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