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Carboxyllic Acids Formation PLease help exam tomorrow watch

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    ah ok, so it says on my spec;

    Recall that carboxyllic acids can also be formed by acid or base catalysed hydrolysis of esters & nitriles

    I fully understand esters.

    never heard of hydrolysis of nitriles??!!!

    Please what does this mean??
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    This page explains it pretty clearly: http://www.chemguide.co.uk/organicpr...ydrolysis.html
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    (Original post by Kyri)
    This page explains it pretty clearly: http://www.chemguide.co.uk/organicpr...ydrolysis.html
    thanks for coming to my rescue again !! oh btw my other chem exam went good the other day. everything i asked on tsr came up !!

    have big module 4 tomorrow.. eeks long nyt tonight !!!Yawn***

    oh kyri, when you are working out experimentally the rate of a reaction. you extrapolate a graph. and then plot rate against time. im finding it very hard to extrapolate... how do you do it.. like where do you draw your triangles to get the gradient?
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    (Original post by MedAmy)
    thanks for coming to my rescue again !! oh btw my other chem exam went good the other day. everything i asked on tsr came up !!

    have big module 4 tomorrow.. eeks long nyt tonight !!!Yawn***

    oh kyri, when you are working out experimentally the rate of a reaction. you extrapolate a graph. and then plot rate against time. im finding it very hard to extrapolate... how do you do it.. like where do you draw your triangles to get the gradient?
    Y2-Y1/X2-X1 = Gradient
    I never bothered with this triangle bs lol

    I'd recommend you sleep, lack of sleep for revision does more harm than good, I say this from personal experience, I'd have a strong A in chemistry if it weren't for getting only 4 hours of sleep lol
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    (Original post by xSkyFire)
    Y2-Y1/X2-X1 = Gradient
    I never bothered with this triangle bs lol

    I'd recommend you sleep, lack of sleep for revision does more harm than good, I say this from personal experience, I'd have a strong A in chemistry if it weren't for getting only 4 hours of sleep lol
    oh but where do you take this ?? im so confused..

    you have a curve.. with time against conc.. so what is first step??
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    (Original post by MedAmy)
    oh but where do you take this ?? im so confused..

    you have a curve.. with time against conc.. so what is first step??
    You just take a tangent to the curve, it's pretty much a line of best fit for the hypotenuse of your triangle that touches the curve, I'd usually do it somewhere in the middle tbh and then take the gradient. This I assume is for first order/second order reactions right ?
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    Glad to hear that the stuff you asked on TSR came up . It's nice to see the time people spend on here actually pays off. Good luck with your next one.

    I assume you're looking for the initial rate of the reaction. That is the rate of the reaction at time = 0 seconds. To do this, you need to draw the tangent of the line at time = 0 seconds. If you're not sure what that means, imagine that the line didn't curve but actually went straight down. You need to put your ruler against the curve at the straight part right at the beginning and draw out this straight line which just touches the curve at that point. You then need to find the gradient of this straight line (I assume you know how to do that), which effectively gives you the rate of change of concentration with respect to time at time = 0 seconds. This will be your rate of reaction in mol dm-3 s-1.
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    (Original post by xSkyFire)
    You just take a tangent to the curve, it's pretty much a line of best fit for the hypotenuse of your triangle that touches the curve, I'd usually do it somewhere in the middle tbh and then take the gradient. This I assume is for first order/second order reactions right ?
    ye 0, 1, 2.

    but that will only give you gradient at one point??any which point??sorry i HATE THIS ********
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    (Original post by Kyri)
    Glad to hear that the stuff you asked on TSR came up . It's nice to see the time people spend on here actually pays off. Good luck with your next one.

    I assume you're looking for the initial rate of the reaction. That is the rate of the reaction at time = 0 seconds. To do this, you need to draw the tangent of the line at time = 0 seconds. If you're not sure what that means, imagine that the line didn't curve but actually went straight down. You need to put your ruler against the curve at the straight part right at the beginning and draw out this straight line which just touches the curve at that point. You then need to find the gradient of this straight line (I assume you know how to do that), which effectively gives you the rate of change of concentration with respect to time at time = 0 seconds. This will be your rate of reaction in mol dm-3 s-1.
    thanks... il give it a try..
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    (Original post by MedAmy)
    ye 0, 1, 2.

    but that will only give you gradient at one point??any which point??sorry i HATE THIS ********
    Oh right, yea you take the tangent from T = 0 for initial rate if that's what you're looking for as stated above ^^
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    That is correct. The point you take the tangent from will give you the rate of reaction at that particular time only. If you want the average rate of reaction if you want that is basically the (concentration at the end - concentration at the start) / change in time. On the graph, that is the same as joining the start and end of the curve with a straight line and finding the gradient of that. When I did my A levels they always asked about initial rates but it's probably changed now.
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    I 've got this tomorrow as well. I'm just glad none of my offers depend on Chemistry so it's going to be for fun.
 
 
 
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