AP Chemistry Chemical Kinetics Problem

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SUMMARY

The discussion focuses on determining the reaction order and rate constant for the reaction O(g) + NO2(g) --> NO(g) + O2(g) using experimental data. The data indicates a first-order reaction with respect to O, and participants confirm that plotting ln[O] against time results in a linear graph. The rate constant is calculated to be +69.3 sec-1 based on the slope derived from the semi-log plot. Excel is recommended for plotting and calculating the rate constant efficiently.

PREREQUISITES
  • Understanding of first-order reaction kinetics
  • Familiarity with integrated rate laws, specifically ln[A] = -kt + ln[A]0
  • Proficiency in using Excel for data analysis and graphing
  • Knowledge of converting units, specifically from atoms/cm3 to mol/L (though not necessary for this problem)
NEXT STEPS
  • Learn how to create semi-log plots in Excel for kinetic data analysis
  • Study the derivation and application of the first-order integrated rate law
  • Explore the concept of reaction order and its significance in chemical kinetics
  • Investigate common pitfalls in experimental data collection and analysis in kinetics
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Chemistry students, educators, and researchers focusing on chemical kinetics, particularly those analyzing reaction rates and orders in laboratory settings.

Hikari
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Homework Statement


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.

Homework Equations


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

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!
 
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Hikari said:

Homework Statement


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.

Homework Equations


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

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!
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
 
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?
 
Hikari said:
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?
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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