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Size:  415.9 KBStruggling to understand how to do these types of Kc questions
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    (Original post by AqsaMx)
    Name:  image.jpeg
Views: 146
Size:  415.9 KBStruggling to understand how to do these types of Kc questions
    Ok this is a mouthful of a question.

    I'm pretty confident you know how to calculate the initial moles of the acid and the alcohol. Divide Mass by the Mr, all the jazz.

    Spoiler:
    Show
    The moles of both should be 0.1 mol.

    The tricky part is understanding how to get the reacted moles, or 'changed' moles in your RICE table.

    The only way to calculate the reacted moles is to use that information about the NaOH neutralisation.

    "The mixture" is going to have all those chemicals INCLUDING the H2SO4 catalyst. Even though it was a catalyst and is not going to appear in our Kc calculation, it's part of the mixture.

    What is the NaOH going to react with. The acids; your ethanoic acid and sulfuric acid (which I think is a nasty trick to catch people out).

    You can figure out the moles of NaOH that was required to react with those acids.
    Spoiler:
    Show
    The answer is 0.035 mol of NaOH.

    You're also told that we used 0.001 mol of H2SO4 . Since it's a catalyst we can assume the moles have been unchanged.
    You can write out the equation for H2SO4 and NaOH to determine the mole ratio. Alternatively, you can realise that H2SO4 is a diprotic acid, so you're going to need 2 NaOH for every 1 H2SO4.

    So if you have 0.001 mol of H2SO4 then ratios tells us that the moles of NaOH you'll need is 0.001 x 2 = 0.002 mol.

    So what that's saying is, of the 0.035 mol of NaOH that we put into the mixture, 0.002 moles of that reacted with H2SO4.

    Therefore, the remaining moles must have reacted with CH3COOH.
    Spoiler:
    Show
    i.e. 0.035 - 0.002 = 0.033 mol
    This tells us that after the esterfication reaction, there was 0.033 mol remaining. Which means, the changed moles must have been 0.1 - 0.033 = 0.067 mol.

    From this point onwards, this becomes a simple RICE table and Kc Calculation.

    ANSWER:
    Spoiler:
    Show
    If you get an answer of 4.12 no units, then congratulations.
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    (Original post by RMNDK)
    Ok this is a mouthful of a question.

    I'm pretty confident you know how to calculate the initial moles of the acid and the alcohol. Divide Mass by the Mr, all the jazz.
    Spoiler:
    Show
    The moles of both should be 0.1 mol.
    The tricky part is understanding how to get the reacted moles, or 'changed' moles in your RICE table.

    The only way to calculate the reacted moles is to use that information about the NaOH neutralisation.

    "The mixture" is going to have all those chemicals INCLUDING the H2SO4 catalyst. Even though it was a catalyst and is not going to appear in our Kc calculation, it's part of the mixture.

    What is the NaOH going to react with. The acids; your ethanoic acid and sulfuric acid (which I think is a nasty trick to catch people out).

    You can figure out the moles of NaOH that was required to react with those acids.
    Spoiler:
    Show
    The answer is 0.035 mol of NaOH.
    You're also told that we used 0.001 mol of H2SO4 . Since it's a catalyst we can assume the moles have been unchanged.
    You can write out the equation for H2SO4 and NaOH to determine the mole ratio. Alternatively, you can realise that H2SO4 is a diprotic acid, so you're going to need 2 NaOH for every 1 H2SO4.

    So if you have 0.001 mol of H2SO4 then ratios tells us that the moles of NaOH you'll need is 0.001 x 2 = 0.002 mol.

    So what that's saying is, of the 0.035 mol of NaOH that we put into the mixture, 0.002 moles of that reacted with H2SO4.

    Therefore, the remaining moles must have reacted with CH3COOH.
    Spoiler:
    Show
    i.e. 0.035 - 0.002 = 0.033 mol
    This tells us that after the esterfication reaction, there was 0.033 mol remaining. Which means, the changed moles must have been 0.1 - 0.033 = 0.067 mol.
    From this point onwards, this becomes a simple RICE table and Kc Calculation.

    ANSWER:
    Spoiler:
    Show
    If you get an answer of 4.12 no units, then congratulations.
    Just wondering if like you said after figure out the equilibrium moles of the reaction. How do you calculate concentration as the volume of container is not mentioned in the question?
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    (Original post by Basdemraj)
    Just wondering if like you said after figure out the equilibrium moles of the reaction. How do you calculate concentration as the volume of container is not mentioned in the question?
    Ahhh good question, you don't need to!

    Here's why.

    You agree that for this particular Kc:

     \mathrm{Kc} = \dfrac{[\mathrm{CH}_3 \mathrm{COOC}_3 \mathrm{H}_7]\times [\mathrm{H}_2 \mathrm{O} ]}{[\mathrm{CH}_3 \mathrm{COOH}] \times [\mathrm{C}_3 \mathrm{H}_7 \mathrm{OH}]}


    Now, as you correctly mentioned, you need volume to calculate concentration:

    i.e.  \mathrm{Conc.} = \dfrac{\mathrm{Moles}}{\mathrm{V  olume (dm^3)}}

    Since we don't know the volume of the container, let's say it's just V dm3

    Now the moles of CH3COOH and C3H7OH are both 0.033 mol.
    The moles of CH3COOC3H7OH and H2O are both 0.067 mol.

    Therefore, I'm going to plug all our values into our Kc, like this.

     \mathrm{Kc} = \dfrac{ \dfrac{0.067}{V} \times \dfrac{0.067}{V} } {\dfrac{0.033}{V} \times \dfrac{0.033}{V}}
    Notice how you're cancelling all the V's out? If you don't see it, here's a perhaps clearer way to see it in the spoiler:
    Spoiler:
    Show
    I rewrite it as:

     \mathrm{Kc} = \dfrac{\dfrac{0.067^2}{V^2}}{ \dfrac{0.033^2}{V^2}}

    Which, in a nicer form, is:

     \mathrm{Kc} = \dfrac{0.067^2}{V^2}\div \dfrac{0.033^2}{V^2}

    And of course, to divide, you flip it and multiply, like so:

     \mathrm{Kc} = \dfrac{0.067^2}{V^2}\times \dfrac{V^2}{0.033^2}

    Now you can see that the  V^2 cancel out.

    So because they cancel out and don't appear in the equation, we don't even need to use in the first place. We can just use the moles!




    Remember: if the moles on both sides of the equation are the same, we don't need volume.

    Hope that makes sense!
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    (Original post by RMNDK)
    Ahhh good question, you don't need to!

    Here's why.

    You agree that for this particular Kc:

     \mathrm{Kc} = \dfrac{[\mathrm{CH}_3 \mathrm{COOC}_3 \mathrm{H}_7]\times [\mathrm{H}_2 \mathrm{O} ]}{[\mathrm{CH}_3 \mathrm{COOH}] \times [\mathrm{C}_3 \mathrm{H}_7 \mathrm{OH}]}


    Now, as you correctly mentioned, you need volume to calculate concentration:

    i.e.  \mathrm{Conc.} = \dfrac{\mathrm{Moles}}{\mathrm{V  olume (dm^3)}}

    Since we don't know the volume of the container, let's say it's just V dm3

    Now the moles of CH3COOH and C3H7OH are both 0.033 mol.
    The moles of CH3COOC3H7OH and H2O are both 0.067 mol.

    Therefore, I'm going to plug all our values into our Kc, like this.

     \mathrm{Kc} = \dfrac{ \dfrac{0.067}{V} \times \dfrac{0.067}{V} } {\dfrac{0.033}{V} \times \dfrac{0.033}{V}}
    Notice how you're cancelling all the V's out? If you don't see it, here's a perhaps clearer way to see it in the spoiler:
    Spoiler:
    Show
    I rewrite it as:

     \mathrm{Kc} = \dfrac{\dfrac{0.067^2}{V^2}}{ \dfrac{0.033^2}{V^2}}

    Which, in a nicer form, is:

     \mathrm{Kc} = \dfrac{0.067^2}{V^2}\div \dfrac{0.033^2}{V^2}

    And of course, to divide, you flip it and multiply, like so:

     \mathrm{Kc} = \dfrac{0.067^2}{V^2}\times \dfrac{V^2}{0.033^2}

    Now you can see that the  V^2 cancel out.
    So because they cancel out and don't appear in the equation, we don't even need to use in the first place. We can just use the moles!




    Remember: if the moles on both sides of the equation are the same, we don't need volume.

    Hope that makes sense!
    Thank you so much!! That makes complete sense! I'm new to TSR so it's great seeing such helpful people explain it so well! Thanks again!
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    (Original post by RMNDK)
    Ok this is a mouthful of a question.

    I'm pretty confident you know how to calculate the initial moles of the acid and the alcohol. Divide Mass by the Mr, all the jazz.
    Spoiler:
    Show
    The moles of both should be 0.1 mol.
    The tricky part is understanding how to get the reacted moles, or 'changed' moles in your RICE table.

    The only way to calculate the reacted moles is to use that information about the NaOH neutralisation.

    "The mixture" is going to have all those chemicals INCLUDING the H2SO4 catalyst. Even though it was a catalyst and is not going to appear in our Kc calculation, it's part of the mixture.

    What is the NaOH going to react with. The acids; your ethanoic acid and sulfuric acid (which I think is a nasty trick to catch people out).

    You can figure out the moles of NaOH that was required to react with those acids.
    Spoiler:
    Show
    The answer is 0.035 mol of NaOH.
    You're also told that we used 0.001 mol of H2SO4 . Since it's a catalyst we can assume the moles have been unchanged.
    You can write out the equation for H2SO4 and NaOH to determine the mole ratio. Alternatively, you can realise that H2SO4 is a diprotic acid, so you're going to need 2 NaOH for every 1 H2SO4.

    So if you have 0.001 mol of H2SO4 then ratios tells us that the moles of NaOH you'll need is 0.001 x 2 = 0.002 mol.

    So what that's saying is, of the 0.035 mol of NaOH that we put into the mixture, 0.002 moles of that reacted with H2SO4.

    Therefore, the remaining moles must have reacted with CH3COOH.
    Spoiler:
    Show
    i.e. 0.035 - 0.002 = 0.033 mol
    This tells us that after the esterfication reaction, there was 0.033 mol remaining. Which means, the changed moles must have been 0.1 - 0.033 = 0.067 mol.
    From this point onwards, this becomes a simple RICE table and Kc Calculation.

    ANSWER:
    Spoiler:
    Show
    If you get an answer of 4.12 no units, then congratulations.
    Thank you so much!! You made it so simple to understand!
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    (Original post by RMNDK)
    Ok this is a mouthful of a question.

    I'm pretty confident you know how to calculate the initial moles of the acid and the alcohol. Divide Mass by the Mr, all the jazz.
    Spoiler:
    Show
    The moles of both should be 0.1 mol.
    The tricky part is understanding how to get the reacted moles, or 'changed' moles in your RICE table.

    The only way to calculate the reacted moles is to use that information about the NaOH neutralisation.

    "The mixture" is going to have all those chemicals INCLUDING the H2SO4 catalyst. Even though it was a catalyst and is not going to appear in our Kc calculation, it's part of the mixture.

    What is the NaOH going to react with. The acids; your ethanoic acid and sulfuric acid (which I think is a nasty trick to catch people out).


    You can figure out the moles of NaOH that was required to react with those acids.
    Spoiler:
    Show
    The answer is 0.035 mol of NaOH.
    You're also told that we used 0.001 mol of H2SO4 . Since it's a catalyst we can assume the moles have been unchanged.
    You can write out the equation for H2SO4 and NaOH to determine the mole ratio. Alternatively, you can realise that H2SO4 is a diprotic acid, so you're going to need 2 NaOH for every 1 H2SO4.

    So if you have 0.001 mol of H2SO4 then ratios tells us that the moles of NaOH you'll need is 0.001 x 2 = 0.002 mol.

    So what that's saying is, of the 0.035 mol of NaOH that we put into the mixture, 0.002 moles of that reacted with H2SO4.

    Therefore, the remaining moles must have reacted with CH3COOH.
    Spoiler:
    Show
    i.e. 0.035 - 0.002 = 0.033 mol
    This tells us that after the esterfication reaction, there was 0.033 mol remaining. Which means, the changed moles must have been 0.1 - 0.033 = 0.067 mol.
    From this point onwards, this becomes a simple RICE table and Kc Calculation.

    ANSWER:
    Spoiler:
    Show
    If you get an answer of 4.12 no units, then congratulations.
    Name:  image.jpg
Views: 79
Size:  318.6 KB You see for this question, would you do the same method but multiply by 5 to get the no of moles in 25cm3 and continue from there?
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    (Original post by AqsaMx)
    Name:  image.jpg
Views: 79
Size:  318.6 KB You see for this question, would you do the same method but multiply by 5 to get the no of moles in 25cm3 and continue from there?
    Correctly spotted.
 
 
 
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