Freedom physics
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#1
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Why does increasing the temperature of an exothermic reaction shift the position of equilibrium to the left?

Eg, with all species in gaseous form:

CO(g)+2H₂(g)⇌CH₃OH(g) ∆H=-94kJmol⁻¹

Please also don't give me similes - I'm trying to use an explanation that involves the increase in energy the system gets to explain this so that it makes full sense to me 🙂.
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username2975726
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temp increases
position of equilibrium needs to shift in a way that will reduce this change to maintain equilibrium
it shifts in the endothermic direction
left side is endothermic so position shifts to the left
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Freedom physics
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#3
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(Original post by amaraub)
temp increases
position of equilibrium needs to shift in a way that will reduce this change to maintain equilibrium
it shifts in the endothermic direction
left side is endothermic so position shifts to the left
Thank you, but I'm asking why it shifts in the wae of the endothermic reaction 🙂
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Pigster
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The forward reaction (being exothermic) involves energy being released (more energy released forming bonds than taken in breaking others). IF you raise the temperature and IF the forward reaction was favoured, the reaction would get even hotter. Not sensible.
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Freedom physics
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(Original post by Pigster)
The forward reaction (being exothermic) involves energy being released (more energy released forming bonds than taken in breaking others). IF you raise the temperature and IF the forward reaction was favoured, the reaction would get even hotter. Not sensible.
That's actually a good point! 😄 ; although, sorry but it still doesn't give a good answer - the reaction doesn't have a brain of its own and can't decide which path to go down 😄
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Freedom physics
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(Original post by Pigster)
The forward reaction (being exothermic) involves energy being released (more energy released forming bonds than taken in breaking others). IF you raise the temperature and IF the forward reaction was favoured, the reaction would get even hotter. Not sensible.
Plus, if it did, the military would already be hot on investigation! 😂
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TODTEMPLE01
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#7
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(Original post by Freedom physics)
Why does increasing the temperature of an exothermic reaction shift the position of equilibrium to the left?

Eg, with all species in gaseous form:

CO(g) 2H₂(g)⇌CH₃OH(g) ∆H=-94kJmol⁻¹

Please also don't give me similes - I'm trying to use an explanation that involves the increase in energy the system gets to explain this so that it makes full sense to me 🙂.
Any eqm reaction wants to oppose any change made to it. As the forward reaction is exothermic (releases energy), the reaction naturally wants to take in energy to oppose the change (and therefore favouring the endothermic/backward reaction). As a result, further increasing the temperature means that the reaction will want to take in even more energy and therefore shifts eqm even further to the side of the endothermic reaction which is left
Last edited by TODTEMPLE01; 1 year ago
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charco
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(Original post by Freedom physics)
Thank you, but I'm asking why it shifts in the way of the endothermic reaction 🙂
It is a function of the Maxwell Boltzmann distribution and the activation energy of both forward and back processes in the equilibrium.

For an endothermic reaction the activation energy of the endothermic process is necessarily lower than that of the exothermic direction.



Hence, the activation energy of the endothermic process on the Maxwell Boltzmann is to the higher energy side.

Equilibrium is the point at which the rate of the forward reaction equals the rate of the back reaction and both of these are a function of the concentration of the reacting species with sufficient activation energy.

An increase in temperature proportionally affects the population of activated particles by a larger percentage at the higher end of the Maxwell Boltmann curve. Hence, the number of particles with sufficient activation energy is affected more for the endothermic change then the exothermic change and the equilibrium moves towards the direction of endothermic change.
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Freedom physics
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(Original post by TODTEMPLE01)
Any eqm reaction wants to oppose any change made to it. As the forward reaction is exothermic (releases energy), the reaction naturally wants to take in energy to oppose the change (and therefore favouring the endothermic/backward reaction). As a result, further increasing the temperature means that the reaction will want to take in even more energy and therefore shifts eqm even further to the side of the endothermic reaction which is left
Thank you; that explanation is partially satisfying and partially explains it and is enough to have me accept it 🙂 - the part which i don't like is just the le chateliers principle part of it which states the reaction wants to oppose any changes and thus keep the system how it is 🙂
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Pigster
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(Original post by Freedom physics)
Thank you; that explanation is partially satisfying and partially explains it and is enough to have me accept it 🙂 - the part which i don't like is just the le chateliers principle part of it which states the reaction wants to oppose any changes and thus keep the system how it is 🙂
As you said earlier, reactions don't 'want' anything. Expressing the lCP in anthropomorphic terms just makes it easier to explain. As charco said, the actual explanation is all to do with the Maxwell Boltzmann distribution. For the endothermic process (with a high Ea), perhaps only 10% of collisions have sufficient energy and for the exothermic process (with lower Ea) maybe 30% of collisions have enough energy. Now, when you raise T, both curves shift right (and down), so for both curves there are now more particles with E>Ea. It might be that now there is 11% and 32% with E>Ea. BUT the 11% represents a 10% increase (from 10 to 11%) and the other represents a 6.7% increase (why did I pick horrible numbers?). The endothermic reaction is proportionally sped up more than the exothermic one.
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