Question 12.8: At elevated temperatures, nitrogen dioxide decomposes to nit......
At elevated temperatures, nitrogen dioxide decomposes to nitric oxide and molecular oxygen:
2 NO_{2} (g) \longrightarrow 2 NO (g) + O_{2} (g)Concentration–time data for consumption of at 300°C are as follows:
(a) Is the reaction first order or second order?
(b) What is the value of the rate constant?
(c) What is the concentration of NO_{2} at t = 20.0 min?
(d) What is the half-life of the reaction when the initial concentration of NO_{2} \text{is} 6.00 \times 10^{-3} M ?
(e) What is t_{1/2} \text{when} [NO_{2}]_{0} \text{is} 3.00 \times 10^{-3} M ?
STRATEGY
To determine whether the reaction is first order or second order, calculate values of \ln [NO_{2}] and 1 / [NO_{2}], and then graph these values versus time. The rate constant can be obtained from the slope of the straight-line plot, and concentrations and half-lives can be calculated from the appropriate equation in Table 12.4.
| Time (s) | [NO_{2}] | Time (s) | [NO_{2}] |
| 0 | 8.00 \times 10^{-3} | 200 | 4.29 \times 10^{-3} |
| 50 | 6.58 \times 10^{-3} | 300 | 3.48 \times 10^{-3} |
| 100 | 5.59 \times 10^{-3} | 400 | 2.93 \times 10^{-3} |
| 150 | 4.85 \times 10^{-3} | 500 | 2.53 \times 10^{-3} |
TABLE 12.4 Characteristics of First- and Second-Order Reactions of the Type A \longrightarrow \text{Products}
| First-Order | Second-Order | |
| Rate law | -\frac{\Delta [A]}{\Delta t} = k [A] | -\frac{\Delta [A]}{\Delta t} = k [A]^{2} |
| Concentration–time equation | \ln [A]_{t} = -kt + \ln [A]_{0} | \frac{1}{[A]_{t}} = kt + \frac{1}{[A]_{0}} |
| Linear graph | \ln [A] \text{versus} t | \frac{1}{[A]} \text{versus} t |
| Graphical determination of k | k = -(Slope) | k = (Slope) |
| Half-life | t_{1/2} = \frac{0.693}{k} | t_{1/2} = \frac{1}{k[A]_{0}} |
| (constant) | (not constant) |
(a) The plot of \ln [NO_{2}] versus time is curved, but the plot of 1 / [NO_{2}] versus time is a straight line. The reaction is therefore second order in NO_{2}
(b) The rate constant equals the slope of the straight line in the plot of 1/[NO_{2}] versus time, which we can estimate from the coordinates of two widely separated points on the line:
k = \text{Slope} = \frac{\Delta y}{\Delta x} = \frac{3.4 M^{-1} – 150 M^{-1}}{400 s – 50 s} = \frac{190 M^{-1}}{350 s} = 0.54 / (M • s)(c) The concentration of NO_{2} at t = 20.0 min (1.20 × 10³ s) can be calculated using the integrated rate law:
\frac{1}{[NO_{2}]_{t}} = kt + \frac{1}{[NO_{2}]_{0}}Substituting the values of k, t, and [NO_{2}]_{0} gives
\frac{1}{[NO_{2}]_{t}} = (\frac{0.54}{M • s}) (1.20 \times 10^{3} s) + \frac{1}{8.00 \times 10^{-3} M} \\ \quad\quad = \frac{648}{M} + \frac{125}{M} = \frac{773}{M} \\ [NO_{2}]_{t} = 1.3 \times 10^{-3} M(d) The half-life of a second–order reaction can be calculated from the rate constant and the initial concentration of NO_{2} (6.00 \times 10^{-3} M) :
t_{1/2} = \frac{1}{k [NO_{2}]_{0}} = \frac{1}{(\frac{0.54}{M • s}) (6.00 \times 10^{-3} M)} = 3.1 \times 10^{2} s(e) When [NO_{2}]_{0} \text{is} 3.00 × 10^{-3} M , t_{1/2} = 6.2 \times 10^{2} s (twice as long as when [NO_{2}]_{0} is because is now smaller by a factor of 2).