Whether you're a student or a professional, IDNLearn.com has answers for everyone. Our experts provide timely and accurate responses to help you navigate any topic or issue with confidence.
Sagot :
Let's tackle this problem step-by-step.
### Step 1: Determine the reaction orders [tex]\( m \)[/tex] and [tex]\( n \)[/tex]
Assume the rate law for the reaction is:
[tex]\[ \text{Rate} = k [\text{HgCl}_2]^m [\text{C}_2\text{O}_4^{2-}]^n \][/tex]
We will use the given experimental data to determine the orders [tex]\( m \)[/tex] and [tex]\( n \)[/tex].
#### Finding the order with respect to [tex]\(\text{C}_2\text{O}_4^{2-}\)[/tex]:
- Experiment 1: [tex]\([\text{HgCl}_2] = 0.164 \, M, \, [\text{C}_2\text{O}_4^{2-}] = 0.15 \, M\)[/tex], Rate = [tex]\(3.2 \times 10^{-1} \, M/s\)[/tex]
- Experiment 2: [tex]\([\text{HgCl}_2] = 0.164 \, M, \, [\text{C}_2\text{O}_4^{2-}] = 0.45 \, M\)[/tex], Rate = [tex]\(2.9 \times 10^{-4} \, M/s\)[/tex]
Since the concentration of [tex]\(\text{HgCl}_2\)[/tex] is the same in both experiments, the change in rate is only due to the change in concentration of [tex]\(\text{C}_2\text{O}_4^{2-}\)[/tex]. We can write:
[tex]\[ \frac{\text{Rate}_2}{\text{Rate}_1} = \left(\frac{[\text{C}_2\text{O}_4^{2-}]_2}{[\text{C}_2\text{O}_4^{2-}]_1}\right)^n \][/tex]
Plugging in the given values:
[tex]\[ \frac{2.9 \times 10^{-4}}{3.2 \times 10^{-1}} = \left(\frac{0.45}{0.15}\right)^n \][/tex]
Solving for [tex]\( n \)[/tex]:
[tex]\[ n = \frac{\log{\left( \frac{2.9 \times 10^{-4}}{3.2 \times 10^{-1}} \right)}}{\log{\left( \frac{0.45}{0.15} \right)}} \][/tex]
[tex]\[ n = -6.38 \][/tex]
#### Finding the order with respect to [tex]\(\text{HgCl}_2\)[/tex]:
- Experiment 2: [tex]\([\text{HgCl}_2] = 0.164 \, M, \, [\text{C}_2\text{O}_4^{2-}] = 0.45 \, M\)[/tex], Rate = [tex]\(2.9 \times 10^{-4} \, M/s\)[/tex]
- Experiment 3: [tex]\([\text{HgCl}_2] = 0.502 \, M, \, [\text{C}_2\text{O}_4^{2-}] = 0.45 \, M\)[/tex], Rate = [tex]\(1.4 \times 10^{-4} \, M/s\)[/tex]
Since the concentration of [tex]\(\text{C}_2\text{O}_4^{2-}\)[/tex] is the same in both experiments, the change in rate is only due to the change in concentration of [tex]\(\text{HgCl}_2\)[/tex]. We can write:
[tex]\[ \frac{\text{Rate}_3}{\text{Rate}_2} = \left(\frac{[\text{HgCl}_2]_3}{[\text{HgCl}_2]_2}\right)^m \][/tex]
Plugging in the given values:
[tex]\[ \frac{1.4 \times 10^{-4}}{2.9 \times 10^{-4}} = \left(\frac{0.502}{0.164}\right)^m \][/tex]
Solving for [tex]\( m \)[/tex]:
[tex]\[ m = \frac{\log{\left( \frac{1.4 \times 10^{-4}}{2.9 \times 10^{-4}} \right)}}{\log{\left( \frac{0.502}{0.164} \right)}} \][/tex]
[tex]\[ m = -0.65 \][/tex]
### Step 2: Determine the rate constant [tex]\( k \)[/tex]
Using any of the experiments:
- Experiment 1: [tex]\([\text{HgCl}_2] = 0.164 \, M, \, [\text{C}_2\text{O}_4^{2-}] = 0.15 \, M\)[/tex], Rate = [tex]\(3.2 \times 10^{-1} \, M/s\)[/tex]
We can write:
[tex]\[ k = \frac{\text{Rate}}{[\text{HgCl}_2]^m [\text{C}_2\text{O}_4^{2-}]^n} \][/tex]
Plugging in the values from Experiment 1:
[tex]\[ k = \frac{3.2 \times 10^{-1}}{(0.164)^{-0.65} \cdot (0.15)^{-6.38}} \][/tex]
[tex]\[ k = 5.49 \times 10^{-7} \, M^{(m+n)} \, s^{-1} \][/tex]
### Step 3: Calculate the reaction rate for given concentrations
For [tex]\([\text{HgCl}_2] = 0.150 \, M\)[/tex] and [tex]\([\text{C}_2\text{O}_4^{2-}] = 0.25 \, M\)[/tex]:
[tex]\[ \text{Rate} = k [\text{HgCl}_2]^m [\text{C}_2\text{O}_4^{2-}]^n \][/tex]
Plugging in the determined constants:
[tex]\[ \text{Rate} = 5.49 \times 10^{-7} \times (0.150)^{-0.65} \times (0.25)^{-6.38} \][/tex]
[tex]\[ \text{Rate} = 0.013 \, M/s \][/tex]
Therefore, the reaction rate when the concentration of [tex]\(\text{HgCl}_2\)[/tex] is 0.150 M and that of [tex]\(\text{C}_2\text{O}_4^{2-}\)[/tex] is 0.25 M is [tex]\(0.013 \, M/s\)[/tex].
### Step 1: Determine the reaction orders [tex]\( m \)[/tex] and [tex]\( n \)[/tex]
Assume the rate law for the reaction is:
[tex]\[ \text{Rate} = k [\text{HgCl}_2]^m [\text{C}_2\text{O}_4^{2-}]^n \][/tex]
We will use the given experimental data to determine the orders [tex]\( m \)[/tex] and [tex]\( n \)[/tex].
#### Finding the order with respect to [tex]\(\text{C}_2\text{O}_4^{2-}\)[/tex]:
- Experiment 1: [tex]\([\text{HgCl}_2] = 0.164 \, M, \, [\text{C}_2\text{O}_4^{2-}] = 0.15 \, M\)[/tex], Rate = [tex]\(3.2 \times 10^{-1} \, M/s\)[/tex]
- Experiment 2: [tex]\([\text{HgCl}_2] = 0.164 \, M, \, [\text{C}_2\text{O}_4^{2-}] = 0.45 \, M\)[/tex], Rate = [tex]\(2.9 \times 10^{-4} \, M/s\)[/tex]
Since the concentration of [tex]\(\text{HgCl}_2\)[/tex] is the same in both experiments, the change in rate is only due to the change in concentration of [tex]\(\text{C}_2\text{O}_4^{2-}\)[/tex]. We can write:
[tex]\[ \frac{\text{Rate}_2}{\text{Rate}_1} = \left(\frac{[\text{C}_2\text{O}_4^{2-}]_2}{[\text{C}_2\text{O}_4^{2-}]_1}\right)^n \][/tex]
Plugging in the given values:
[tex]\[ \frac{2.9 \times 10^{-4}}{3.2 \times 10^{-1}} = \left(\frac{0.45}{0.15}\right)^n \][/tex]
Solving for [tex]\( n \)[/tex]:
[tex]\[ n = \frac{\log{\left( \frac{2.9 \times 10^{-4}}{3.2 \times 10^{-1}} \right)}}{\log{\left( \frac{0.45}{0.15} \right)}} \][/tex]
[tex]\[ n = -6.38 \][/tex]
#### Finding the order with respect to [tex]\(\text{HgCl}_2\)[/tex]:
- Experiment 2: [tex]\([\text{HgCl}_2] = 0.164 \, M, \, [\text{C}_2\text{O}_4^{2-}] = 0.45 \, M\)[/tex], Rate = [tex]\(2.9 \times 10^{-4} \, M/s\)[/tex]
- Experiment 3: [tex]\([\text{HgCl}_2] = 0.502 \, M, \, [\text{C}_2\text{O}_4^{2-}] = 0.45 \, M\)[/tex], Rate = [tex]\(1.4 \times 10^{-4} \, M/s\)[/tex]
Since the concentration of [tex]\(\text{C}_2\text{O}_4^{2-}\)[/tex] is the same in both experiments, the change in rate is only due to the change in concentration of [tex]\(\text{HgCl}_2\)[/tex]. We can write:
[tex]\[ \frac{\text{Rate}_3}{\text{Rate}_2} = \left(\frac{[\text{HgCl}_2]_3}{[\text{HgCl}_2]_2}\right)^m \][/tex]
Plugging in the given values:
[tex]\[ \frac{1.4 \times 10^{-4}}{2.9 \times 10^{-4}} = \left(\frac{0.502}{0.164}\right)^m \][/tex]
Solving for [tex]\( m \)[/tex]:
[tex]\[ m = \frac{\log{\left( \frac{1.4 \times 10^{-4}}{2.9 \times 10^{-4}} \right)}}{\log{\left( \frac{0.502}{0.164} \right)}} \][/tex]
[tex]\[ m = -0.65 \][/tex]
### Step 2: Determine the rate constant [tex]\( k \)[/tex]
Using any of the experiments:
- Experiment 1: [tex]\([\text{HgCl}_2] = 0.164 \, M, \, [\text{C}_2\text{O}_4^{2-}] = 0.15 \, M\)[/tex], Rate = [tex]\(3.2 \times 10^{-1} \, M/s\)[/tex]
We can write:
[tex]\[ k = \frac{\text{Rate}}{[\text{HgCl}_2]^m [\text{C}_2\text{O}_4^{2-}]^n} \][/tex]
Plugging in the values from Experiment 1:
[tex]\[ k = \frac{3.2 \times 10^{-1}}{(0.164)^{-0.65} \cdot (0.15)^{-6.38}} \][/tex]
[tex]\[ k = 5.49 \times 10^{-7} \, M^{(m+n)} \, s^{-1} \][/tex]
### Step 3: Calculate the reaction rate for given concentrations
For [tex]\([\text{HgCl}_2] = 0.150 \, M\)[/tex] and [tex]\([\text{C}_2\text{O}_4^{2-}] = 0.25 \, M\)[/tex]:
[tex]\[ \text{Rate} = k [\text{HgCl}_2]^m [\text{C}_2\text{O}_4^{2-}]^n \][/tex]
Plugging in the determined constants:
[tex]\[ \text{Rate} = 5.49 \times 10^{-7} \times (0.150)^{-0.65} \times (0.25)^{-6.38} \][/tex]
[tex]\[ \text{Rate} = 0.013 \, M/s \][/tex]
Therefore, the reaction rate when the concentration of [tex]\(\text{HgCl}_2\)[/tex] is 0.150 M and that of [tex]\(\text{C}_2\text{O}_4^{2-}\)[/tex] is 0.25 M is [tex]\(0.013 \, M/s\)[/tex].
Thank you for being part of this discussion. Keep exploring, asking questions, and sharing your insights with the community. Together, we can find the best solutions. Thank you for visiting IDNLearn.com. We’re here to provide clear and concise answers, so visit us again soon.
