# Need help intuiting a few chemistry concepts (rate equation and equilibrium) **URL:** \ **Category:** Factual Questions\ **Created:** [January 30, 2020, 8:54pm UTC](https://boards.straightdope.com/t/need-help-intuiting-a-few-chemistry-concepts-rate-equation-and-equilibrium/847246 "2020-01-30T20:54:05Z")\ **Posts on this page:** 17\ **Page:** 2
**Author:** ![Chronos](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/chronos/32/134_2.png) [@Chronos](https://boards.straightdope.com/u/Chronos)\ **Post date:** [February 2, 2020, 8:20pm UTC](https://boards.straightdope.com/t/need-help-intuiting-a-few-chemistry-concepts-rate-equation-and-equilibrium/847246/21 "2020-02-02T20:20:15Z")
The general principle is that if you disturb any system, the equilibrium will shift to favor the opposite of your disturbance (though usually to a lesser degree). If you increase temperature, then you’ll get more reactions that absorb energy. If you add more of some reagent, then you’ll get more reactions that use up that reagent. And if you increase the pressure, you’ll get more reactions that decrease the pressure (by reducing the number of moles). ---
**Author:** ![Ruken](https://avatars.discourse-cdn.com/v4/letter/r/f475e1/32.png) [@Ruken](https://boards.straightdope.com/u/Ruken)\ **Post date:** [February 2, 2020, 8:34pm UTC](https://boards.straightdope.com/t/need-help-intuiting-a-few-chemistry-concepts-rate-equation-and-equilibrium/847246/22 "2020-02-02T20:34:17Z")
I think OP gets that but wants to dig into why that happens. ---
**Author:** ![am77494](https://avatars.discourse-cdn.com/v4/letter/a/d2c977/32.png) [@am77494](https://boards.straightdope.com/u/am77494)\ **Post date:** [February 3, 2020, 1:29am UTC](https://boards.straightdope.com/t/need-help-intuiting-a-few-chemistry-concepts-rate-equation-and-equilibrium/847246/23 "2020-02-03T01:29:27Z")
> [@Ruken](#): > > I think OP gets that but wants to dig into why that happens. I don’t think that’s what the OP is asking. I think the OP is intuiting that the reaction RATE favors the product at higher pressure not the EQUILIBRIUM favors the product at higher pressure. ---
**Author:** ![Ruken](https://avatars.discourse-cdn.com/v4/letter/r/f475e1/32.png) [@Ruken](https://boards.straightdope.com/u/Ruken)\ **Post date:** [February 3, 2020, 11:38am UTC](https://boards.straightdope.com/t/need-help-intuiting-a-few-chemistry-concepts-rate-equation-and-equilibrium/847246/24 "2020-02-03T11:38:57Z")
> [@am77494](#): > > I don’t think that’s what the OP is asking. I think the OP is intuiting that the reaction RATE favors the product at higher pressure not the EQUILIBRIUM favors the product at higher pressure. The reaction rate _does_ favor the product if perturbing the system to higher pressure, until the new equilibrium is reached. ---
**Author:** ![am77494](https://avatars.discourse-cdn.com/v4/letter/a/d2c977/32.png) [@am77494](https://boards.straightdope.com/u/am77494)\ **Post date:** [February 3, 2020, 12:28pm UTC](https://boards.straightdope.com/t/need-help-intuiting-a-few-chemistry-concepts-rate-equation-and-equilibrium/847246/25 "2020-02-03T12:28:22Z")
> [@Ruken](#): > > The reaction rate _does_ favor the product if perturbing the system to higher pressure, until the new equilibrium is reached. Okay, lets get this sorted out. There are two questions, one about the reaction rate and the second about equilibrium. Here is the detailed question : Q1 \> With a reaction at equilibrium at a given pressure, why is the forward reaction initially faster, when the pressure is raised ? The answer follows from the rate equations : Forward reaction rate = kf_PP\_Red_PP\_Blue but PP\_Red = X\_Red \* P where X\_Red is the mole fraction of Red, and PP\_Blue = X\_Blue \* P (P is the total pressure) So Forward Reaction Rate = kf \* X\_Red \* X\_Blue \* P^2 Backward Reaction Rate = kr \* PP\_Green = kr \* X\_Green \* P So forward reaction rate has a P^2 term in it and the reverse reaction has only P in it. Does the OP want to intuit why one has P^2 term in it and the other P in it ? Q2 \> Why does the reaction favor Green at equilibrium at higher pressure ? At equilibrium, Forward Reaction Rate = Backward Reaction Rate =\> kf_X\_Red_X\_Blue_P^2 = kr_X\_Green\*P or (X\_Green/ X\_Red\*X\_Blue) = (kf/kr)\*P = Equilibrium constant So the equilibrium constant has the P term in favor of Green. Is this the question the OP wants to intuit ? ---
**Author:** ![Ruken](https://avatars.discourse-cdn.com/v4/letter/r/f475e1/32.png) [@Ruken](https://boards.straightdope.com/u/Ruken)\ **Post date:** [February 3, 2020, 1:07pm UTC](https://boards.straightdope.com/t/need-help-intuiting-a-few-chemistry-concepts-rate-equation-and-equilibrium/847246/26 "2020-02-03T13:07:33Z")
It’s pretty clear the hangup isn’t the math. He even wrote that. ---
**Author:** ![am77494](https://avatars.discourse-cdn.com/v4/letter/a/d2c977/32.png) [@am77494](https://boards.straightdope.com/u/am77494)\ **Post date:** [February 3, 2020, 1:52pm UTC](https://boards.straightdope.com/t/need-help-intuiting-a-few-chemistry-concepts-rate-equation-and-equilibrium/847246/27 "2020-02-03T13:52:10Z")
> [@Ruken](#): > > It’s pretty clear the hangup isn’t the math. He even wrote that. The Hangup for the explanation is the question not the math. I am not sure your post clarifies if the question is about equilibrium or kinetics. ---
**Author:** ![Ruken](https://avatars.discourse-cdn.com/v4/letter/r/f475e1/32.png) [@Ruken](https://boards.straightdope.com/u/Ruken)\ **Post date:** [February 3, 2020, 2:05pm UTC](https://boards.straightdope.com/t/need-help-intuiting-a-few-chemistry-concepts-rate-equation-and-equilibrium/847246/28 "2020-02-03T14:05:20Z")
> [@am77494](#): > > The Hangup for the explanation is the question not the math. I am not sure your post clarifies if the question is about equilibrium or kinetics. They’re the same thing. ---
**Author:** ![am77494](https://avatars.discourse-cdn.com/v4/letter/a/d2c977/32.png) [@am77494](https://boards.straightdope.com/u/am77494)\ **Post date:** [February 3, 2020, 2:19pm UTC](https://boards.straightdope.com/t/need-help-intuiting-a-few-chemistry-concepts-rate-equation-and-equilibrium/847246/29 "2020-02-03T14:19:19Z")
> [@Ruken](#): > > They’re the same thing. How are they the same thing ? Do you have a cite or explanation? ---
**Author:** ![robby](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/robby/32/11048_2.png) [@robby](https://boards.straightdope.com/u/robby)\ **Post date:** [February 3, 2020, 4:07pm UTC](https://boards.straightdope.com/t/need-help-intuiting-a-few-chemistry-concepts-rate-equation-and-equilibrium/847246/30 "2020-02-03T16:07:56Z")
> [@KidCharlemagne](#): > > So you’re saying the moles/liter value used for B in A+2B -\> C is only for 1B (and therefore must be included twice)? Not exactly. I’m saying that changing the concentration of B has twice the effect of changing the concentration of A. > [@KidCharlemagne](#): > > Well if you increase the pressure and the volume is kept constant, then the temperature goes up, no? You kind of have it backwards. How exactly are you increasing the pressure with constant volume? The only way to increase pressure with constant volume is to either increase the number of moles (i.e. add more stuff) or increase the temperature. ---
**Author:** ![KidCharlemagne](https://avatars.discourse-cdn.com/v4/letter/k/ed8c4c/32.png) [@KidCharlemagne](https://boards.straightdope.com/u/KidCharlemagne)\ **Post date:** [February 6, 2020, 2:33pm UTC](https://boards.straightdope.com/t/need-help-intuiting-a-few-chemistry-concepts-rate-equation-and-equilibrium/847246/31 "2020-02-06T14:33:11Z")
> [@robby](#): > > Not exactly. I’m saying that changing the concentration of B has twice the effect of changing the concentration of A. > > You kind of have it backwards. How exactly are you increasing the pressure with constant volume? > > The only way to increase pressure with constant volume is to either increase the number of moles (i.e. add more stuff) or increase the temperature. I guess I was assuming that the reaction was happening in a container with fixed volume. ---
**Author:** ![KidCharlemagne](https://avatars.discourse-cdn.com/v4/letter/k/ed8c4c/32.png) [@KidCharlemagne](https://boards.straightdope.com/u/KidCharlemagne)\ **Post date:** [February 6, 2020, 2:38pm UTC](https://boards.straightdope.com/t/need-help-intuiting-a-few-chemistry-concepts-rate-equation-and-equilibrium/847246/32 "2020-02-06T14:38:39Z")
Been busy but I’m back now. Perhaps I should attempt to understand why a system in equilibrium seeks to “undo” the added stress. Is it because the system is seeking the lowest energy state? If that’s the case, I’m not quite sure why the energy state would necessarily be best lowered by the pressure vs whatever side of the reaction was more endothermic and therefore absorbed the most heat. ---
**Author:** ![am77494](https://avatars.discourse-cdn.com/v4/letter/a/d2c977/32.png) [@am77494](https://boards.straightdope.com/u/am77494)\ **Post date:** [February 6, 2020, 2:55pm UTC](https://boards.straightdope.com/t/need-help-intuiting-a-few-chemistry-concepts-rate-equation-and-equilibrium/847246/33 "2020-02-06T14:55:38Z")
> [@KidCharlemagne](#): > > Perhaps I should attempt to understand why a system in equilibrium seeks to “undo” the added stress. Using terms like stress makes me think you are thinking chemical reactions like a beam balance which is in static equilibrium. Chemical reactions are not in static equilibrium: they are in dynamic equilibrium I.e. forward reactions keep happening at the same rate as reverse reactions. > [@KidCharlemagne](#): > > Is it because the system is seeking the lowest energy state? That is very insightful and you are in the right track. It’s not energy though but Gibbs free energy - [Gibbs free energy - Wikipedia](https://en.m.wikipedia.org/wiki/Gibbs_free_energy) In my opinion Gibbs was one of the genius scientists to have realized this. We use Gibbs free energy equations to calculate equilibrium constants when we know nothing about the reaction rates. ---
**Author:** ![KidCharlemagne](https://avatars.discourse-cdn.com/v4/letter/k/ed8c4c/32.png) [@KidCharlemagne](https://boards.straightdope.com/u/KidCharlemagne)\ **Post date:** [February 6, 2020, 4:54pm UTC](https://boards.straightdope.com/t/need-help-intuiting-a-few-chemistry-concepts-rate-equation-and-equilibrium/847246/34 "2020-02-06T16:54:38Z")
> [@am77494](#): > > Using terms like stress makes me think you are thinking chemical reactions like a beam balance which is in static equilibrium. Chemical reactions are not in static equilibrium: they are in dynamic equilibrium I.e. forward reactions keep happening at the same rate as reverse reactions. > That is very insightful and you are in the right track. It’s not energy though but Gibbs free energy - [Gibbs free energy - Wikipedia](https://en.m.wikipedia.org/wiki/Gibbs_free_energy) > > In my opinion Gibbs was one of the genius scientists to have realized this. We use Gibbs free energy equations to calculate equilibrium constants when we know nothing about the reaction rates. “Stress” seemed awfully imprecise to me as well, but it’s been the term used in 2 different textbooks and many an online source. That’s why I was seeking clarification. So if it’s “seeking “ the lowest free energy state then how can we necessarily say that favoring the higher mole reactant is the best way to reduce that energy if we don’t know the thermodynamics of that particular reaction? Maybe the decrease in entropy dwarfs any possible enthalpy effects? I literally just read about this so what I just said is probably idiotic. ---
**Author:** ![am77494](https://avatars.discourse-cdn.com/v4/letter/a/d2c977/32.png) [@am77494](https://boards.straightdope.com/u/am77494)\ **Post date:** [February 6, 2020, 5:57pm UTC](https://boards.straightdope.com/t/need-help-intuiting-a-few-chemistry-concepts-rate-equation-and-equilibrium/847246/35 "2020-02-06T17:57:39Z")
> [@KidCharlemagne](#): > > …then how can we necessarily say that favoring the higher mole reactant is the best way to reduce that energy … I am not sure what you are talking about. Who is we ? When you say “favoring” are you talking about reaction rates or equilibrium concentration? Can you illustrate your question with an example ? ---
**Author:** ![KidCharlemagne](https://avatars.discourse-cdn.com/v4/letter/k/ed8c4c/32.png) [@KidCharlemagne](https://boards.straightdope.com/u/KidCharlemagne)\ **Post date:** [February 6, 2020, 8:21pm UTC](https://boards.straightdope.com/t/need-help-intuiting-a-few-chemistry-concepts-rate-equation-and-equilibrium/847246/36 "2020-02-06T20:21:27Z")
> [@am77494](#): > > I am not sure what you are talking about. > > Who is we ? When you say “favoring” are you talking about reaction rates or equilibrium concentration? Can you illustrate your question with an example ? “We” meaning humanity in general and scientists in particular. I’m speaking of LeChatelier’s principle and equilibrium. By “favoring“ I mean which side of the reaction accelerates in response to an increase in pressure. I can’t quite figure out why the side of the reaction with more moles necessarily increases. If the system is attempting to reduce Gibbs free energy, like you said (I think), then it can change entropy, enthalpy, and/or temp (according to that Gibbs equation). I would think that decreasing the number of moles would decrease entropy since there are fewer possible states. But don’t you need to take into account the change in enthalpy in a reaction to determine if it will proceed? LeChatelier seems to be saying that if you increase pressure, entropy is the only thing that matters because the reaction will always favors the side with more moles. I’m getting way ahead of myself here by bringing this Gibbs free energy into the equation, but thus far, no one has been able to tell me, without resorting solely to math, why an increase in pressure favors the reaction direction that reduces the total number of moles in the system. ---
**Author:** ![am77494](https://avatars.discourse-cdn.com/v4/letter/a/d2c977/32.png) [@am77494](https://boards.straightdope.com/u/am77494)\ **Post date:** [February 7, 2020, 4:21pm UTC](https://boards.straightdope.com/t/need-help-intuiting-a-few-chemistry-concepts-rate-equation-and-equilibrium/847246/37 "2020-02-07T16:21:57Z")
> [@KidCharlemagne](#): > > “We” meaning humanity in general and scientists in particular. I’m speaking of \*\*LeChatelier’s principle and equilibrium. \*\* By “favoring“ I mean which side of the reaction \*\*accelerates \*\*in response to an increase in … {Bolding mine above} I think you are mixing up concepts. Equilibrium is not a time dependent thing - equilibrium quantities are not “accelerated” by change in pressure. To help me understand your question, please take a look at the Red + Blue —\> Green reaction that I took some time to put together. Please take a look at that and tell me which part of it is not intuitive to you. We can then widen that to general reactions. [Previous page](https://boards.straightdope.com/t/need-help-intuiting-a-few-chemistry-concepts-rate-equation-and-equilibrium/847246.md?page=1)