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Pls help with Kc Watch

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Size:  112.5 KB 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
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    Will the concentration of C increase or decrease overall if you add more?
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    (Original post by alow)
    Will the concentration of C increase or decrease overall if you add more?
    I think it will increase ?
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    (Original post by coconut64)
    I think it will increase ?
    Yes, it will. If you write the equation for Kc you can see why this is.

    So what will the colour change be?
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    (Original post by alow)
    Yes, it will. If you write the equation for Kc you can see why this is.

    So what will the colour change be?
    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
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    (Original post by 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
    Not quite.

    You add C, which means that some of this will revert to A + B (Le Chatelier). However as Kc 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 Kc.
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    (Original post by alow)
    Not quite.

    You add C, which means that some of this will revert to A + B (Le Chatelier). However as Kc 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 Kc.
    I have written the expression. So I guess the conc of c will decrease as a result?
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    (Original post by coconut64)
    I have written the expression. So I guess the conc of c will decrease as a result?
    So the denominator increases ([A][B]2) but Kc remains constant. What happens to the numerator ([C])?
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    (Original post by alow)
    So the denominator increases ([A][B]2) but Kc remains constant. What happens to the numerator ([C])?
    C has to decrease as ([A][B]2) increases ... No?
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    (Original post by coconut64)
    C has to decrease as ([A][B]2) increases ... No?
    If [C] decreases Kc becomes smaller value/bigger value, which is smaller. But we know Kc is constant.
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    (Original post by alow)
    If [C] decreases Kc becomes smaller value/bigger value, which is smaller. But we know Kc is constant.
    I don't quite get this .... But I suppose what you are trying to say is that C remains the same?
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    (Original post by coconut64)
    I don't quite get this .... But I suppose what you are trying to say is that C remains the same?
    K_c = \dfrac{[C]}{[A][B]^2}

    Right?

    We know [A][B]^2 increases because of Le Chatelier's principle. We know that K_c stays the same.

    What must happen to [C] for those two things to be true?
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    (Original post by alow)
    K_c = \dfrac{[C]}{[A][B]^2}

    Right?

    We know [A][B]^2 increases because of Le Chatelier's principle. We know that K_c stays the same.

    What must happen to [C] for those two things to be true?
    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 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...
    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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    (Original post by alow)
    But you've added some C. Only some of this is used up to produce enough A and B to restore the equilibrium.
    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....
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    (Original post by 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....
    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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    (Original post by 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?
    light purple?
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    (Original post by coconut64)
    light purple?
    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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    (Original post by 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.
    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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    (Original post by 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
    The water was initially colourless. You said it goes purple, which is more purple than colourless. We don't care about the colour of the solid/liquid (of potassium permanganate or C) we're adding, we care about the colour of the solution which we are adding it to.
 
 
 
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