Chapter 12: Q35 E (page 706)

Pure ozone decomposes slowly to oxygen,\({\bf{2}}{{\bf{O}}_{\bf{3}}}{\bf{(g)}} \to {\bf{3}}{{\bf{O}}_{\bf{2}}}{\bf{(g)}}\). Use the data provided in a graphical method and determine the order and rate constant of the reaction.
Time(hr)
0
2.0x10³
7.6x 10⁴
1.00x10⁴
1.23x10⁴
1.43x10⁴
1.70x10⁴
(O₃) (M)
1.0x10^-5
4.98x10^-6
2.07x10^-6
1.66x10^-6
1.39x10^-6
1.22x10^-6
1.05x10^-6

Short Answer

Expert verified

The order of the reaction is second order. The reaction's rate constant is \({\bf{50}}{\bf{.21Lmo}}{{\bf{l}}^{{\bf{ - 1}}}}{{\bf{s}}^{{\bf{ - 1}}}}\).

Step by step solution

Definition

A chemical reaction in which the rate of reaction is proportional to the concentration of two reacting molecules is called a second-order reaction.

Step 2: Calculation of rate Constant

For a second-order reaction, the rate constant is given as

\(\frac{1}{{(A)}} - \frac{1}{{{{(A)}_0}}} = kt\)

where (A) is the concentration of ozone at time t. (A) is the ozone concentration at time t=0.

Given table,

Time(hr)	\({\bf{(}}{{\bf{O}}_{\bf{3}}}{\bf{)}}\,{\bf{Mol }}{{\bf{L}}^{{\bf{ - 1}}}}\)	\({\bf{1/(}}{{\bf{O}}_{\bf{3}}}{\bf{)}}\,{\bf{L Mo}}{{\bf{l}}^{{\bf{ - 1}}}}\)
0	\({\bf{1}}{\bf{.0x1}}{{\bf{0}}^{{\bf{ - 5}}}}\)	\({\bf{1x1}}{{\bf{0}}^{\bf{5}}}\)
\({\bf{2x1}}{{\bf{0}}^{\bf{3}}}\)	\({\bf{4}}{\bf{.98x1}}{{\bf{0}}^{{\bf{ - 6}}}}\)	\({\bf{2}}{\bf{.00x1}}{{\bf{0}}^{\bf{5}}}\)
\({\bf{7}}{\bf{.6x1}}{{\bf{0}}^{\bf{3}}}\)	\({\bf{2}}{\bf{.07 x1}}{{\bf{0}}^{{\bf{ - 6}}}}\)	\({\bf{4}}{\bf{.83 x1}}{{\bf{0}}^{\bf{5}}}\)
\({\bf{1}}{\bf{.0 x1}}{{\bf{0}}^{\bf{4}}}\)	\({\bf{1}}{\bf{.66 x1}}{{\bf{0}}^{{\bf{ - 6}}}}\)	\({\bf{6}}{\bf{.02 x1}}{{\bf{0}}^{\bf{5}}}\)
\({\bf{1}}{\bf{.23 x1}}{{\bf{0}}^{\bf{4}}}\)	\({\bf{1}}{\bf{.39 x1}}{{\bf{0}}^{{\bf{ - 6}}}}\)	\({\bf{7}}{\bf{.19 x1}}{{\bf{0}}^{\bf{5}}}\)
\({\bf{1}}{\bf{.43 x1}}{{\bf{0}}^{\bf{4}}}\)	\({\bf{1}}{\bf{.22 x1}}{{\bf{0}}^{{\bf{ - 6}}}}\)	\({\bf{8}}{\bf{.19 x1}}{{\bf{0}}^{\bf{5}}}\)
\({\bf{1}}{\bf{.7 x1}}{{\bf{0}}^{\bf{4}}}\)	\({\bf{1}}{\bf{.05 x1}}{{\bf{0}}^{{\bf{ - 6}}}}\)	\({\bf{9}}{\bf{.52 x1}}{{\bf{0}}^{\bf{5}}}\)

Now, the slope of the plot is given as

\(\begin{align}Slope &= \frac{{(7.19 \times {{10}^5}) - (4.83 \times {{10}^5})}}{{(1.23 - 0.76) \times {{10}^4}}}\\ &= 50.21Lmo{l^{ - 1}}h{r^{ - 1}}\end{align}\)

Rate calculation

For a second-order reaction, the rate constant can be determined from the slope of the line, which is equal to k.

So,

\({\bf{Rate = 50}}{\bf{.21Lmo}}{{\bf{l}}^{{\bf{ - 1}}}}{\bf{h}}{{\bf{r}}^{{\bf{ - 1}}}}\)

Plotting of graph

The plot of ln\({\bf{1/}}{{\bf{O}}_{\bf{3}}}\) vs time is a linear plot with a positive slope. This indicates a second-order reaction.

Thus, the order of the reaction is second order, and the rate constant is \({\bf{50}}{\bf{.21Lmo}}{{\bf{l}}^{{\bf{ - 1}}}}{\bf{h}}{{\bf{r}}^{{\bf{ - 1}}}}\)