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    CH3COOH + C2H5OH >>>>>>>>>> CH3COOC2H5 + H20

    30g of ethanoic acid and 23g of ethanol were mixed and allowed to reach EQUiLIBRIUM at 60 degrees. At EQUiLIBRIUM the number of moles of ester was 0.33.
    Calculate Kc.
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    (Original post by Lets get close...)
    CH3COOH + C2H5OH >>>>>>>>>> CH3COOC2H5 + H20

    30g of ethanoic acid and 23g of ethanol were mixed and allowed to reach EQUiLIBRIUM at 60 degrees. At EQUiLIBRIUM the number of moles of ester was 0.33.
    Calculate Kc.
    First of all this forum is for discussion of the uni courses not academic help, look here:
    http://www.thestudentroom.co.uk/forumdisplay.php?f=130
    for that.

    ok now the question, you should know that

    $

K_c=\frac{[Ester][Water]}{[Alcohol][Acid]}

    as Kc is unit-less the volume of liquid can be ignored as it will have no effect on the outcome, so we can use the number of moles at equilibrium rather than the concs of each of the substances.

    The RMM of Ethanoic acid is 60 g mol^{-1} so at the start we have \frac{30g}{60 g mol^{-1}}=0.5mol of acid.

    There is a 1:1 ratio of acid to ester so at equilibrium there will be 0.5-0.33=0.17mol of the acid.

    Using a similar argument for the alcohol gives us also 0.17mol of alcohol at equilibrium.

    For the water we recognise that for every mole of ester that forms a mole of water also forms, therefore there will be 0.33 moles of water at equilibrium

    plugging all these values into our expression for Kc gives us

    K_c=\frac{0.33\times0.33}{0.17\t  imes0.17}=3.77 (3sf)

    ask if you have any questions about this
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    (Original post by thelostchild)
    First of all this forum is for discussion of the uni courses not academic help, look here:
    http://www.thestudentroom.co.uk/forumdisplay.php?f=130
    for that.

    ok now the question, you should know that

    $

K_c=\frac{[Ester][Water]}{[Alcohol][Acid]}

    as Kc is unit-less the volume of liquid can be ignored as it will have no effect on the outcome, so we can use the number of moles at equilibrium rather than the concs of each of the substances.

    The RMM of Ethanoic acid is 60 g mol^{-1} so at the start we have \frac{30g}{60 g mol^{-1}}=0.5mol of acid.
    There is a 1:1 ratio of acid to ester so at equilibrium there will be 0.5-0.33=0.17mol of the acid.

    Using a similar argument for the alcohol gives us also 0.17mol of alcohol at equilibrium.

    For the water we recognise that for every mole of ester that forms a mole of water also forms, therefore there will be 0.33 moles of water at equilibrium

    plugging all these values into our expression for Kc gives us

    K_c=\frac{0.33\times0.33}{0.17\t  imes0.17}=3.77 (3sf)

    ask if you have any questions about this
    can u go over that highlighted bit please i dont get the bit the 60gmol
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    (Original post by Lets get close...)
    can u go over that highlighted bit please i dont get the bit the 60gmol
    ok you know what the Relative Molecular Mass of a molecule is right (essentially the mass of 1 mole of it where 1 mole is number of atoms in \frac{1}{12} of a gram of Carbon-12)

    using a periodic table we can work out the RMM of Ethanoic acid from its molecular formula (C_{2}H_4O_2)

    from the periodic table we can see the molecular weight of Carbon is 12, Hydrogen is 1 and Oxygen is 16

    so the RMM of Ethanoic acid will be
    (12 \times 2) + (1 \times 4) + (16 \times 2)=60

    the g mol^{-1} is just the unit of the RMM (its read as grams per mole) meaning you would have that mass of compound if you had 1 mole of it

    for the last bit you just need to remember that \text{Moles}=\frac{\text{Mass}}{  \text{RMM}}


    hope that helps
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    (Original post by thelostchild)
    ok you know what the Relative Molecular Mass of a molecule is right (essentially the mass of 1 mole of it where 1 mole is number of atoms in \frac{1}{12} of a gram of Carbon-12)

    using a periodic table we can work out the RMM of Ethanoic acid from its molecular formula (C_{2}H_4O_2)

    from the periodic table we can see the molecular weight of Carbon is 12, Hydrogen is 1 and Oxygen is 16

    so the RMM of Ethanoic acid will be
    (12 \times 2) + (1 \times 4) + (16 \times 2)=60

    the g mol^{-1} is just the unit of the RMM (its read as grams per mole) meaning you would have that mass of compound if you had 1 mole of it

    for the last bit you just need to remember that \text{Moles}=\frac{\text{Mass}}{  \text{RMM}}


    hope that helps
    ohh yh i didnt know wt you RMM stood for thanks alot....u got msn by any chance i could use your wisdom on there lol if thats ok....
 
 
 
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