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AP Chemistry Chemical Kinetics Problem

  1. Dec 9, 2014 #1
    1. The problem statement, all variables and given/known data
    The rate of the reaction was studied at a certain temperature.
    O(g) + NO2(g) --> NO(g) + O2(g)



    In the first set of experiments, NO2 was in large excess, at a concentration of 1.0 * 10^ 13 molecules/cm3 with the following data collected.



    time (s) [O] (atoms/cm3)
    0 .......................2400
    0.01.....................1200
    0.02.....................600
    0.03.......................0

    Part (a). Find the reaction order with respect to O. (Explain with graphs)

    Part (b) Find the rate constant with respect to O.


    2. Relevant equations
    Rate = K[A]
    First Order Reaction Integrated Rate Law: ln[A] = -kt + ln[A]0

    3. The attempt at a solution
    I want to know how to transform the graph/data set into a linear line to find the reaction order (I think it is first order). I think I ln[O] and plot it against time (s) to achieve a linear graph for a first order graph. How do I do that? Do I need to convert atoms/cm3 to mol/L? I tried doing that and it is not looking right. Help would be appreciated!
     
  2. jcsd
  3. Dec 9, 2014 #2
    You don't have to convert to mol/L. The first three data points are OK, but the 4th data point should be 300 atoms/cc (there must be a typo, or maybe you're supposed to think that the measurement is in the noise at 300 atoms/cc). You can plot the first three points, and leave out the 4th, or you can plot all four, with 300 as the last point. If you plot the first 3 points, they will lie on a straight line on a semi-log plot. If you have Excel, just change the vertical scale to a log scale (which Excel will automatically do for you). You can get the first order rate constant from the slope. Excel will even do a curve fit for you, and provide you with the slope, so that you can know the rate constant immediately.

    Chet
     
  4. Dec 9, 2014 #3
    Thanks you! I got the equation to be y = -69.315x + 7.7832 without plotting the last point. This means the rate constant is -69.3 respect to [O]. It is negative because the concentration is going down as time increases. Is that correct?
     
  5. Dec 9, 2014 #4
    In your equation, there is a negative sign in front of k. So the rate constant should be +69.3/sec.

    Chet
     
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