## 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)

## 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