# Pls help with Kc

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Can someone help me out with part c please? I don't really understand the answer which suggests that the colour will be 'more blue than before the addition of C'.... I wrote that the colour will at first changes from deep blue to colourless as the increase of conc of c will shift the position of equilibrium to the left, this produces more A and B. Since the reaction is still at room temperature, the temp is constant, Kc will be restored. Surely at this point, the colour will lighten and the final colour change should be light blue....

Which part have I gone wrong??

Thanks

Which part have I gone wrong??

Thanks

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(Original post by

Will the concentration of C increase or decrease overall if you add more?

**alow**)Will the concentration of C increase or decrease overall if you add more?

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#4

(Original post by

I think it will increase ?

**coconut64**)I think it will increase ?

_{c}you can see why this is.

So what will the colour change be?

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(Original post by

Yes, it will. If you write the equation for K

So what will the colour change be?

**alow**)Yes, it will. If you write the equation for K

_{c}you can see why this is.So what will the colour change be?

thx

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#6

(Original post by

from what I understand, if the conc of c increases, this means the conc of a and b decreases. Hence postion of equilibrium shifts the to left to make more A & B. This has no effect on Kc which is only affected by temperature change. So I reckon the colour will change from deep blue to colourless as more reactants are made.

thx

**coconut64**)from what I understand, if the conc of c increases, this means the conc of a and b decreases. Hence postion of equilibrium shifts the to left to make more A & B. This has no effect on Kc which is only affected by temperature change. So I reckon the colour will change from deep blue to colourless as more reactants are made.

thx

You add C, which means that some of this will revert to A + B (Le Chatelier). However as K

_{c}stays constant, and [A] and [B] have increased, what does this mean for [C]? Will it stay the same, increase or decrease?

Write the expression for K

_{c}.

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(Original post by

Not quite.

You add C, which means that some of this will revert to A + B (Le Chatelier). However as K

Write the expression for K

**alow**)Not quite.

You add C, which means that some of this will revert to A + B (Le Chatelier). However as K

_{c}stays constant, and [A] and [B] have increased, what does this mean for [C]? Will it stay the same, increase or decrease?Write the expression for K

_{c}.
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#8

(Original post by

I have written the expression. So I guess the conc of c will decrease as a result?

**coconut64**)I have written the expression. So I guess the conc of c will decrease as a result?

^{2}) but K

_{c}remains constant. What happens to the numerator ([C])?

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(Original post by

So the denominator increases ([A][B]

**alow**)So the denominator increases ([A][B]

^{2}) but K_{c}remains constant. What happens to the numerator ([C])?^{2}) increases ... No?

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#10

_{c}becomes smaller value/bigger value, which is smaller. But we know K

_{c}is constant.

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(Original post by

If [C] decreases K

**alow**)If [C] decreases K

_{c}becomes smaller value/bigger value, which is smaller. But we know K_{c}is constant.
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#12

(Original post by

I don't quite get this .... But I suppose what you are trying to say is that C remains the same?

**coconut64**)I don't quite get this .... But I suppose what you are trying to say is that C remains the same?

Right?

We know increases because of Le Chatelier's principle. We know that stays the same.

What must happen to for those two things to be true?

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(Original post by

Right?

We know increases because of Le Chatelier's principle. We know that stays the same.

What must happen to for those two things to be true?

**alow**)Right?

We know increases because of Le Chatelier's principle. We know that stays the same.

What must happen to for those two things to be true?

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#14

(Original post by

From what I know, the position of equilibrium shifts to the left, so more A and B are made. This means that more c is used up, meaning the conc of c will decrease...

**coconut64**)From what I know, the position of equilibrium shifts to the left, so more A and B are made. This means that more c is used up, meaning the conc of c will decrease...

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(Original post by

But you've added some C. Only some of this is used up to produce enough A and B to restore the equilibrium.

**alow**)But you've added some C. Only some of this is used up to produce enough A and B to restore the equilibrium.

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#16

(Original post by

Oh, I had no idea as I always assume that most of it would be used up.... But since some A and B are produced, surely the colour will lighten somehow....

**coconut64**)Oh, I had no idea as I always assume that most of it would be used up.... But since some A and B are produced, surely the colour will lighten somehow....

I'll give you an example: Take a cup of water. Add some table salt to it (colourless in solution) and some potassium permanganate to it (deep purple in solution). What colour change do you get?

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(Original post by

The volume of solvent doesn't change. All you're effectively doing is dissolving some colourless molecules and some strongly coloured molecules.

I'll give you an example: Take a cup of water. Add some table salt to it (colourless in solution) and some potassium permanganate to it (deep purple in solution). What colour change do you get?

**alow**)The volume of solvent doesn't change. All you're effectively doing is dissolving some colourless molecules and some strongly coloured molecules.

I'll give you an example: Take a cup of water. Add some table salt to it (colourless in solution) and some potassium permanganate to it (deep purple in solution). What colour change do you get?

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#18

(Original post by

light purple?

**coconut64**)light purple?

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(Original post by

Yes. So it would become more purple than before. Just like your problem, the solution becomes more blue than before, the addition of a colourless solute does not matter.

**alow**)Yes. So it would become more purple than before. Just like your problem, the solution becomes more blue than before, the addition of a colourless solute does not matter.

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#20

(Original post by

I said it will get lighter actually .... I guess I am wrong again since you have misread.... If you are putting a darker solution in water, how on earth will it get darker? So dark purple combine with water, I believe would be light purple right, not dark purple.... I am still lost

**coconut64**)I said it will get lighter actually .... I guess I am wrong again since you have misread.... If you are putting a darker solution in water, how on earth will it get darker? So dark purple combine with water, I believe would be light purple right, not dark purple.... I am still lost

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