Question 12.6: Experimental concentration-versus-time data for the decompos......
Experimental concentration-versus-time data for the decomposition of gaseous N_{2}O_{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 is N_{2}O_{5} a first-order reaction. What is the value of the rate constant for consumption of N_{2}O_{5} ?
STRATEGY
To confirm that the reaction is first order, check to see if a plot of \ln [N_{2}O_{5}] versus time gives a straight line. The rate constant for a first-order reaction equals minus the slope of the straight line.
TABLE 12.1 Concentrations as a Function of Time at 55°C for the Reaction 2 N_{2}O_{5} (g) \longrightarrow 4 NO_{2} (g) + O_{2} (g)
| Concentration (M) | |||
| Time (s) | N_{2}O_{5} | NO_{2} | O_{2} |
| 0 | 0.0200 | 0 | 0 |
| 100 | 0.0169 | 0.0063 | 0.0016 |
| 200 | 0.0142 | 0.0115 | 0.0029 |
| 300 | 0.0120 | 0.016 | 0.0040 |
| 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 | 0.0070 |
Values of \ln [N_{2}O_{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_{2}O_{5} The slope of the line can be determined from the coordinates of any two widely sepa-rated points on the line, and the rate constant k can be calculated from the slope:
\text{Slope} = \frac{\Delta y}{\Delta x} = \frac{(-5.02) – (-4.17)}{650 s – 150 s} = \frac{-0.85}{500 s} = -1.7 \times 10^{-3} s^{-1} \\ k = – (\text{Slope}) = 1.7 \times 10^{-3} s^{-1}Note that the slope is negative, k is positive, and the value of k agrees with the value obtained earlier in Worked Example 12.3 by the method of initial rates.
| Time (s) | [N_{2}O_{5}] | \ln [N_{2}O_{5}] |
| 0 | 0.02 | -3.912 |
| 100 | 0.0169 | -4.08 |
| 200 | 0.0142 | -4.255 |
| 300 | 0.012 | -4.423 |
| 400 | 0.0101 | -4.595 |
| 500 | 0.0086 | -4.756 |
| 600 | 0.0072 | -4.934 |
| 700 | 0.0061 | -5.099 |
