Reconciling Ea with G

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Hello. Must be having a brainfart here, sure the answer is easy for most of you. So many bonds generally require activation energy. This is because if the molecules aren't moving fast enough, the old bonds won't be able to break so the new ones won't form.

I just watched a video on gibbs free energy. And the reason that T is muliplied by S is because at high temperatures, molecules will smash into each other and fall apart.

let us take the example of an alkene becoming hydrogenated through an addition reaction. Two molecules become one, so entropy is decreased. If you need to put energy into this system in order for it to happen (activation energy), then how can adding energy (Raising the temperature) make it LESS spontaneous?
 

mjc123

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H
And the reason that T is muliplied by S is because at high temperatures, molecules will smash into each other and fall apart.
This is nonsense. ΔG = ΔH - TΔS even at low temperatures.

Thermodynamics and kinetics are different things. There is no relation between the magnitude or sign of ΔG and the size of Ea. All reactions have an activation energy, and will get faster at higher temperatures. That is different from the effect of temperature on the position of equilibrium. In the example you give, the reaction gets faster at higher T (by Arrhenius: lnk = lnA - Ea/RT) but ΔG becomes less negative/more positive by ΔG = ΔH - TΔS (with negative ΔS).
In fact that doesn't say it all. Because lnK = -ΔG°/RT = -ΔH°/RT + ΔS°/R
then d(lnK)/dT = ΔH°/RT2
so the temperature variation of the equilibrium constant depends on ΔH°, not ΔS°.
 

TeethWhitener

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f you need to put energy into this system in order for it to happen (activation energy), then how can adding energy (Raising the temperature) make it LESS spontaneous?
You're making a common mistake here: confusing kinetics with thermodynamics. The activation energy is a kinetic term: it dictates the bare minimum energy required for a reaction to happen. This in turn impacts the rate of the reaction. The Gibbs free energy is a thermodynamic term. It dictates (regardless of the rate of reaction) whether the reaction is thermodynamically favorable--essentially what the mixture of reactants and products looks like at equilibrium.

I know a lot of people use the word "spontaneous" to describe reactions that have a negative free energy, but it's a terrible term in my opinion. The conversion of diamond to graphite at room temperature has a negative free energy, but the lay person would hardly describe it as spontaneous: the activation energy is far too high (rate is far too slow) to see this conversion take place.
 

mjc123

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I know a lot of people use the word "spontaneous" to describe reactions that have a negative free energy, but it's a terrible term in my opinion.
I heartily agree, but there seem to be many people (on chemicalforums, for example) who will defend the use of "spontaneous" to describe a reaction that doesn't happen (or only with great difficulty) but has negative ΔG. Of course, you can define "spontaneous" to mean "negative ΔG" if you want, on the Humpty Dumpty principle, but it doesn't seem a very useful choice.
 
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Semantics over the word spontaneous aside, I still don’t understand how to reconcile the two. In my example Adding Energy is required for the reaction to take place, but the more energy you ad, the more likely the products are to break apart. So how does adding energy to the system make products more likely to form if it also makes products more likely to break apart?
 

TeethWhitener

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Raising the temperature increases the rates of both the forward and the back reactions.
 
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Got it. So the temperature increases the rates of reaction both ways. And the temperature also shifts the equilibrium of the reaction. Right?

So adding heat to a system might kick off a reaction, but if u add too much heat you will shift the equilibrium and possibly reverse the reaction?
 
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another question (thank u all for your help). Okay so let’s say a combustion reaction. Thermodynamically favorable but has a high activation energy. But on a Boltzmann distribution, even at room temperature, at least a few molecules have enough energy to activate. So why don’t those few molecules start the reaction, and therefore make the log just start burning on its own, albeit very very slowly?
 

TeethWhitener

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And the temperature also shifts the equilibrium of the reaction. Right?
Yes, this comes from Le Chatelier’s principle (or the van’t Hoff equation from @mjc123 ’s post). But it’s important to keep in mind that equilibrium is a thermodynamic concept, while rates are a kinetics concept.

another question (thank u all for your help). Okay so let’s say a combustion reaction. Thermodynamically favorable but has a high activation energy. But on a Boltzmann distribution, even at room temperature, at least a few molecules have enough energy to activate. So why don’t those few molecules start the reaction, and therefore make the log just start burning on its own, albeit very very slowly?
Combustion is kind of a bad example because it’s ridiculously complex at the microscopic level. Even something simple like the reaction of methane with oxygen requires on the order of 50 or so intermediate reactions to model really accurately. Intermediates make things more complicated.

But even if we assume a simple, highly exothermic reaction, it’s only self-propagating if the rate of reaction is fast enough to generate heat more rapidly than heat is dissipated. The temperature where this occurs is called the autoignition temperature. It’s probably quite high for something like wood (edit: Ray Bradbury would say that paper’s autoignition temperature is about 451 deg F, and it turns out that’s pretty close to true), but in some reactions it can be less than room temperature. These substances are called pyrophoric, and they tend to be good for things like rocket fuels.
 
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