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The Learning Objective of this Module is to use graphs to analyze the kinetics of a reaction.
In Section 14.3, 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:
Experimental data for this reaction at 330°C are listed in Table 14.5; 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 14.15. 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 14.15) 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 14.15). This plot is a straight line, indicating that the reaction is second order in NO_{2}.
Table 14.5 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 14.15 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 14.16, the graphs can be used to determine the reaction order of an unknown reaction. In contrast, the method described in Section 14.3 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.
Figure 14.16 Properties of Reactions That Obey Zeroth, First, and SecondOrder Rate Laws
Example 9  

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}(soln) → 4NO_{2}(soln) + O_{2}(g) 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.
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 14.16 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[N_{2}O_{5}] 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, \(\textrm{slope}=\dfrac{\ln[\mathrm{N_2O_5}]_{3000}\ln[\mathrm{N_2O_5}]_0}{3000\textrm{ s}0\textrm{ s}}=\dfrac{(4.756)(3.310)}{3000\textrm{ s}}=4.820\times10^{4}\textrm{ s}^{1}\) 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.
Answer: second order in C_{4}H_{6}; rate = k[C_{4}H_{6}]^{2}; k = 1.3 × 10^{−2} M^{−1}·s^{−1} 
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.
Conceptual Problems 


Answers 


Numerical Problems  

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?

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