How would you explain the effect of a decrease in temperature using collision theory for 2no<>n2o2 (1 Viewer)

[=)(=]

^^^

Thanks

 

[Bob99]

^^^

Thanks

Hey man are you sure its supposed to be N2O2 and not N2 + O2???

 

[=)(=]

yes its the equilibrium reaction where n2o2 is colourless and no is brown

 

[Bob99]

yes its the equilibrium reaction where n2o2 is colourless and no is brown

oh right cuz the one i have seen usually is N2O4(g) ⇌ 2NO2(g) so i was a bit confused.
But for collision theory decreasing the temperature will remove energy from the system, and according to Le chateliers principle. it will favour the exothermic reaction. I have yet to revise my collision theory but from the top of my head i think it has to do with activation energy or something.

 

[carrotsss]

If the temperature is increased, there is more (heat) energy. This means that out of the collisions that occur, there is an increased likelihood that each collision will have sufficient activation energy for the reaction to succeed, leading to an overall increase in reaction rate.

The reaction which has a higher activation energy in the reversible reaction (typically the endothermic reaction) will therefore be more likely to have successful reactions (even though the exo will also have an increase, just not as much), leading to an increase in reaction rate of the endothermic reaction and a shift towards the other side’s concentration.

 

[Bob99]

If the temperature is increased, there is more (heat) energy. This means that out of the collisions that occur, there is an increased likelihood that each collision will have sufficient activation energy for the reaction to succeed, leading to an overall increase in reaction rate.

yes that is right, but how would you explain it for a decrease in temperature, because its gonna favour the exothermic reaction, so how would you use collision theory to explain that.

 

[carrotsss]

yes that is right, but how would you explain it for a decrease in temperature, because its gonna favour the exothermic reaction, so how would you use collision theory to explain that.

Oh shoot misread the question lol.

In that case the opposite happens, less reactions have sufficient activation energy, disproportionately affecting the endothermic reaction (which has higher activation energy), leading to a decrease in its reaction rate and hence a shift in concentration towards it.

Sorry for the disjointed explanation, only just learnt this

 

[Bob99]

Oh shoot misread the question lol.

In that case the opposite happens, less reactions have sufficient activation energy, disproportionately affecting the endothermic reaction (which has higher activation energy), leading to a decrease in its reaction rate and hence a shift in concentration towards it.

Sorry for the disjointed explanation, only just learnt this

nah your explanation is good. It's concerning for me considering my exam is on monday and i have not revised collision theory yet

 

[=)(=]

thank yall

 

[wizzkids]

For a full discussion of Collision Theory and equilibrium, see this previous post:

equilibrium constant and temperature

I understand that with increasing temperature in exothermic reactions, the equilibrium constant decreases, while in endothermic the constant increases. But why is that the case (like what does the increasing temperature do to the products/reactants of the reaction causing the constant to...

www.boredofstudies.org