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To predict the rate for Experiment 4, we need to determine the rate law for the reaction. The rate law has the general form:
[tex]\[ \text{rate} = k \times [\text{NO}]^m \times [\text{Cl}_2]^n \][/tex]
where:
- [tex]\( k \)[/tex] is the rate constant,
- [tex]\( m \)[/tex] is the order of reaction with respect to NO,
- [tex]\( n \)[/tex] is the order of reaction with respect to Cl[tex]\(_2\)[/tex].
We can determine the values of [tex]\( m \)[/tex], [tex]\( n \)[/tex], and [tex]\( k \)[/tex] using the provided experimental data.
### Step-by-Step Solution
1. Determine the reaction order with respect to Cl[tex]\(_2\)[/tex] ( [tex]\( n \)[/tex] ):
Use data from Experiments 1 and 2, where NO concentration is constant and Cl[tex]\(_2\)[/tex] concentration varies.
[tex]\[ \frac{\text{rate}_2}{\text{rate}_1} = \left( \frac{[\text{NO}]_2}{[\text{NO}]_1} \right)^m \times \left( \frac{[\text{Cl}_2]_2}{[\text{Cl}_2]_1} \right)^n \][/tex]
Given:
[tex]\[ \frac{5.70}{0.63} = \left( \frac{0.20}{0.20} \right)^m \times \left( \frac{0.30}{0.10} \right)^n \][/tex]
Simplifying:
[tex]\[ 9 = 1 \times 3^n \][/tex]
Therefore:
[tex]\[ 9 = 3^n \implies n = 2 \][/tex]
2. Determine the reaction order with respect to NO ( [tex]\( m \)[/tex] ):
Use data from Experiments 1 and 3, where Cl[tex]\(_2\)[/tex] concentration is constant and NO concentration varies.
[tex]\[ \frac{\text{rate}_3}{\text{rate}_1} = \left( \frac{[\text{NO}]_3}{[\text{NO}]_1} \right)^m \times \left( \frac{[\text{Cl}_2]_3}{[\text{Cl}_2]_1} \right)^n \][/tex]
Given:
[tex]\[ \frac{2.58}{0.63} = \left( \frac{0.80}{0.20} \right)^m \times \left( \frac{0.10}{0.10} \right)^2 \][/tex]
Simplifying:
[tex]\[ 4.1 = 4^m \times 1 \][/tex]
Therefore:
[tex]\[ 4.1 = 4^m \implies m = 1 \][/tex]
3. Calculate the rate constant ([tex]\( k \)[/tex]) using data from one experiment:
Using data from Experiment 1:
[tex]\[ \text{rate} = k \times [\text{NO}]^1 \times [\text{Cl}_2]^2 \][/tex]
[tex]\[ 0.63 = k \times (0.20)^1 \times (0.10)^2 \][/tex]
Simplifying:
[tex]\[ 0.63 = k \times 0.20 \times 0.01 \][/tex]
[tex]\[ 0.63 = k \times 0.002 \][/tex]
Therefore:
[tex]\[ k = \frac{0.63}{0.002} = 315 \][/tex]
4. Predict the rate for Experiment 4:
Using the rate law:
[tex]\[ \text{rate}_4 = k \times [\text{NO}]^1 \times [\text{Cl}_2]^2 \][/tex]
Given concentrations for Experiment 4:
[tex]\[ [\text{NO}] = 0.40 \quad \text{and} \quad [\text{Cl}_2] = 0.20 \][/tex]
[tex]\[ \text{rate}_4 = 315 \times (0.40)^1 \times (0.20)^2 \][/tex]
Calculating:
[tex]\[ \text{rate}_4 = 315 \times 0.40 \times 0.04 \][/tex]
[tex]\[ \text{rate}_4 = 315 \times 0.016 \][/tex]
[tex]\[ \text{rate}_4 = 5.04 \][/tex]
So, the predicted rate for Experiment 4 is:
[tex]\[ \boxed{5.04 \, \text{M/s}} \][/tex]
[tex]\[ \text{rate} = k \times [\text{NO}]^m \times [\text{Cl}_2]^n \][/tex]
where:
- [tex]\( k \)[/tex] is the rate constant,
- [tex]\( m \)[/tex] is the order of reaction with respect to NO,
- [tex]\( n \)[/tex] is the order of reaction with respect to Cl[tex]\(_2\)[/tex].
We can determine the values of [tex]\( m \)[/tex], [tex]\( n \)[/tex], and [tex]\( k \)[/tex] using the provided experimental data.
### Step-by-Step Solution
1. Determine the reaction order with respect to Cl[tex]\(_2\)[/tex] ( [tex]\( n \)[/tex] ):
Use data from Experiments 1 and 2, where NO concentration is constant and Cl[tex]\(_2\)[/tex] concentration varies.
[tex]\[ \frac{\text{rate}_2}{\text{rate}_1} = \left( \frac{[\text{NO}]_2}{[\text{NO}]_1} \right)^m \times \left( \frac{[\text{Cl}_2]_2}{[\text{Cl}_2]_1} \right)^n \][/tex]
Given:
[tex]\[ \frac{5.70}{0.63} = \left( \frac{0.20}{0.20} \right)^m \times \left( \frac{0.30}{0.10} \right)^n \][/tex]
Simplifying:
[tex]\[ 9 = 1 \times 3^n \][/tex]
Therefore:
[tex]\[ 9 = 3^n \implies n = 2 \][/tex]
2. Determine the reaction order with respect to NO ( [tex]\( m \)[/tex] ):
Use data from Experiments 1 and 3, where Cl[tex]\(_2\)[/tex] concentration is constant and NO concentration varies.
[tex]\[ \frac{\text{rate}_3}{\text{rate}_1} = \left( \frac{[\text{NO}]_3}{[\text{NO}]_1} \right)^m \times \left( \frac{[\text{Cl}_2]_3}{[\text{Cl}_2]_1} \right)^n \][/tex]
Given:
[tex]\[ \frac{2.58}{0.63} = \left( \frac{0.80}{0.20} \right)^m \times \left( \frac{0.10}{0.10} \right)^2 \][/tex]
Simplifying:
[tex]\[ 4.1 = 4^m \times 1 \][/tex]
Therefore:
[tex]\[ 4.1 = 4^m \implies m = 1 \][/tex]
3. Calculate the rate constant ([tex]\( k \)[/tex]) using data from one experiment:
Using data from Experiment 1:
[tex]\[ \text{rate} = k \times [\text{NO}]^1 \times [\text{Cl}_2]^2 \][/tex]
[tex]\[ 0.63 = k \times (0.20)^1 \times (0.10)^2 \][/tex]
Simplifying:
[tex]\[ 0.63 = k \times 0.20 \times 0.01 \][/tex]
[tex]\[ 0.63 = k \times 0.002 \][/tex]
Therefore:
[tex]\[ k = \frac{0.63}{0.002} = 315 \][/tex]
4. Predict the rate for Experiment 4:
Using the rate law:
[tex]\[ \text{rate}_4 = k \times [\text{NO}]^1 \times [\text{Cl}_2]^2 \][/tex]
Given concentrations for Experiment 4:
[tex]\[ [\text{NO}] = 0.40 \quad \text{and} \quad [\text{Cl}_2] = 0.20 \][/tex]
[tex]\[ \text{rate}_4 = 315 \times (0.40)^1 \times (0.20)^2 \][/tex]
Calculating:
[tex]\[ \text{rate}_4 = 315 \times 0.40 \times 0.04 \][/tex]
[tex]\[ \text{rate}_4 = 315 \times 0.016 \][/tex]
[tex]\[ \text{rate}_4 = 5.04 \][/tex]
So, the predicted rate for Experiment 4 is:
[tex]\[ \boxed{5.04 \, \text{M/s}} \][/tex]
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