Answer :
Let's analyze the given data and determine the required parameters step-by-step.
### Experimental Data
[tex]\[ \begin{array}{|l|l|l|l|} \hline \text{Exp.} & [A] \, (\text{mol/dm}^3) & [B] \, (\text{mol/dm}^3) & \text{Rate} \, (\text{mol/dm}^3/\text{s}) \\ \hline 1 & 6.0 \times 10^{-3} & 1.0 \times 10^{-3} & 0.012 \\ \hline 2 & 6.0 \times 10^{-3} & 2.0 \times 10^{-3} & 0.024 \\ \hline 3 & 2.0 \times 10^{-3} & 1.5 \times 10^{-3} & 0.002 \\ \hline 4 & 4.0 \times 10^{-3} & 1.5 \times 10^{-3} & 0.008 \\ \hline \end{array} \][/tex]
The general rate law for the reaction is given by:
[tex]\[ r = k \cdot [A]^a \cdot [B]^b \][/tex]
We'll determine each component systematically.
### (i) Order of the reaction with respect to [tex]\(A\)[/tex]:
To find the order with respect to [tex]\(A\)[/tex] [tex]\((a)\)[/tex], we need to compare experiments where [tex]\( [B] \)[/tex] is constant.
Comparing Experiment 3 and Experiment 4:
[tex]\[ \frac{\text{Rate}_4}{\text{Rate}_3} = \frac{0.008}{0.002} = 4 \][/tex]
[tex]\[ \frac{[A]_4}{[A]_3} = \frac{4.0 \times 10^{-3}}{2.0 \times 10^{-3}} = 2 \][/tex]
[tex]\[ 4 = 2^a \\ \text{From this, } a = \log_2{4} = 2 \][/tex]
So, the order with respect to [tex]\(A\)[/tex] is:
[tex]\[ a = 2 \][/tex]
### (ii) Order of the reaction with respect to [tex]\(B\)[/tex]:
To find the order with respect to [tex]\(B\)[/tex] [tex]\((b)\)[/tex], we need to compare experiments where [tex]\( [A] \)[/tex] is constant.
Comparing Experiment 1 and Experiment 2:
[tex]\[ \frac{\text{Rate}_2}{\text{Rate}_1} = \frac{0.024}{0.012} = 2 \][/tex]
[tex]\[ \frac{[B]_2}{[B]_1} = \frac{2.0 \times 10^{-3}}{1.0 \times 10^{-3}} = 2 \][/tex]
[tex]\[ 2 = 2^b \\ \text{From this, } b = \log_2{2} = 1 \][/tex]
So, the order with respect to [tex]\(B\)[/tex] is:
[tex]\[ b = 1 \][/tex]
### (iii) Overall order of the reaction:
The overall order is the sum of the individual orders:
[tex]\[ a + b = 2 + 1 = 3 \][/tex]
### (iv) Rate law equation:
Substituting the orders into the rate law, we get:
[tex]\[ r = k \cdot [A]^2 \cdot [B]^1 \][/tex]
### (v) Calculation of the rate constant [tex]\(k\)[/tex]:
We can use data from any experiment to calculate [tex]\(k\)[/tex]. Let's use the data from Experiment 1:
[tex]\[ \text{Rate} = 0.012 \text{ mol/dm}^3\text{/s} \\ [A] = 6.0 \times 10^{-3} \text{ mol/dm}^3 \\ [B] = 1.0 \times 10^{-3} \text{ mol/dm}^3 \][/tex]
Substitute these values into the rate law equation:
[tex]\[ 0.012 = k \cdot (6.0 \times 10^{-3})^2 \cdot (1.0 \times 10^{-3}) \][/tex]
[tex]\[ 0.012 = k \cdot 36 \times 10^{-12} \][/tex]
[tex]\[ k = \frac{0.012}{36 \times 10^{-12}} = 3.3333 \times 10^{5} \][/tex]
So, the rate constant [tex]\(k\)[/tex] is approximately [tex]\( 333333.3333 \)[/tex] [tex]\(\text{dm}^9/\text{mol}^3/\text{s} \)[/tex].
Therefore, the detailed solution to the problem is as follows:
1. Order with respect to [tex]\(A\)[/tex]: 2.
2. Order with respect to [tex]\(B\)[/tex]: 1.
3. Overall order: 3.
4. Rate law: [tex]\( r = k \cdot [A]^2 \cdot [B]^1 \)[/tex].
5. Rate constant [tex]\(k\)[/tex]: Approximately [tex]\( 333333.3333 \)[/tex] [tex]\(\text{dm}^9/\text{mol}^3/\text{s} \)[/tex].
### Experimental Data
[tex]\[ \begin{array}{|l|l|l|l|} \hline \text{Exp.} & [A] \, (\text{mol/dm}^3) & [B] \, (\text{mol/dm}^3) & \text{Rate} \, (\text{mol/dm}^3/\text{s}) \\ \hline 1 & 6.0 \times 10^{-3} & 1.0 \times 10^{-3} & 0.012 \\ \hline 2 & 6.0 \times 10^{-3} & 2.0 \times 10^{-3} & 0.024 \\ \hline 3 & 2.0 \times 10^{-3} & 1.5 \times 10^{-3} & 0.002 \\ \hline 4 & 4.0 \times 10^{-3} & 1.5 \times 10^{-3} & 0.008 \\ \hline \end{array} \][/tex]
The general rate law for the reaction is given by:
[tex]\[ r = k \cdot [A]^a \cdot [B]^b \][/tex]
We'll determine each component systematically.
### (i) Order of the reaction with respect to [tex]\(A\)[/tex]:
To find the order with respect to [tex]\(A\)[/tex] [tex]\((a)\)[/tex], we need to compare experiments where [tex]\( [B] \)[/tex] is constant.
Comparing Experiment 3 and Experiment 4:
[tex]\[ \frac{\text{Rate}_4}{\text{Rate}_3} = \frac{0.008}{0.002} = 4 \][/tex]
[tex]\[ \frac{[A]_4}{[A]_3} = \frac{4.0 \times 10^{-3}}{2.0 \times 10^{-3}} = 2 \][/tex]
[tex]\[ 4 = 2^a \\ \text{From this, } a = \log_2{4} = 2 \][/tex]
So, the order with respect to [tex]\(A\)[/tex] is:
[tex]\[ a = 2 \][/tex]
### (ii) Order of the reaction with respect to [tex]\(B\)[/tex]:
To find the order with respect to [tex]\(B\)[/tex] [tex]\((b)\)[/tex], we need to compare experiments where [tex]\( [A] \)[/tex] is constant.
Comparing Experiment 1 and Experiment 2:
[tex]\[ \frac{\text{Rate}_2}{\text{Rate}_1} = \frac{0.024}{0.012} = 2 \][/tex]
[tex]\[ \frac{[B]_2}{[B]_1} = \frac{2.0 \times 10^{-3}}{1.0 \times 10^{-3}} = 2 \][/tex]
[tex]\[ 2 = 2^b \\ \text{From this, } b = \log_2{2} = 1 \][/tex]
So, the order with respect to [tex]\(B\)[/tex] is:
[tex]\[ b = 1 \][/tex]
### (iii) Overall order of the reaction:
The overall order is the sum of the individual orders:
[tex]\[ a + b = 2 + 1 = 3 \][/tex]
### (iv) Rate law equation:
Substituting the orders into the rate law, we get:
[tex]\[ r = k \cdot [A]^2 \cdot [B]^1 \][/tex]
### (v) Calculation of the rate constant [tex]\(k\)[/tex]:
We can use data from any experiment to calculate [tex]\(k\)[/tex]. Let's use the data from Experiment 1:
[tex]\[ \text{Rate} = 0.012 \text{ mol/dm}^3\text{/s} \\ [A] = 6.0 \times 10^{-3} \text{ mol/dm}^3 \\ [B] = 1.0 \times 10^{-3} \text{ mol/dm}^3 \][/tex]
Substitute these values into the rate law equation:
[tex]\[ 0.012 = k \cdot (6.0 \times 10^{-3})^2 \cdot (1.0 \times 10^{-3}) \][/tex]
[tex]\[ 0.012 = k \cdot 36 \times 10^{-12} \][/tex]
[tex]\[ k = \frac{0.012}{36 \times 10^{-12}} = 3.3333 \times 10^{5} \][/tex]
So, the rate constant [tex]\(k\)[/tex] is approximately [tex]\( 333333.3333 \)[/tex] [tex]\(\text{dm}^9/\text{mol}^3/\text{s} \)[/tex].
Therefore, the detailed solution to the problem is as follows:
1. Order with respect to [tex]\(A\)[/tex]: 2.
2. Order with respect to [tex]\(B\)[/tex]: 1.
3. Overall order: 3.
4. Rate law: [tex]\( r = k \cdot [A]^2 \cdot [B]^1 \)[/tex].
5. Rate constant [tex]\(k\)[/tex]: Approximately [tex]\( 333333.3333 \)[/tex] [tex]\(\text{dm}^9/\text{mol}^3/\text{s} \)[/tex].
