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Sagot :
Certainly! Let's solve the given problem step by step.
### (a) Determining the Order of the Reaction
We need to determine whether the reaction is first-order or second-order. The concentration data at different times given are:
| Time (min) | [H](M) |
|------------|--------|
| 0 | 0.500 |
| 20 | 0.400 |
| 40 | 0.300 |
| 60 | 0.200 |
For a first-order reaction, the relationship is given by:
[tex]\[ \ln([H]) = \ln([H]_0) - kt \][/tex]
where [tex]\([H]_0\)[/tex] is the initial concentration, [tex]\(k\)[/tex] is the rate constant, and [tex]\(t\)[/tex] is the time.
For a second-order reaction, the relationship is given by:
[tex]\[ \frac{1}{[H]} = \frac{1}{[H]_0} + kt \][/tex]
We can conclude that the reaction is first-order because the slope of the plot [tex]\( \ln([H]) \)[/tex] versus time is linear with a negative slope.
### (b) The Rate Constant
Given our calculations, the rate constant [tex]\(k\)[/tex] for this reaction is:
[tex]\[ k = 0.015182771340371236 \, \text{min}^{-1} \][/tex]
### (c) Time to Reach a Concentration of 0.200 M from Initial Concentration 0.500 M
To find the time when the concentration reaches 0.200 M, we use the first-order reaction formula:
[tex]\[ \ln(\frac{[H]_0}{[H]}) = kt \][/tex]
Substituting the known values:
[tex]\[ \ln(\frac{0.500}{0.200}) = 0.015182771340371236 \, t \][/tex]
Solving for [tex]\(t\)[/tex]:
[tex]\[ t \approx 60.35 \, \text{minutes} \][/tex]
### (d) How Many Minutes Does It Take for the Concentration to Change from 0.400 M to 0.200 M
Use the same formula for a first-order reaction:
[tex]\[ \ln(\frac{[H]_1}{[H]_2}) = k \Delta t \][/tex]
Substituting the concentrations from 0.400 M to 0.200 M:
[tex]\[ \ln(\frac{0.400}{0.200}) = 0.015182771340371236 \, \Delta t \][/tex]
Solving for the time interval [tex]\( \Delta t \)[/tex]:
[tex]\[ \Delta t \approx \frac{\ln(2)}{0.015182771340371236} \approx 45.67 \, \text{minutes} \][/tex]
Therefore, it takes approximately 45.67 minutes for the concentration to change from 0.400 M to 0.200 M.
As a summary:
(a) The reaction is first-order.
(b) The rate constant [tex]\(k\)[/tex] is 0.015182771340371236 min⁻¹.
(c) The concentration will reach 0.200 M from an initial concentration of 0.500 M in approximately 60.35 minutes.
(d) It takes about 45.67 minutes for the concentration to drop from 0.400 M to 0.200 M.
### (a) Determining the Order of the Reaction
We need to determine whether the reaction is first-order or second-order. The concentration data at different times given are:
| Time (min) | [H](M) |
|------------|--------|
| 0 | 0.500 |
| 20 | 0.400 |
| 40 | 0.300 |
| 60 | 0.200 |
For a first-order reaction, the relationship is given by:
[tex]\[ \ln([H]) = \ln([H]_0) - kt \][/tex]
where [tex]\([H]_0\)[/tex] is the initial concentration, [tex]\(k\)[/tex] is the rate constant, and [tex]\(t\)[/tex] is the time.
For a second-order reaction, the relationship is given by:
[tex]\[ \frac{1}{[H]} = \frac{1}{[H]_0} + kt \][/tex]
We can conclude that the reaction is first-order because the slope of the plot [tex]\( \ln([H]) \)[/tex] versus time is linear with a negative slope.
### (b) The Rate Constant
Given our calculations, the rate constant [tex]\(k\)[/tex] for this reaction is:
[tex]\[ k = 0.015182771340371236 \, \text{min}^{-1} \][/tex]
### (c) Time to Reach a Concentration of 0.200 M from Initial Concentration 0.500 M
To find the time when the concentration reaches 0.200 M, we use the first-order reaction formula:
[tex]\[ \ln(\frac{[H]_0}{[H]}) = kt \][/tex]
Substituting the known values:
[tex]\[ \ln(\frac{0.500}{0.200}) = 0.015182771340371236 \, t \][/tex]
Solving for [tex]\(t\)[/tex]:
[tex]\[ t \approx 60.35 \, \text{minutes} \][/tex]
### (d) How Many Minutes Does It Take for the Concentration to Change from 0.400 M to 0.200 M
Use the same formula for a first-order reaction:
[tex]\[ \ln(\frac{[H]_1}{[H]_2}) = k \Delta t \][/tex]
Substituting the concentrations from 0.400 M to 0.200 M:
[tex]\[ \ln(\frac{0.400}{0.200}) = 0.015182771340371236 \, \Delta t \][/tex]
Solving for the time interval [tex]\( \Delta t \)[/tex]:
[tex]\[ \Delta t \approx \frac{\ln(2)}{0.015182771340371236} \approx 45.67 \, \text{minutes} \][/tex]
Therefore, it takes approximately 45.67 minutes for the concentration to change from 0.400 M to 0.200 M.
As a summary:
(a) The reaction is first-order.
(b) The rate constant [tex]\(k\)[/tex] is 0.015182771340371236 min⁻¹.
(c) The concentration will reach 0.200 M from an initial concentration of 0.500 M in approximately 60.35 minutes.
(d) It takes about 45.67 minutes for the concentration to drop from 0.400 M to 0.200 M.
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