Industrial Chemistry, Equillibrium Constant (1 Viewer)

[Yindi]

Does anyone 'understand' why temperature is the only factor that affects K for a given reaction.

thanks.

 

[louielouiee]

If volumes, concentrations or pressure change, then both the products & reactants used in the calculation of K will shift correspondingly, effectively cancelling out the effect of one another (meaning they don't affect the overall equilibrium expression)

However, temperature DOES affect K as per Le Chatelier. Think of temperature as essentially adding another product/ reactant.

So, in terms of endothermic reactions, if temperature increases then K increases (vice versa for lower temperature)
If it is exothermic, if temperature increases then K decreases (vice versa)

This can be seen once K has been calculated!
Is K>1 then it lies towards the products.
If K ~1 then you can say there are equal amounts of both.
If K<1 then it lies towards the reactants.

Taa daa

 

[Yindi]

If volumes, concentrations or pressure change, then both the products & reactants used in the calculation of K will shift correspondingly, effectively cancelling out the effect of one another (meaning they don't affect the overall equilibrium expression)

However, temperature DOES affect K as per Le Chatelier. Think of temperature as essentially adding another product/ reactant.

So, in terms of endothermic reactions, if temperature increases then K increases (vice versa for lower temperature)
If it is exothermic, if temperature increases then K decreases (vice versa)

This can be seen once K has been calculated!
Is K>1 then it lies towards the products.
If K ~1 then you can say there are equal amounts of both.
If K<1 then it lies towards the reactants.

Taa daa

Okay thanks for the help.
But... you say when concentration/volume/pressure change, reactant and products shift accordingly, this is exactly the same case with temperature.
For example with the Haber process, Increasign the pressure increases products and decreases reactants, also, similarly decreasing the temperature increases the products and decreses the reactants. Don't really see a difference.

I asked if anyone understood why K is affected by temperature, all you gave me was this "However, temperature DOES affect K as per Le Chatelier."

If an icrease in products/decrease in reactants (and vice versa) is all that effects K, such as with temperature, then pressure would also affect K, so would volume, I haven't explored concentrations of products and reactants too much to be sure that K is/is not affected by them. But it seems like pressure/temperature/volume have very similar effects...

The only theory I have is that equillibrium occurs in a closed system, and increasing/decreasing temperature affects the total enthalpy in the thermodynamic system whcih would somehow affect K.

Also this is wrong:
"This can be seen once K has been calculated!
Is K>1 then it lies towards the products.
If K ~1 then you can say there are equal amounts of both.
If K<1 then it lies towards the reactants."

 

[study-freak]

Does anyone 'understand' why temperature is the only factor that affects K for a given reaction.

thanks.

I believe the answer is way beyond the scope of HSC.

But the gist (as I understand it) is that:
it comes from what's known as the Boltzmann distribution, which may approximately describe the number/proportion of particles occupying different energy levels in an equilibrium configuration. It mathematically just happens that when this Boltzmann distribution is adapted, the number of possible microstates is maximised (sorry for the jargon but it just means that the configuration is the most statistically probable). Temperature is the sole parameter of this distribution and determines the spread (in simple terms, how 'excited' the particles are in general). The greater the temperature, the greater the proportion of particles occupying higher energy levels.

Pressure and the total number of particles (related to conc) do not affect the ratio of occupation numbers in any particular excited state and the ground state. This also means that the equilibrium constant (the ratio between the product of concentrations of the products and the product of concentrations of reactants in an equilibrium system) is unchanged (can mathematically prove this).

If you increase the temperature, the Boltzmann distribution is changed and the species with higher energy becomes more thermally accessible than before. The new equilibrium configuration will be such that compared to before, there is proportionally more of the higher energy species (product in the case of an endothermic reaction, reactant in the case of an exothermic reaction). This is equivalent to saying that the equilibrium shifts in the direction of an endothermic reaction and the equilibrium constant will change to reflect this change.

Not sure if this made any sense to you, but if you are interested in these things, you might want to study statistical thermodynamics.

 

[Yindi]

I believe the answer is way beyond the scope of HSC.

But the gist (as I understand it) is that:
it comes from what's known as the Boltzmann distribution, which may approximately describe the number/proportion of particles occupying different energy levels in an equilibrium configuration. It mathematically just happens that when this Boltzmann distribution is adapted, the number of possible microstates is maximised (sorry for the jargon but it just means that the configuration is the most statistically probable). Temperature is the sole parameter of this distribution and determines the spread (in simple terms, how 'excited' the particles are in general). The greater the temperature, the greater the proportion of particles occupying higher energy levels.

Pressure and the total number of particles (related to conc) do not affect the ratio of occupation numbers in any particular excited state and the ground state. This also means that the equilibrium constant (the ratio between the product of concentrations of the products and the product of concentrations of reactants in an equilibrium system) is unchanged (can mathematically prove this).

If you increase the temperature, the Boltzmann distribution is changed and the species with higher energy becomes more thermally accessible than before. The new equilibrium configuration will be such that compared to before, there is proportionally more of the higher energy species (product in the case of an endothermic reaction, reactant in the case of an exothermic reaction). This is equivalent to saying that the equilibrium shifts in the direction of an endothermic reaction and the equilibrium constant will change to reflect this change.

Not sure if this made any sense to you, but if you are interested in these things, you might want to study statistical thermodynamics.

Thank you, that's exactly what I was looking for