual forward- and back-reactions proceed are altered, but this does not mean that the overall rate must also change.

And I don't really understand your question. What do you mean by "normally"? And "the same" as compared to what?