Chapter 12
Q. 12.12
Determining Reaction Order Graphically
At elevated temperatures, nitrogen dioxide decomposes to nitric oxide and molecular oxygen:
2 NO_2(g) → 2 NO(g) + O_2(g)
Concentration–time data for the consumption of NO_2 at 300 °C are as follows: (table below)
(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 is 6.00 × 10^{-3} M?
(e) What is t_{1/2} when [ NO_2]_0 is 3.00 × 10^{-3} M?
STRATEGY
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 S Products | ||
| First-Order | Second-Orde | |
| Rate law | -\frac{\Delta[A]}{\Delta t}=k[A] | -\frac{\Delta[A]}{\Delta t}=k[A]^2 |
| Concentration–time equation | \ln [A]=-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[A]_0}
(constant) |
t_{1/2}=\frac{1}{k}
(not constant) |
Step-by-Step
Verified Solution
(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{340 \mathrm{M}^{-1}-150 \mathrm{M}^{-1}}{400 \mathrm{~s}-50 \mathrm{~s}}=\frac{190 \mathrm{M}^{-1}}{350 \mathrm{~s}}=0.54 /(\mathrm{M} \cdot \mathrm{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}{\left[\mathrm{NO}_2\right]_t}=k t+\frac{1}{\left[\mathrm{NO}_2\right]_0}Substituting the values of k, t, and [ NO_2]_0 gives
\begin{aligned}\frac{1}{\left[\mathrm{NO}_2\right]_t} & =\left(\frac{0.54}{\mathrm{M} \cdot \mathrm{s}}\right)\left(1.20 \times 10^3 \mathrm{~s}\right)+\frac{1}{8.00 \times 10^{-3} \mathrm{M}} \\& =\frac{648}{\mathrm{M}}+\frac{125}{\mathrm{M}}=\frac{773}{\mathrm{M}} \\{\left[\mathrm{NO}_2\right]_t } & =1.3 \times 10^{-3} \mathrm{M}\end{aligned}(d) The half-life of this second-order reaction when the initial concentration of NO_2 is 6.00 × 10^{-3} M can be calculated from the rate constant and the initial concentration:
t_{1 / 2}=\frac{1}{k\left[\mathrm{NO}_2\right]_0}=\frac{1}{\left(\frac{0.54}{\mathrm{M} \cdot \mathrm{s}}\right)\left(6.00 \times 10^{-3} \mathrm{M}\right)}=3.1 \times 10^2 \mathrm{~s}(e) When [ NO_2]_0 is 3.00 × 10^{-3} M, t_{1/2} = 6.2 × 10² s (twice as long as when [ NO_2]_0 is 6.00 × 10^{-3} M because [ NO_2]_0 is now smaller by a factor of 2).
| Time (s) | [ NO_2 ] | ln [ NO_2 ] | 1/[ NO_2 ] |
| 0 | 8.00 \times 10^{-3} | -4.828 | 125 |
| 50 | 6.58 \times 10^{-3} | -5.024 | 152 |
| 100 | 5.59 \times 10^{-3} | -5.187 | 179 |
| 150 | 4.85 \times 10^{-3} | -5.329 | 206 |
| 200 | 4.29 \times 10^{-3} | -5.451 | 233 |
| 300 | 3.48 \times 10^{-3} | -5.661 | 287 |
| 400 | 2.93 \times 10^{-3} | -5.833 | 341 |
| 500 | 2.53 \times 10^{-3} | -5.98 | 395 |
