Get the answers you've been looking for with the help of IDNLearn.com's expert community. Discover trustworthy solutions to your questions quickly and accurately with help from our dedicated community of experts.
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.
Your participation means a lot to us. Keep sharing information and solutions. This community grows thanks to the amazing contributions from members like you. IDNLearn.com is committed to providing the best answers. Thank you for visiting, and see you next time for more solutions.
