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    in the iodine clock reaction (tbh most reactions)

    H2O2 + 2I- +2H+ -----> I2 +2h2o

    2s2o3 (thiosulpahte) + I2 ------> 2I- + S4ob

    how is the amount of iodine produced in a measure time proportional to the volume of sodium thiosulphte used/taken and then rate = 1/T...?

    pleae someone explain in standard english
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    (Original post by Rimmiie)
    in the iodine clock reaction (tbh most reactions)

    H2O2 + 2I- +2H+ -----> I2 +2h2o

    2s2o3 (thiosulpahte) + I2 ------> 2I- + S4ob

    how is the amount of iodine produced in a measure time proportional to the volume of sodium thiosulphte used/taken and then rate = 1/T...?

    pleae someone explain in standard english
    As the iodine is produced it would become coloured (from colourless), BUT the thiosulphate ions present absorb the iodine as it comes into existence.

    This can only happen as long while there remains thiosulphate ions in the solution. At the instant all of the thiosulphate ions are used up (by the iodine), BANG, the iodine colour appears.

    The time taken for all of the thiosulphate to be used up is inversely proportional to the rate. This is because time for any procedure is inversely proportional to the rate of the procedure.

    Thnk about a car travelling from A to B. The longer the time taken the slower the car - this is inverse proportionality, i.e. Large Time = Small Rate
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    (Original post by charco)
    As the iodine is produced it would become coloured (from colourless), BUT the thiosulphate ions present absorb the iodine as it comes into existence.

    This can only happen as long while there remains thiosulphate ions in the solution. At the instant all of the thiosulphate ions are used up (by the iodine), BANG, the iodine colour appears.

    The time taken for all of the thiosulphate to be used up is inversely proportional to the rate. This is because time for any procedure is inversely proportional to the rate of the procedure.

    Thnk about a car travelling from A to B. The longer the time taken the slower the car - this is inverse proportionality, i.e. Large Time = Small Rate
    ohhhhh myyy! thankyou thankyouu you gorgeouss person!
    ill repp you after my 24hrs
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    (Original post by Rimmiie)
    ...
    Have you been reading A2 Chemistry by George Facer? I only ask because I was reading it this morning and found myself confused by a passage incredibly similar to that in your OP.
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    (Original post by Rimmiie)
    ohhhhh myyy! thankyou thankyouu you gorgeouss person!
    ill repp you after my 24hrs
    forget the rep..

    .. just send your youngest daughter :teeth:
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    As I understood it, the amount of iodine produced in the time it took for the solution to go blue (the measured time) is proportional to the volume of thiosulfate used, since the more thiosulfate used, the more iodine would need to be produced in order to react with all of it (which must occur before the iodine can start reacting with starch and produce the blue-black colour).

    Because the same volume of thiosulfate is used in each experiment, the same amount of iodine must be produced in each experiment in the time it takes for the solution to go blue (the measured time). If rate = change in concentration / time, and we know that the change in concentration (of iodine) is constant from experiment to experiment, then the rate must = k / time. In other words, rate is proportional to 1 / time.
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    (Original post by charco)
    forget the rep..

    .. just send your youngest daughter :teeth:
    haaahaa i would but im 18
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    (Original post by porkstein)
    Have you been reading A2 Chemistry by George Facer? I only ask because I was reading it this morning and found myself confused by a passage incredibly similar to that in your OP.
    yepp!
    he doesnt explain rates that well
 
 
 
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