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Magazine Chapter 12: Q36 E (page 706)
From the given data, use a graphical method to determine the order and rate constant of the following reaction: 2X鉄禮 + Z
Time(s) | 5.0 | 10.0 | 15.0 | 20.0 | 25.0 | 30.0 | 35.0. | 40.0 |
(X)(M) | 0.0990 | 0.0497 | 0.0332 | 0.0249 | 0.0200 | 0.0166 | 0.0143 | 0.0125 |
Short Answer
The order of the reaction is second order. The rate constant of the reaction is \(50.21Lmo{l^{ - 1}}{s^{ - 1}}\).
Step by step solution
Plotting of graph
The plot of ln1/\({{\bf{O}}_{\bf{3}}}\) vs time is linear. This indicates a second-order reaction.
Step 2: Calculation of rate
The constant rate of a second-order reaction is as follows:
\(\frac{1}{{(A)}} - \frac{1}{{{{(A)}_0}}} = kt\)
where (A) is the concentration of ozone at different times. (A) is the initial ozone concentration, i.e., at time t=0.
Time(hr) | (X) Mol L-1 | 1/(X)L Mol-1 |
5.0 | 0.0990 | 10.10 |
10.0 | 0.0497 | 20.12 |
15.0 | 0.0332 | 30.12 |
20.0 | 0.0249 | 40.16 |
25.0 | 0.0200 | 50.00 |
30.0 | 0.0166 | 60.24 |
35.0 | 0.0143 | 69.93 |
40.0 | 0.0125 | 80.00 |
Now, the slope of the plot is given as
\(Slope = \frac{{\Delta y}}{{\Delta x}}\)
Where \({\bf{\Delta y}}\) is the change in values of (1/(X)) at two different times t and is the corresponding difference in times
\(Slope = \frac{{(50.00) - (40.16)}}{{(25.0 - 20.0)}} = 1.968Lmo{l^{ - 1}}{s^{ - 1}}\)
Rate calculation
Rate = (slope)
= \({\bf{1}}{\bf{.986Lmo}}{{\bf{l}}^{{\bf{ - 1}}}}{{\bf{s}}^{{\bf{ - 1}}}}\)
Thus, the order of the reaction is second order with a rate constant of \({\bf{1}}{\bf{.986Lmo}}{{\bf{l}}^{{\bf{ - 1}}}}{{\bf{s}}^{{\bf{ - 1}}}}\)
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