Question 12.6: Plotting Data for a First-Order Reaction Experimental concen......
Plotting Data for a First-Order Reaction
Experimental concentration-versus-time data for the decomposition of gaseous N_2O_5 at 55 °C are listed in Table 12.1 and are plotted in Figure 12.1. Use those data to confirm that the decomposition of N_2O_5 is a first-order reaction. What is the value of the rate constant for the consumption of N_2O_5including the proper units?
STRATEGY
To confirm that the reaction is first order, check to see whether a plot of ln [ N_2O_5 ] versus time gives a straight line. The rate constant for a first-order reaction equals -1 times the slope of the straight line.
Table 12.1 Concentrations as a Function of Time at 55° C for the Reaction 2 N_2O_5 (g) → 4 NO_2 (g) + O_2 (g)
| Concentration (M) | |||
| Time (s) | N_2O_5 | NO_2 | O_2 |
| 0 | 0.02 | 0 | 0 |
| 100 | 0.0169 | 0.0063 | 0.0016 |
| 200 | 0.0142 | 0.0115 | 0.0029 |
| 300 | 0.012 | 0.016 | 0.004 |
| 400 | 0.0101 | 0.0197 | 0.0049 |
| 500 | 0.0086 | 0.0229 | 0.0057 |
| 600 | 0.0072 | 0.0256 | 0.0064 |
| 700 | 0.0061 | 0.0278 | 0.007 |
Values of ln [ N_2O_5 ] are listed in the following table and are plotted versus time in the graph:
Because the data points lie on a straight line, the reaction is first order in N_2O_5. The slope of the line can be determined from the coordinates of any two widely separated points on the line and the rate constant k can be calculated from the slope:
\begin{aligned}& \text { Slope }=\frac{\Delta y}{\Delta x}=\frac{(-5.02)-(-4.17)}{650 \mathrm{~s}-150 \mathrm{~s}}=\frac{-0.85}{500 \mathrm{~s}}=-1.7 \times 10^{-3} \mathrm{~s}^{-1} \\& k=-(\text { Slope })=1.7 \times 10^{-3} \mathrm{~s}^{-1}\end{aligned}Note that the units of k are the units expected for a frst-order reaction.
