Answer :
To determine the rate law and the rate constant for the reaction [tex]\( 2 A + B + 2 C \rightarrow 3 X \)[/tex] using the provided experimental data, follow these steps:
### Step 1: Determine Reaction Orders with Respect to Each Reactant
We use the experimental data to compare the changes in the initial rates with the changes in the initial concentrations of the reactants.
#### Determine the Order with Respect to [tex]\( A \)[/tex]
Compare Experiments 1 and 2:
- In Experiment 1: [tex]\([A] = 0.10 \, \text{mol/L}\)[/tex]
- In Experiment 2: [tex]\([A] = 0.20 \, \text{mol/L}\)[/tex]
- [tex]\([B]\)[/tex] and [tex]\([C]\)[/tex] remain the same in both experiments ([tex]\(0.10 \, \text{mol/L}\)[/tex] each).
The rates are given as:
- Rate in Experiment 1: [tex]\(0.0003 \, \text{mol/(L·s)}\)[/tex]
- Rate in Experiment 2: [tex]\(0.0012 \, \text{mol/(L·s)}\)[/tex]
Calculate the order with respect to [tex]\( A \)[/tex]:
[tex]\[ \frac{\text{Rate}_2}{\text{Rate}_1} = \left( \frac{[ A ]_2}{[ A ]_1} \right)^{\text{order}_A} \][/tex]
[tex]\[ \frac{0.0012}{0.0003} = \left( \frac{0.20}{0.10} \right)^{\text{order}_A} \][/tex]
[tex]\[ 4 = 2^{\text{order}_A} \][/tex]
[tex]\[ \text{order}_A = 2 \][/tex]
#### Determine the Order with Respect to [tex]\( B \)[/tex]
Compare Experiments 1 and 3:
- In Experiment 1: [tex]\([B] = 0.10 \, \text{mol/L}\)[/tex]
- In Experiment 3: [tex]\([B] = 0.30 \, \text{mol/L}\)[/tex]
- [tex]\([A]\)[/tex] and [tex]\([C]\)[/tex] remain the same in both experiments ([tex]\(0.10 \, \text{mol/L}\)[/tex] each).
The rates are given as:
- Rate in Experiment 1: [tex]\(0.0003 \, \text{mol/(L·s)}\)[/tex]
- Rate in Experiment 3: [tex]\(0.0003 \, \text{mol/(L·s)}\)[/tex]
Calculate the order with respect to [tex]\( B \)[/tex]:
[tex]\[ \frac{\text{Rate}_3}{\text{Rate}_1} = \left( \frac{[ B ]_3}{[ B ]_1} \right)^{\text{order}_B} \][/tex]
[tex]\[ \frac{0.0003}{0.0003} = \left( \frac{0.30}{0.10} \right)^{\text{order}_B} \][/tex]
[tex]\[ 1 = 3^{\text{order}_B} \][/tex]
[tex]\[ \text{order}_B = 0 \][/tex]
#### Determine the Order with Respect to [tex]\( C \)[/tex]
Compare Experiments 2 and 4:
- In Experiment 2: [tex]\([C] = 0.10 \, \text{mol/L}\)[/tex]
- In Experiment 4: [tex]\([C] = 0.20 \, \text{mol/L}\)[/tex]
- [tex]\([A]\)[/tex] remains the same in both experiments ([tex]\(0.20 \, \text{mol/L}\)[/tex]) and [tex]\([B]\)[/tex] is [tex]\(0.10 \, \text{mol/L}\)[/tex] in both.
The rates are given as:
- Rate in Experiment 2: [tex]\(0.0012 \, \text{mol/(L·s)}\)[/tex]
- Rate in Experiment 4: [tex]\(0.0024 \, \text{mol/(L·s)}\)[/tex]
Calculate the order with respect to [tex]\( C \)[/tex]:
[tex]\[ \frac{\text{Rate}_4}{\text{Rate}_2} = \left( \frac{[ C ]_4}{[ C ]_2} \right)^{\text{order}_C} \][/tex]
[tex]\[ \frac{0.0024}{0.0012} = \left( \frac{0.20}{0.10} \right)^{\text{order}_C} \][/tex]
[tex]\[ 2 = 2^{\text{order}_C} \][/tex]
[tex]\[ \text{order}_C = 1 \][/tex]
### Step 2: Write the Rate Law
The determined orders are:
- Order with respect to [tex]\( A \)[/tex]: [tex]\( 2 \)[/tex]
- Order with respect to [tex]\( B \)[/tex]: [tex]\( 0 \)[/tex]
- Order with respect to [tex]\( C \)[/tex]: [tex]\( 1 \)[/tex]
Thus, the rate law is:
[tex]\[ \text{Rate} = k [A]^2 [B]^0 [C]^1 \][/tex]
[tex]\[ \text{Rate} = k [A]^2 [C] \][/tex]
### Step 3: Calculate the Rate Constant [tex]\( k \)[/tex]
Use the data from Experiment 1 to determine [tex]\( k \)[/tex].
[tex]\[ \text{Rate} = k [A]^2 [C] \][/tex]
[tex]\[ 0.0003 \, \text{mol/(L·s)} = k (0.10 \, \text{mol/L})^2 (0.10 \, \text{mol/L}) \][/tex]
[tex]\[ 0.0003 \, \text{mol/(L·s)} = k (0.001) \, \text{mol}^3 / \text{L}^3 \][/tex]
[tex]\[ k = \frac{0.0003 \, \text{mol/(L·s)}}{0.001 \, \text{mol}^3 / \text{L}^3} \][/tex]
[tex]\[ k = 0.3 \, \text{L}^2 / \text{(mol^2·s)} \][/tex]
### Final Rate Law and Rate Constant
The rate law for the reaction is:
[tex]\[ \text{Rate} = 0.3 \, \text{L}^2 / \text{(mol^2·s)} [A]^2 [C] \][/tex]
### Units:
The units of the rate constant [tex]\( k \)[/tex] are [tex]\( \text{L}^2 / \text{(mol^2·s)} \)[/tex].
### Step 1: Determine Reaction Orders with Respect to Each Reactant
We use the experimental data to compare the changes in the initial rates with the changes in the initial concentrations of the reactants.
#### Determine the Order with Respect to [tex]\( A \)[/tex]
Compare Experiments 1 and 2:
- In Experiment 1: [tex]\([A] = 0.10 \, \text{mol/L}\)[/tex]
- In Experiment 2: [tex]\([A] = 0.20 \, \text{mol/L}\)[/tex]
- [tex]\([B]\)[/tex] and [tex]\([C]\)[/tex] remain the same in both experiments ([tex]\(0.10 \, \text{mol/L}\)[/tex] each).
The rates are given as:
- Rate in Experiment 1: [tex]\(0.0003 \, \text{mol/(L·s)}\)[/tex]
- Rate in Experiment 2: [tex]\(0.0012 \, \text{mol/(L·s)}\)[/tex]
Calculate the order with respect to [tex]\( A \)[/tex]:
[tex]\[ \frac{\text{Rate}_2}{\text{Rate}_1} = \left( \frac{[ A ]_2}{[ A ]_1} \right)^{\text{order}_A} \][/tex]
[tex]\[ \frac{0.0012}{0.0003} = \left( \frac{0.20}{0.10} \right)^{\text{order}_A} \][/tex]
[tex]\[ 4 = 2^{\text{order}_A} \][/tex]
[tex]\[ \text{order}_A = 2 \][/tex]
#### Determine the Order with Respect to [tex]\( B \)[/tex]
Compare Experiments 1 and 3:
- In Experiment 1: [tex]\([B] = 0.10 \, \text{mol/L}\)[/tex]
- In Experiment 3: [tex]\([B] = 0.30 \, \text{mol/L}\)[/tex]
- [tex]\([A]\)[/tex] and [tex]\([C]\)[/tex] remain the same in both experiments ([tex]\(0.10 \, \text{mol/L}\)[/tex] each).
The rates are given as:
- Rate in Experiment 1: [tex]\(0.0003 \, \text{mol/(L·s)}\)[/tex]
- Rate in Experiment 3: [tex]\(0.0003 \, \text{mol/(L·s)}\)[/tex]
Calculate the order with respect to [tex]\( B \)[/tex]:
[tex]\[ \frac{\text{Rate}_3}{\text{Rate}_1} = \left( \frac{[ B ]_3}{[ B ]_1} \right)^{\text{order}_B} \][/tex]
[tex]\[ \frac{0.0003}{0.0003} = \left( \frac{0.30}{0.10} \right)^{\text{order}_B} \][/tex]
[tex]\[ 1 = 3^{\text{order}_B} \][/tex]
[tex]\[ \text{order}_B = 0 \][/tex]
#### Determine the Order with Respect to [tex]\( C \)[/tex]
Compare Experiments 2 and 4:
- In Experiment 2: [tex]\([C] = 0.10 \, \text{mol/L}\)[/tex]
- In Experiment 4: [tex]\([C] = 0.20 \, \text{mol/L}\)[/tex]
- [tex]\([A]\)[/tex] remains the same in both experiments ([tex]\(0.20 \, \text{mol/L}\)[/tex]) and [tex]\([B]\)[/tex] is [tex]\(0.10 \, \text{mol/L}\)[/tex] in both.
The rates are given as:
- Rate in Experiment 2: [tex]\(0.0012 \, \text{mol/(L·s)}\)[/tex]
- Rate in Experiment 4: [tex]\(0.0024 \, \text{mol/(L·s)}\)[/tex]
Calculate the order with respect to [tex]\( C \)[/tex]:
[tex]\[ \frac{\text{Rate}_4}{\text{Rate}_2} = \left( \frac{[ C ]_4}{[ C ]_2} \right)^{\text{order}_C} \][/tex]
[tex]\[ \frac{0.0024}{0.0012} = \left( \frac{0.20}{0.10} \right)^{\text{order}_C} \][/tex]
[tex]\[ 2 = 2^{\text{order}_C} \][/tex]
[tex]\[ \text{order}_C = 1 \][/tex]
### Step 2: Write the Rate Law
The determined orders are:
- Order with respect to [tex]\( A \)[/tex]: [tex]\( 2 \)[/tex]
- Order with respect to [tex]\( B \)[/tex]: [tex]\( 0 \)[/tex]
- Order with respect to [tex]\( C \)[/tex]: [tex]\( 1 \)[/tex]
Thus, the rate law is:
[tex]\[ \text{Rate} = k [A]^2 [B]^0 [C]^1 \][/tex]
[tex]\[ \text{Rate} = k [A]^2 [C] \][/tex]
### Step 3: Calculate the Rate Constant [tex]\( k \)[/tex]
Use the data from Experiment 1 to determine [tex]\( k \)[/tex].
[tex]\[ \text{Rate} = k [A]^2 [C] \][/tex]
[tex]\[ 0.0003 \, \text{mol/(L·s)} = k (0.10 \, \text{mol/L})^2 (0.10 \, \text{mol/L}) \][/tex]
[tex]\[ 0.0003 \, \text{mol/(L·s)} = k (0.001) \, \text{mol}^3 / \text{L}^3 \][/tex]
[tex]\[ k = \frac{0.0003 \, \text{mol/(L·s)}}{0.001 \, \text{mol}^3 / \text{L}^3} \][/tex]
[tex]\[ k = 0.3 \, \text{L}^2 / \text{(mol^2·s)} \][/tex]
### Final Rate Law and Rate Constant
The rate law for the reaction is:
[tex]\[ \text{Rate} = 0.3 \, \text{L}^2 / \text{(mol^2·s)} [A]^2 [C] \][/tex]
### Units:
The units of the rate constant [tex]\( k \)[/tex] are [tex]\( \text{L}^2 / \text{(mol^2·s)} \)[/tex].
