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Straightfoward chemistry question.

  1. Mar 11, 2005 #1

    ash

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    Could someone please explain to me why it is valid to use the inverse of time as a measure of the rate of a reaction? Thanks for any help!
     
  2. jcsd
  3. Mar 11, 2005 #2

    Borek

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    Any spped has time in denominator, they differ by numerators only. And numerators are selected so that definition is general and convinient to use.


    Chemical calculators for labs and education
    BATE - pH calculations, titration curves, hydrolisis
     
  4. Mar 11, 2005 #3

    ash

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    Thanks for replying Borek but I still don't really understand.
     
  5. Mar 11, 2005 #4

    Borek

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    Speed of the car is a distance covered in time, so it is km/h (or m/s or anything else) - which is length times inverse of time. Length has unit that is in the numerator.

    Now, for the chemical reaction distance to cover can be expressed as fraction - 0 for the starting point, 1 for the and. Speed of the reaction is "fraction of the reaction/time" - fraction time inverse of time. But fraction of the reaction is unitless - so what is left is just inverse of time.

    Somebody correct me if I am wrong, I have not touched kinetics for 20 years :surprised


    Chemical calculators for labs and education
    BATE - pH calculations, titration curves, hydrolisis
     
  6. Mar 11, 2005 #5

    Moonbear

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    Any rate is measured per unit time. This can be anything from kilometers per hour to revolutions per minute (rpm) to dollars earned per hour. To turn any of those words into math symbols, you would write the event you're measuring (kilometers traveled, revolutions turned, dollars earned) as the numerator and the time elapsed as the denominator, such as km/h. So, reaction rate is just another rate. The event being measured is the change in concentration of a product and you are measuring that per unit time. This is a simplified way of looking at it.

    Since I'm not sure whether you are learning this in high school or college, have you had any calculus yet? If so, reaction rate is really determined using derivatives to calculate the slope of the tangent to the plot of concentration of product over time. If you haven't had calculus yet and that sentence sounds like it was written in a foreign language, just ignore it for now, you'll get to it in a more advanced class.
     
  7. Mar 12, 2005 #6
    Let's say A --> B. We can measure the rates of rxn by the amount time this rxn is in state A. If it's in state A for 5 sec and the reaction rate is 1 process / 5 sec or 0.2 / sec.
     
  8. Mar 12, 2005 #7

    GCT

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    Rate of reaction in this case refers to instantaneous rate. The instantaneous rate at any time for a curve where the y-axis represents change in concentration and the x-axis represents time, is simply the tangent line of the curve at the specific point in time; now the smallest change in the concentration value that can be achieved pertains to one reaction.

    [tex]f^{'} (x)=lim_{h->0} \frac{(f(x+h)-f(x)}{ [\Delta x] } [/tex]
     
    Last edited: Mar 12, 2005
  9. Mar 12, 2005 #8

    Gokul43201

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    For an n-th order reaction :

    [tex]\frac {dx}{dt} = -kx^n [/tex]

    [tex]=>~\frac{dx}{x^n} = -kdt [/tex]

    [tex]=>~\frac{x^{1-n}}{1-n} = -kt + C~,~~n \not{=} 1 [/tex]

    [tex]=>~kt = [/tex]some number

    [tex]Or,~1/t ~~\alpha ~~k [/tex]

    [tex]n=1~=>~kt = ln(x_0/x)~=>1/t~~ \alpha~~ k [/tex]

    To sum up, 1/t is always proportional to the rate constant (being loose with additive constants).
     
    Last edited: Mar 12, 2005
  10. Mar 13, 2005 #9

    ash

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    Thanks for your help everyone!:)
     
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