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    Hi, would anybody be able to help me try to figure this out?

    "Work out the equilibrium constant for the reaction of glacial ethanoic acid with ethanol, given that 60g of the acid mixed with 46g ethanol produces (at equilibrium) 58.7g ethyl ethanoate and 12g water..."

    I have worked out the equation as

    CH3COOH + CH3CH2OH ----> CH3COOCH2CH3 + H20 (is that right?)

    I have been given the answer (it's 1.33333333) but could anyone help me understand how to work it out?

    Thank you so much.
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    (Original post by Silver_Stardust)
    Hi, would anybody be able to help me try to figure this out?

    "Work out the equilibrium constant for the reaction of glacial ethanoic acid with ethanol, given that 60g of the acid mixed with 46g ethanol produces (at equilibrium) 57.8g ethyl ethanoate and 12g water..."

    I have worked out the equation as

    CH3COOH + CH3CH2OH ----> CH3COOCH2CH3 + H20 (is that right?)

    I have been given the answer (it's 1.33333333) but could anyone help me understand how to work it out?

    Thank you so much.
    You sure that's right, from my working out I get 3.77
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    (Original post by T kay)
    You sure that's right, from my working out I get 3.77

    Ahhh sorry, i've just checked the question, It's actually 58.7g ethyl ethanoate... (i'll just go and edit my 1st post!) does that come out right now?

    How do you go about working it out? Thanks for your help!
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    The answer is 4 (you can obtain 1.333 if you absorb the [H2O] into the Kc, but since there is no water to start with (i.e. not in high excess) you really shouldn't)
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    (Original post by EierVonSatan)
    The answer is 4 (you can obtain 1.333 if you absorb the [H2O] into the Kc, but since there is no water to start with (i.e. not in high excess) you really shouldn't)
    So the real answer is 4? Thank you! How do you work it out? I don't even know where to start!
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    (Original post by Silver_Stardust)
    Ahhh sorry, i've just checked the question, It's actually 58.7g ethyl ethanoate... (i'll just go and edit my 1st post!) does that come out right now?

    How do you go about working it out? Thanks for your help!
    That doesn't make a big difference, I get 4.12.

    I'll go through what I did:

    First we need to know the moles of the reactants at the start... (moles = mass/Mr

    For the acid: 60/60 = 1
    For the alcohol: 46/46 = 1


    Then we work out the equilibrium moles of the reactant...

    For the ether: 58.7/88 = 0.67
    For the water: 12/18 = 0.67


    Now you work out the moles at equilibrium for the reactants. (1-x) x being the equilibrium moles of the product:

    For the acid: 1 - 0.67 = 0.33
    For the alcohol: 1 - 0.67 = 0.33


    Now to calculate the constant: Kc = products/reactants

    Kc=0.67(squared)/0.33(squared)

    Kc= 4.12
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    (Original post by Silver_Stardust)
    So the real answer is 4? Thank you! How do you work it out? I don't even know where to start!
    ^^ T kay's method is the same as mine
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    (Original post by Silver_Stardust)
    So the real answer is 4? Thank you! How do you work it out? I don't even know where to start!
    4.12

    Working out above
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    (Original post by T kay)
    4.12

    Working out above
    You're method is spot on but you're carrying through an error:

    x = 12/18 = 2/3
    1-x = 1/3

    Kc = [2/3]^2/[1/3]^2 = 4

    only round at the end of a calculation
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    (Original post by T kay)
    That doesn't make a big difference, I get 4.12.

    I'll go through what I did:

    First we need to know the moles of the reactants at the start... (moles = mass/Mr

    For the acid: 60/60 = 1
    For the alcohol: 46/46 = 1


    Then we work out the equilibrium moles of the reactant...

    For the ether: 58.7/88 = 0.67
    For the water: 12/18 = 0.67


    Now you work out the moles at equilibrium for the reactants. (1-x) x being the equilibrium moles of the product:

    For the acid: 1 - 0.67 = 0.33
    For the alcohol: 1 - 0.67 = 0.33


    Now to calculate the constant: Kc = products/reactants

    Kc=0.67(squared)/0.33(squared)

    Kc= 4.12

    Thank you so so so so much!! That's so helpful!

    I just have 2 small questions.... just so that i know i understand it properly...

    When you work out the moles at eqilibrium for the reactants, you did (1-x) Was the 1 the moles of reactants at the start? (i.e. if you had 2 moles of ethanol it would be 2-x)?

    Also, when you are doing this part of the question how do you know which product to minus from each reactant??
    e.g. for the acid it was 1-0.67, but was it the 0.67 from the ether or the 0.67 from the water?? Does my question make sense? Sorry if it's a really stupid question!

    Thankyou again for your help!
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    (Original post by EierVonSatan)
    Thanks

    You're right you should give your final answer to three significant figures.
    Ah this helps as well for when i do calculations, thank you as well for your help!!!
    As you can see, calculations in chemistry are not my strong point :p:
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    (Original post by Silver_Stardust)
    Ah this helps as well for when i do calculations, thank you as well for your help!!!
    As you can see, calculations in chemistry are not my strong point :p:
    welcome - it's just practice, I'm sure you'll get the hang of it
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    (Original post by Silver_Stardust)
    Thank you so so so so much!! That's so helpful!

    I just have 2 small questions.... just so that i know i understand it properly...

    When you work out the moles at eqilibrium for the reactants, you did (1-x) Was the 1 the moles of reactants at the start? (i.e. if you had 2 moles of ethanol it would be 2-x)?

    Also, when you are doing this part of the question how do you know which product to minus from each reactant??
    e.g. for the acid it was 1-0.67, but was it the 0.67 from the ether or the 0.67 from the water?? Does my question make sense? Sorry if it's a really stupid question!

    Thankyou again for your help!
    Yes the one is the moles at the start, but there's no such thing as 2-x if there's two moles. (I think)

    As for the second question, on all of the acid + alcohol = ether + water the values have been the same, so it doesn't matter really.
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    (Original post by Silver_Stardust)
    Ah this helps as well for when i do calculations, thank you as well for your help!!!
    As you can see, calculations in chemistry are not my strong point :p:
    Are you with AQA?

    If so, wait until you get to acids and bases! To be honest, I found it easy enough but there's soooo much types of question, calculations etc it's so easy to get mixed up.
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    (Original post by T kay)
    Are you with AQA?

    If so, wait until you get to acids and bases! To be honest, I found it easy enough but there's soooo much types of question, calculations etc it's so easy to get mixed up.
    Hehe, it's actully kind of embarassing but... i'm 1st year uni! All through A level i managed to ignore calculations and i shomehow still passed and got into uni... however, now its got the the point where i can't ignore them any more! I may be on here quite a bit asking for help

    p.s. i was with OCR for a level
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    (Original post by Silver_Stardust)
    Hehe, it's actully kind of embarassing but... i'm 1st year uni! All through A level i managed to ignore calculations and i shomehow still passed and got into uni... however, now its got the the point where i can't ignore them any more! I may be on here quite a bit asking for help

    p.s. i was with OCR for a level
    Same here! I'm going to post some questions that I've struggled with later.
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    (Original post by T kay)
    Same here! I'm going to post some questions that I've struggled with later.
    Yeah, it's really useful to be able to get help straight away from people who know their stuff! Well if you need any help with non-calculation related chemistry then i might be able to help
 
 
 
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