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Bemused

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  1. Funnily enough I thought that was exactly what I did - what I wrote is analogous to the spreadsheet you (kindly) provided. Much obliged.
  2. Thanks for your efforts, so my initial post was essentially correct, the rate of reaction for two reacting species (assuming not zero order for either) will change according to adding different volumes of fixed concentration. If the initial concentration of the reactant whose volume is held constant is larger then that of the reactant you add in increasing volume the rate will decrease and vice-versa.
  3. "If the concentration is fixed (ie constant) and the rate is proportional to the concentration how can the rate vary" Maybe I'm not explaining myself well. I am starting with fixed concentrations and then adding varying volumes of one to a fixed volume of the other. Since the volumes are changing then the concentrations of the reactants must also be changing according to how I tried to describe it above. Maybe I'm not framing the problem well or not following something in your answers to date.
  4. Ok so assuming the reaction rate is 1st order with respect to both reactants then the rate should vary according to how I have described it in the original post right?
  5. Well increasing concentration increases the number of reactant particles, increasing the number that have the required activation energy and increasing the frequency of collisions per unit volume. The rate is a constant times the concentrations raised to powers dependent on the reaction mechanism so is therefore proportional to the concentration. Could you elucidate why using the example I provided? In the first case the concentration of reactants (moles/unit volume) increases which means the rate should increase. When reversing the initial concentrations the rate decrease with increasing volume. Thanks.
  6. It's obvious why increasing the concentration of reactants increases rate. What if you add simply more of a given reactant at a fixed concentration? Should this also increase the rate? Example: To 20ml of 0.1M A make three tests with 5ml, 10ml and 15ml of 1M B Case 1: 2 x10-3 moles A and 5 x 10-3 moles B in 25ml = 0.28 moles reactants/L Case 2: 2x10-3 moles A and 0,01moles B in 30 ml = 0.4 moles reactant/L Case 3: 2x 10-3 moles A and 0.015moles B in 35ml = 0.485 moles reactant/L So since we have more reactant moles/L as we add volume of reactant B at a fixed concentration the rate should increase, right? Actually just thought, if you reversed the concentrations i.e. 20ml of 1M A and add 5ml, 10ml, 15ml of 0.1M B the rate should decrease as the reactants/L decrease when adding more volume. So rate will either increase or decrease on adding more volume of a given reactant depending on relative concentrations, right?
  7. Does your analogy with a measuring tape not break down though in that the "error" is only one-way and fixed? Each burette measurement could be on either side of the true measurement and to a different degree. Hence the tolerance can influence the measurement. Also, thinking about glassware with just a single mark, you mentioned in your initial post that tolerance affects accuracy while reading resolution affects precision. Taking the case of a pipette though (say 25ml and +/-0.05ml) you would report the measurement as 25.00 +/- 0.05ml right? In this case the tolerance also defines or at least bounds the precision, no? Finally, can you confirm that the tolerance listed on a burette refers to the delivery of the total volume e.g. 50ml +/- 0.1ml (for Class B) and not to each individual delivery e.g. 1ml +/- 0.1ml? Appreciate your help with this.
  8. Actually, thinking about this some more, the tolerance must refer to the delivery of the total volume (50ml), right? Otherwise delivery of small volumes, say 0.1ml would have 100% uncertainty! If the tolerance applies over the delivery of the full volume then the uncertainty per division is 0.1/500 = 0.0002ml for a class B and 0.05/500=0.0001ml for Class A.
  9. Thanks Studiot, So the measurement ranges I gave in my example for a Class A or B burette are correct then? Or does the printed tolerance on a burette only refer to the full scale nominal delivery you mention above (going from 0 to 50)? Essentially what I am asking is how would you report a titre result for a class A (+/-0.05 ml) and class B (+/-0.1ml) burette.
  10. Hi all, Class A burettes (50ml) have a stated tolerance of +/- 0.05ml while class B are +/- 0.1ml. What are the practical implications of this in recording measurements? Typical practice is to estimate a second decimal place which seems wrong with class B burettes which are most commonly in use in schools - a titre would be recorded as e.g. 25.55 +/- 0.1 ml. Here the number of significant digits are outside the tolerance or is the actual reading to be taken as somewhere between 25.45 and 25.65? In the case of a class A then this would be between 25.50 and 25.60. In essence, it seems wrong to record more significant digits than the tolerance allows - this would be rejected in engineering. I'd appreciate thoughts, thanks.
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