Using Graphs to Determine Integrated Rate Laws
 Page ID
 6426
Skills to Develop

To use graphs to analyze the kinetics of a reaction.
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:
\[\mathrm{2NO_2(g)}\xrightarrow{\Delta}\mathrm{2NO(g)}+\mathrm{O_2(g)} \label{14.26}\]
Experimental data for this reaction at 330°C are listed in Table \(\PageIndex{1}\); 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.
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 
The actual concentrations of NO_{2} are plotted versus time in part (a) in Figure \(\PageIndex{1}\). 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 \(\PageIndex{1}\)) 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 \(\PageIndex{1}\)). This plot is a straight line, indicating that the reaction is second order in NO_{2}.
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 \(\PageIndex{2}\), the graphs can be used to determine the reaction order of an unknown reaction. In contrast, the method of initial rates 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 \(\PageIndex{1}\)
Dinitrogen pentoxide (N_{2}O_{5}) decomposes to NO_{2} and O_{2} at relatively low temperatures in the following reaction:
\[2N_2O_5(soln) → 4NO_2(soln) + O_2(g)\]
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.
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 \(\PageIndex{1}\) 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,
Thus k = 4.820 × 10^{−4} s^{−1}.
Exercise \(\PageIndex{1}\)
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.