equilibrium questions PLEASE HELP (1 Viewer)

[xisti]

guys pls help me i am having a diabolical brain fart plsplspls i have two questions

1. How can the position of equilibrium shift if the equilibrium constant has to stay the same? Because if the equilibrium constant is the ratio of products to reactants, how can the position of equilibrium (which i know to be the relative concentrations of each) be different?
2. (this ties in with the first question a bit), for the iron thiocyanate reaction, why does the colour of the solution change when i add/remove reactants?? if keq is staying the same then isnt the ratio of products to reactants the same?? like if i have 2 red and 4 blue shouldn't that make the same colour as 4 red and 8 blue? or is it like dependent on the position of equilibrium which changes (but idk why)
   
   
   
   
   
   

 

[quarkkkk]

How can the position of equilibrium shift if the equilibrium constant has to stay the same? Because if the equilibrium constant is the ratio of products to reactants, how can the position of equilibrium (which i know to be the relative concentrations of each) be different?

an equilibrium only shifts when it is disturbed (ie. the ratio of products to reactants changes). it shifts to re-establish the correct ratio of products to reactants (ie. maintain its keq). it's BECAUSE the keq has to remain constant that the equilibrium shifts
(with the exception of temperature changes obviously where keq does change)

(this ties in with the first question a bit), for the iron thiocyanate reaction, why does the colour of the solution change when i add/remove reactants?? if keq is staying the same then isnt the ratio of products to reactants the same?? like if i have 2 red and 4 blue shouldn't that make the same colour as 4 red and 8 blue? or is it like dependent on the position of equilibrium which changes (but idk why)

it's because one of the reactants (SCN-) is colourless. consider this: if you add SCN-, equilibrium will shift right to reduce the [SCN-] and will thus produce more iron thiocyanate which is blood red. because you are adding a colourless solution and the equilibrium hence shifts to become more blood red, you'd expect a more blood red colour.

 

[kkk579]

I think it's cos changing equilibrium constant changes equilibrium position but changing equilibrium position doesn't necessarily change the constant. I asked basically this same q to my tutor a few months ago and that's what he said

 

[xisti]

an equilibrium only shifts when it is disturbed (ie. the ratio of products to reactants changes). it shifts to re-establish the correct ratio of products to reactants (ie. maintain its keq). it's BECAUSE the keq has to remain constant that the equilibrium shifts
(with the exception of temperature changes obviously where keq does change)


it's because one of the reactants (SCN-) is colourless. consider this: if you add SCN-, equilibrium will shift right to reduce the [SCN-] and will thus produce more iron thiocyanate which is blood red. because you are adding a colourless solution and the equilibrium hence shifts to become more blood red, you'd expect a more blood red colour.

omg ok thank you!!
what about when theres a system in equilibrium with a keq of 2, and then like theres one system with like 8mol/L products and 4 reactants, and then another one thats 4 and 2? the keq is the same, but my teacher said that the system with 8mol/L of products would be further to the right and i dont really understand that? is it because theres a difference of 4mol/L rather than 2mol/L?

THAT MAKES SO MUCH SENSE THANK YOU SO MUCH!!

 

[xisti]

also like what is the position of equilibrium actually referring to? is it the relative difference between the amounts of products and reactants or is it just the ratio as well? im kinda confused

 

[anonymoushehe]

also like what is the position of equilibrium actually referring to? is it the relative difference between the amounts of products and reactants or is it just the ratio as well? im kinda confused

im pretty the position of the equilibrium is referring the relative magnitudes of the amount of products and reactants. if the position of the equilibrium lies on the right, it means that the amount of product is much larger than the reactants because the forward reaction is more strongly favoured; if the position of the equilibrium lies on the left, it means that the amount of product is relatively small compared to the reactants since the reverse reaction is more strongly favoured - because since the equilbrium constant is just [p]/[r], a large K value means that the amount of products much be larger than the reactants at equilibrium and vice versa.

 

[Eagle Mum]

also like what is the position of equilibrium actually referring to? is it the relative difference between the amounts of products and reactants or is it just the ratio as well? im kinda confused

The direction of the reactions is all about getting towards the specific ratio for the particular reactants & products.



For the above reaction, if either C, D or both are added, the ratio of these species would increase, so the reaction will favour towards the left to produce more A & B and reduce C & D, so that the ratio will approach the value of K again.
If either A, B or both are added, the reaction will be towards the right to produce more C & D and reduce A & B.
If either C, D or both are removed, the reaction will be towards the right to produce more C & D which will reduce A & B.
If either A, B or both are removed, the reaction will be towards the left to produce more A & B which will reduce C &c D.


The relative differences will also change and it should be possible to mathematically calculate these values but they aren’t used to predict the direction of the reactions since it is fundamentally about the equilibrium ratio.

 

[wizzkids]

How can the position of equilibrium shift if the equilibrium constant has to stay the same? Because if the equilibrium constant is the ratio of products to reactants, how can the position of equilibrium (which i know to be the relative concentrations of each) be different?

A good way to illustrate your question is to use the example of the dissociation of a weak acid in aqueous solution.
We commonly refer to acids as weak because the percentage of dissociation is low in aqueous solution. The percent dissociation is a good measure of the position of equilibrium. Let's look at ethanoic (acetic) acid which is a monoprotic acid.
The acid constant for this reaction CH₃COOH -> CH₃COO^- + H⁺ is K_a =1.73 x 10^-4
If I start with 1 M ethanoic acid (acetic acid) solution then at 25^oC the hydrogen ion concentration will be 0.004 mol/L and the percentage dissociation will be 0.4 per cent. However if I start with 0.001M ethanoic acid solution, the hydrogen ion concentration will be 0.0004 mol/L and the percentage dissociation will now be about 40 per cent. Huh? How does that work? The K_a value has not changed, but the position of equilibrium has changed because the acid molecules are 1000 times more dilute and they dissociate more in dilute solution.
Does that help your understanding?