# how does increasing concentration of products not affect equilibrium constant?

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#1

Ok so I don't get this paragraph.
Instead of representing it as ethanol and ethanoic acid, I try to understand this by just writing it as 1/1 (and x being any change in concentration). If you make the bottom line of the expression larger, it is then 1/1+x .

To restore the situation, the concentration of the bottom line of the fraction is reduced as the reactants react to make more product. so remove the x from the bottom to show a decrease in reactants (now 1/1) and then add an x ontop to show increase in products (now 1+x/1).

so in my head, the final result of this shenanigan is 1+x/1 whereas the original was 1/1 showing that the fraction has changed.

?!??!

(thanks for any help)
0
6 months ago
#2
I'm a bit rusty with my chemistry as, well, covid19 (Y13), I think the error in your methodology comes to the fact that you assume the end value of concentrations is a complete conversion from the excess denominator (reactant = 1+x) to the numerator (products = 1+x) whereas the concentrations of the reactant will gradually transfer from 1+x ---> a value in which product production is in equilibrium to reactant backwards production as that will be the point at which equilibrium is met and thus the end value will mathematically be (1+(x/2)/1+(x/2))
I kept this in mathematical terms as I guess that's how you'd prefer, hopefully this makes sense.
1
6 months ago
#3
Forgot to add, but I'm guessing you see it aswell, this results in it ending at a 1:1 ratio, just of greater volumes
0
6 months ago
#4
(Original post by pondering-soul)

Ok so I don't get this paragraph.
Instead of representing it as ethanol and ethanoic acid, I try to understand this by just writing it as 1/1 (and x being any change in concentration). If you make the bottom line of the expression larger, it is then 1/1+x .

To restore the situation, the concentration of the bottom line of the fraction is reduced as the reactants react to make more product. so remove the x from the bottom to show a decrease in reactants (now 1/1) and then add an x ontop to show increase in products (now 1+x/1).

so in my head, the final result of this shenanigan is 1+x/1 whereas the original was 1/1 showing that the fraction has changed.

?!??!

(thanks for any help)
Equilibrium constant is a CONSTANT (i.e. it doesn't change, as long as temperature doesn't change). If you add lots of product, then the system is no longer in equilibrium. Some product must turn into reactant in order for the system to reach equilibrium. The ratio of products to reactants at equilibrium will always reach the same value as it was before, as long as temperature didn't change.

Consider the extremely simple reaction A --> B
K = [B]/[A]

Suppose K = 2 at a temperature of 273K.

Then if [A] = 1 at equilibrium, [B] must = 2 because the constant K = 2 for that reaction at the given temperature.

Now we add some product, so [B] = 2+x

New ratio immediately after adding x product is (2+x):1. But we are not in equilibrium. The system will now spontaneously change itself so that the ratio [B]/[A] becomes K = 2 (assuming negligible temperature change).

(2+x)/1 > 2 = K, so some product has to turn into reactant (as predicted by le Chatelier's principle).
Once the system reaches equilibrium, the new ratio will be K = 2 again.

For example, if x = 1, then we would've had [B] = 3 and [A]=1 immediately after adding 1 unit of [B], but in order to get back to the 2:1 ratio, we need to solve (3-y)/(1+y) = 2

(since the reaction is A-->B, so for every y moles of A that reacts, y moles of B is formed, or conversely, for every y moles of B that reacts, y moles of A are formed).

(3-y) = 2+2y, so y = 1/3, hence the final equilibrium concentrations after adding one mole of [B] is

[A] = 1.33333...
[B] = 2.66666...

then K = [B]/[A] = 2, as required since K is constant at constant temperature.
Last edited by K-Man_PhysCheM; 6 months ago
1
6 months ago
#5
(Original post by K-Man_PhysCheM)
Equilibrium constant is a CONSTANT (i.e. it doesn't change, as long as temperature doesn't change). If you add lots of product, then the system is no longer in equilibrium. Some product must turn into reactant in order for the system to reach equilibrium. The ratio of products to reactants at equilibrium will always reach the same value as it was before, as long as temperature didn't change.

Consider the extremely simple reaction A --> B
K = [B]/[A]

Suppose K = 2 at a temperature of 273K.

Then if [A] = 1 at equilibrium, [B] must = 2 because the constant K = 2 for that reaction at the given temperature.

Now we add some product, so [B] = 2+x

New ratio immediately after adding x product is (2+x):1. But we are not in equilibrium. The system will now spontaneously change itself so that the ratio [B]/[A] becomes K = 2 (assuming negligible temperature change).

(2+x)/1 > 2 = K, so some product has to turn into reactant (as predicted by le Chatelier's principle).
Once the system reaches equilibrium, the new ratio will be K = 2 again.

For example, if x = 1, then we would've had [B] = 3 and [A]=1 immediately after adding 1 unit of [B], but in order to get back to the 2:1 ratio, we need to solve (3-x)/(1+x) = 2
(since the reaction is A-->B, so for every x moles of A that reacts, x moles of B is formed).

(3-x) = 2+2x, so x = 1/3, hence the final equilibrium concentrations after adding one mole of [B] is

[A] = 1.33333...
[B] = 2.66666...

then K = [B]/[A] = 2, as required since K is constant at constant temperature.
PRSOM. Very well explained to the OP and others
0
6 months ago
#6
(Original post by pondering-soul)

Ok so I don't get this paragraph.
Instead of representing it as ethanol and ethanoic acid, I try to understand this by just writing it as 1/1 (and x being any change in concentration). If you make the bottom line of the expression larger, it is then 1/1+x .

To restore the situation, the concentration of the bottom line of the fraction is reduced as the reactants react to make more product. so remove the x from the bottom to show a decrease in reactants (now 1/1) and then add an x ontop to show increase in products (now 1+x/1).

so in my head, the final result of this shenanigan is 1+x/1 whereas the original was 1/1 showing that the fraction has changed.

?!??!

(thanks for any help)
To answer your question more specifically, suppose initial ratio is 1:1 and then we add reactant so ratio becomes 1/(1+x) as you said. Then y amount of reactant reacts to reach equilibrium. So the final ratio will be
(1+y)/(1+x-y) = 1

You would need to solve for y there to find out how much reacted to force the same ratio as before.
1
6 months ago
#7
(Original post by Reality Check)
PRSOM. Very well explained to the OP and others
0
#8
(Original post by K-Man_PhysCheM)
To answer your question more specifically, suppose initial ratio is 1:1 and then we add reactant so ratio becomes 1/(1+x) as you said. Then y amount of reactant reacts to reach equilibrium. So the final ratio will be
(1+y)/(1+x-y) = 1

You would need to solve for y there to find out how much reacted to force the same ratio as before.
ohh ok my mistake was assuming that (according to this) y is the same as x (the amount of reactant reacting to reach equilibrium is not the same as x, hence why you put y)

thnx bro
0
#9
(Original post by Ricco.af)
I'm a bit rusty with my chemistry as, well, covid19 (Y13), I think the error in your methodology comes to the fact that you assume the end value of concentrations is a complete conversion from the excess denominator (reactant = 1+x) to the numerator (products = 1+x) whereas the concentrations of the reactant will gradually transfer from 1+x ---> a value in which product production is in equilibrium to reactant backwards production as that will be the point at which equilibrium is met and thus the end value will mathematically be (1+(x/2)/1+(x/2))
I kept this in mathematical terms as I guess that's how you'd prefer, hopefully this makes sense.
yeah it does thanks alot
0
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