Answer :
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].
