Dynamic Equilibrium??!

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tetris179
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#1
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#1
Can somebody please how different factors affect the shift in the simplest possible way?
I understand how pressure affects the shift
I understand how catalysts affect the shift
I have some understanding of how concentration affects the shift
I have zero understanding of how temperature affect the shift.
Seriously, I can't get my head around why it does what it does?

So if the forward reaction is exothermic (a negative enthalpy change as the product has less energy than the reactants).
And we decrease the temperature of the system...
The system wants to increase the temperature yes (Le Chatelier's Principle - system counteracts).
It does this by favouring the exothermic reaction (forward).
But why? Why does favouring the exothermic forward reaction help to increase the overall temperature?

Honestly if you could explain i'd be overjoyed, i never understood this at GCSE either and i'm in year 12 and i still don't.
Any other examples would be much appreciated
Last edited by tetris179; 2 years ago
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Peter.G
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#2
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#2
Remember this trick: LEX and HEND.

LEX: Lowering the temperature favours the Exothermic reaction.

HEND: Higher temperatures favour the endothermic reaction.

If the deltaH given is negative, then the forward reaction would be exothermic, and using LEX we can see that lowering the temperature of the system would favour the forward reaction, and more products would be made. The backward reaction would be endothermic (always opposite of forward reaction) and thus, using HEND, making the temperature higher would favour the backward reaction and more reactants would be formed.

It’s hard to get your head around but hopefully that helped
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Peter.G
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(Original post by Peter.G)
Remember this trick: LEX and HEND.

LEX: Lowering the temperature favours the Exothermic reaction.

HEND: Higher temperatures favour the endothermic reaction.

If the deltaH given is negative, then the forward reaction would be exothermic, and using LEX we can see that lowering the temperature of the system would favour the forward reaction, and more products would be made. The backward reaction would be endothermic (always opposite of forward reaction) and thus, using HEND, making the temperature higher would favour the backward reaction and more reactants would be formed.

It’s hard to get your head around but hopefully that helped
That is literally all you need to know for temperature
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tetris179
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(Original post by Peter.G)
That is literally all you need to know for temperature
can't thank you enough!
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charco
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#5
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#5
(Original post by tetris179)
can't thank you enough!
You have a rule of thumb but you don't have a reason!

The rate of a chemical reaction depends on the number of particles that successfully collide.
At equilibrium, the rate of the forward reaction is equal to the rate of the reverse reaction.

For an exothermic reaction (forward) the activation energy is less than the back (endothermic) reaction. This can be easily shown using a reaction energy profile.

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The number of particles with the required activation energy is given by the Maxwell Boltzmann energy distribution and from that, the Arrhenius relationship at Temperature = T1:

k = Ae-Ea/RT1

where k is the rate constant.

Hence, for the forward reaction:

k(f) = Ae-Ea(f)/RT1

And for the backward reaction:

k(b) = Ae-Ea(b)/RT1

And for a first order reaction (simplest case)
rate(f) = k(f) x [reactant]

At equilibrium:

rate(forward) = rate(backward)

hence:

k(f) x [reactant] = k(b) x [product]

Using the natural log form of the Arrhenius relationship:

lnk(f) = lnA - Ea(f)/RT1
lnk(b) = lnA - Ea(b)/RT1
----------------------------- subtract
lnk(f) - lnk(b) = [Ea(b) - Ea(f)] x 1/RT1

We have already established that the Ea(b) is greater than Ea(f) for an exothermic (forward) reaction.

Hence, [Ea(b) - Ea(f)] is a number greater than 0, i.e. a positive number.

Hence, increasing Temperature from T1 to T2 will make the positive number smaller.

Therefore, the difference lnk(f) - lnk(b) must also become smaller. This effectively means that k(f) decreases with respect to k(b). (or k(b) increases more than k(f))

And as,

k(f) x [reactant] = k(b) x [product]

at equilibrium. To maintain equilibrium, [reactant] must increase with respect to [product]

To do this the reaction shifts towards the left hand side.

Summary:
Increasing the temperature of an exothermic reaction drives it backwards, towards the side of the reactants in order to re-establish the equilibrium condition.
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Pigster
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#6
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#6
(Original post by charco)
You have a rule of thumb but you don't have a reason!
Stolen.

(By me, not you. Well, not to my knowledge)
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charco
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(Original post by Pigster)
Stolen.

(By me, not you. Well, not to my knowledge)
During the hundreds of years that I've taught the effect of temperature on equilibrium, I've relied on Maxwell-Boltmann diagrams and simply stated that increasing the temperature of an exothermic (forward) process has more effect on the reverse rate than the forward rate, but I've never felt comfortable with this simple assertion. While teaching and looking at the MB curves, I wasn't convinced myself 100%, and if the teacher isn't convinced ...

So this morning I thought that I'd demonstrate it. The proof above is actually for me, not the OP.
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Pigster
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#8
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(Original post by charco)
During the hundreds of years that I've taught the effect of temperature on equilibrium, I've relied on Maxwell-Boltmann diagrams and simply stated that increasing the temperature of an exothermic (forward) process has more effect on the reverse rate than the forward rate, but I've never felt comfortable with this simple assertion. While teaching and looking at the MB curves, I wasn't convinced myself 100%, and if the teacher isn't convinced ...

So this morning I thought that I'd demonstrate it. The proof above is actually for me, not the OP.
What you showed is what I reckoned would be what's going on, but it is loverly to see it laid out so nicely.

Ta muchly, my students are usually math genii, so it'll be interesting to see what they make of it.
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