Chapter 13.4: Using Graphs to Determine Rate Laws, Rate Constants and Reaction Orders
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 42111
Prince George's Community College 

Unit I: Atoms Unit II: Molecules Unit III: States of Matter Unit IV: Reactions Unit V: Kinetics & Equilibrium Unit VI: Thermo & Electrochemistry Unit VII: Nuclear Chemistry 
Learning Objective
 To use graphs to analyze the kinetics of a reaction.
In Section 13.3, you learned that the integrated rate law for each common type of reaction (zeroth, first, or second order in a single reactant) can be plotted as a straight line. Using these plots offers an alternative to the methods described for showing how reactant concentration changes with time and determining reaction order.
We will illustrate the use of these graphs by considering the thermal decomposition of NO_{2} gas at elevated temperatures, which occurs according to the following reaction:
\( 2NO_{2}\left ( g \right ) \overset{\Delta}{\rightarrow} 2NO\left ( g \right )+O_{2}\left ( g \right ) \tag{13.4.1}\)
Experimental data for this reaction at 330°C are listed in Table 13.4.1; they are provided as [NO_{2}], ln[NO_{2}], and 1/[NO_{2}] versus time to correspond to the integrated rate laws for zeroth, first, and secondorder reactions, respectively. The actual concentrations of NO_{2} are plotted versus time in part (a) in Figure 13.4.1. Because the plot of [NO_{2}] versus t is not a straight line, we know the reaction is not zeroth order in NO_{2}. A plot of ln[NO_{2}] versus t (part (b) in Figure 13.4.1) shows us that the reaction is not first order in NO_{2} because a firstorder reaction would give a straight line. Having eliminated zerothorder and firstorder behavior, we construct a plot of 1/[NO_{2}] versus t (part (c) in Figure 13.4.1). This plot is a straight line, indicating that the reaction is second order in NO_{2}.
Table 13.4.1 Concentration of NO_{2} as a Function of Time at 330°C
Time (s)  [NO_{2}] (M)  ln[NO_{2}]  1/[NO_{2}] (M^{−1}) 

0  1.00 × 10^{−2}  −4.605  100 
60  6.83 × 10^{−3}  −4.986  146 
120  5.18 × 10^{−3}  −5.263  193 
180  4.18 × 10^{−3}  −5.477  239 
240  3.50 × 10^{−3}  −5.655  286 
300  3.01 × 10^{−3}  −5.806  332 
360  2.64 × 10^{−3}  −5.937  379 
Figure 13.4.1 The Decomposition of NO_{2} These plots show the decomposition of a sample of NO_{2} at 330°C as (a) the concentration of NO_{2} versus t, (b) the natural logarithm of [NO_{2}] versus t, and (c) 1/[NO_{2}] versus t.
We have just determined the reaction order using data from a single experiment by plotting the concentration of the reactant as a function of time. Because of the characteristic shapes of the lines shown in Figure 13.4.2, the graphs can be used to determine the reaction order of an unknown reaction. In contrast, the method described in Section 13.3 required multiple experiments at different NO_{2} concentrations as well as accurate initial rates of reaction, which can be difficult to obtain for rapid reactions.
Example 13.4.1
Dinitrogen pentoxide (N_{2}O_{5}) decomposes to NO_{2} and O_{2} at relatively low temperatures in the following reaction:
\( 2N_{2}O_{5}\left ( soln \right ) \rightarrow 4NO_{2}\left ( soln \right )+O_{2}\left ( g \right ) \)
This reaction is carried out in a CCl_{4} solution at 45°C. The concentrations of N_{2}O_{5} as a function of time are listed in the following table, together with the natural logarithms and reciprocal N_{2}O_{5} concentrations. Plot a graph of the concentration versus t, ln concentration versus t, and 1/concentration versus t and then determine the rate law and calculate the rate constant.
Time (s)  [N_{2}O_{5}] (M)  ln[N_{2}O_{5}]  1/[N_{2}O_{5}] (M^{−1}) 

0  0.0365  −3.310  27.4 
600  0.0274  −3.597  36.5 
1200  0.0206  −3.882  48.5 
1800  0.0157  −4.154  63.7 
2400  0.0117  −4.448  85.5 
3000  0.00860  −4.756  116 
3600  0.00640  −5.051  156 
Given: balanced chemical equation, reaction times, and concentrations
Asked for: graph of data, rate law, and rate constant
Strategy:
A Use the data in the table to separately plot concentration, the natural logarithm of the concentration, and the reciprocal of the concentration (the vertical axis) versus time (the horizontal axis). Compare the graphs with those in Figure 13.4.2 to determine the reaction order.
B Write the rate law for the reaction. Using the appropriate data from the table and the linear graph corresponding to the rate law for the reaction, calculate the slope of the plotted line to obtain the rate constant for the reaction.
Solution:
A Here are plots of [N_{2}O_{5}] versus t, ln[N_{2}O_{5}] versus t, and 1/[N_{2}O_{5}] versus t:
The plot of ln[N_{2}O_{5}] versus t gives a straight line, whereas the plots of [N_{2}O_{5}] versus t and 1/[N_{2}O_{5}] versus t do not. This means that the decomposition of N_{2}O_{5} is first order in [N_{2}O_{5}].
B The rate law for the reaction is therefore
\( rate = k \left [N_{2}O_{5} \right ]\)
Calculating the rate constant is straightforward because we know that the slope of the plot of ln[A] versus t for a firstorder reaction is −k. We can calculate the slope using any two points that lie on the line in the plot of ln[N_{2}O_{5}] versus t. Using the points for t = 0 and 3000 s,
\( slope= \dfrac{ln\left [N_{2}O_{5} \right ]_{3000}ln\left [N_{2}O_{5} \right ]_{0}}{3000\;s0\;s} = \dfrac{\left [4.756 \right ]\left [3.310 \right ]}{3000\;s} =4.820\times 10^{4}\;s^{1} \)
Thus k = 4.820 × 10^{−4} s^{−1}.
Exercise
1,3Butadiene (CH_{2}=CH—CH=CH_{2;} C_{4}H_{6}) is a volatile and reactive organic molecule used in the production of rubber. Above room temperature, it reacts slowly to form products. Concentrations of C_{4}H_{6} as a function of time at 326°C are listed in the following table along with ln[C_{4}H_{6}] and the reciprocal concentrations. Graph the data as concentration versus t, ln concentration versus t, and 1/concentration versus t. Then determine the reaction order in C_{4}H_{6}, the rate law, and the rate constant for the reaction.
Time (s)  [C_{4}H_{6}] (M)  ln[C_{4}H_{6}]  1/[C_{4}H_{6}] (M^{−1}) 

0  1.72 × 10^{−2}  −4.063  58.1 
900  1.43 × 10^{−2}  −4.247  69.9 
1800  1.23 × 10^{−2}  −4.398  81.3 
3600  9.52 × 10^{−3}  −4.654  105 
6000  7.30 × 10^{−3}  −4.920  137 
Answer:
second order in C_{4}H_{6}; rate = k[C_{4}H_{6}]^{2}; k = 1.3 × 10^{−2} M^{−1}·s^{−1}
Summary
For a zerothorder reaction, a plot of the concentration of any reactant versus time is a straight line with a slope of −k. For a firstorder reaction, a plot of the natural logarithm of the concentration of a reactant versus time is a straight line with a slope of −k. For a secondorder reaction, a plot of the inverse of the concentration of a reactant versus time is a straight line with a slope of k.
Key Takeaway
 Plotting the concentration of a reactant as a function of time produces a graph with a characteristic shape that can be used to identify the reaction order in that reactant.
Conceptual Problems

Compare firstorder differential and integrated rate laws with respect to the following. Is there any information that can be obtained from the integrated rate law that cannot be obtained from the differential rate law?
 the magnitude of the rate constant
 the information needed to determine the order
 the shape of the graphs

In the singlestep, secondorder reaction 2A → products, how would a graph of [A] versus time compare to a plot of 1/[A] versus time? Which of these would be the most similar to the same set of graphs for A during the singlestep, secondorder reaction A + B → products? Explain.

For reactions of the same order, what is the relationship between the magnitude of the rate constant and the reaction rate? If you were comparing reactions with different orders, could the same arguments be made? Why?
Answers

 For a given reaction under particular conditions, the magnitude of the firstorder rate constant does not depend on whether a differential rate law or an integrated rate law is used.
 The differential rate law requires multiple experiments to determine reactant order; the integrated rate law needs only one experiment.
 Using the differential rate law, a graph of concentration versus time is a curve with a slope that becomes less negative with time, whereas for the integrated rate law, a graph of ln[reactant] versus time gives a straight line with slope = −k. The integrated rate law allows you to calculate the concentration of a reactant at any time during the reaction; the differential rate law does not.

The reaction rate increases as the rate constant increases. We cannot directly compare reaction rates and rate constants for reactions of different orders because they are not mathematically equivalent.
Numerical Problems

One method of using graphs to determine reaction order is to use relative rate information. Plotting the log of the relative rate versus log of relative concentration provides information about the reaction. Here is an example of data from a zerothorder reaction:
Relative [A] (M) Relative Rate (M/s) 1 1 2 1 3 1 Varying [A] does not alter the reaction rate. Using the relative rates in the table, generate plots of log(rate) versus log(concentration) for zeroth, first and secondorder reactions. What does the slope of each line represent?

The table below follows the decomposition of N_{2}O_{5} gas by examining the partial pressure of the gas as a function of time at 45°C. What is the reaction order? What is the rate constant? How long would it take for the pressure to reach 105 mmHg at 45°C?
Time (s) Pressure (mmHg) 0 348 400 276 1600 156 3200 69 4800 33
Contributors
 Anonymous
Modified by Joshua Halpern (Howard University), Scott Sinex, and Scott Johnson (PGCC)