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Sure! Let’s solve this step-by-step to find the rate of appearance of [tex]\( \text{CO}_2 \)[/tex] given the concentrations [tex]\([ \text{NO}_2 ] = [ \text{CO} ] = 0.495 \, \text{M}\)[/tex].
### Step 1: Determine the Reaction Orders
Given the rate law in the form:
[tex]\[ \text{rate} = k [\text{NO}_2]^x [\text{CO}]^y \][/tex]
We have the experimental data from 3 experiments:
1. Experiment 1:
- [tex]\([ \text{NO}_2 ]_1 = 0.263 \, \text{M}\)[/tex]
- [tex]\([ \text{CO} ]_1 = 0.826 \, \text{M}\)[/tex]
- [tex]\(\text{rate}_1 = 1.44 \times 10^{-5} \, \text{M/s}\)[/tex]
2. Experiment 2:
- [tex]\([ \text{NO}_2 ]_2 = 0.263 \, \text{M}\)[/tex]
- [tex]\([ \text{CO} ]_2 = 0.413 \, \text{M}\)[/tex]
- [tex]\(\text{rate}_2 = 1.44 \times 10^{-5} \, \text{M/s}\)[/tex]
3. Experiment 3:
- [tex]\([ \text{NO}_2 ]_3 = 0.526 \, \text{M}\)[/tex]
- [tex]\([ \text{CO} ]_3 = 0.413 \, \text{M}\)[/tex]
- [tex]\(\text{rate}_3 = 5.76 \times 10^{-5} \, \text{M/s}\)[/tex]
#### Finding the order with respect to [tex]\( \text{CO} \)[/tex] [tex]\((y)\)[/tex]:
Comparing Experiment 1 and Experiment 2 where [tex]\([ \text{NO}_2 ]\)[/tex] is constant:
[tex]\[ \frac{\text{rate}_2}{\text{rate}_1} = \left( \frac{[\text{CO}]_2}{[\text{CO}]_1} \right)^y \][/tex]
[tex]\[ \frac{1.44 \times 10^{-5}}{1.44 \times 10^{-5}} = \left( \frac{0.413}{0.826} \right)^y \][/tex]
[tex]\[ 1 = \left( 0.5 \right)^y \][/tex]
Here, [tex]\( y \)[/tex] would be 0 (since the ratio results in 1, indicating a zero-order dependence on [tex]\( [\text{CO}] \)[/tex]).
#### Finding the order with respect to [tex]\( \text{NO}_2 \)[/tex] [tex]\((x)\)[/tex]:
Comparing Experiment 1 and Experiment 3 where [tex]\([ \text{CO} ]\)[/tex] is constant:
[tex]\[ \frac{\text{rate}_3}{\text{rate}_1} = \left( \frac{[\text{NO}_2]_3}{[\text{NO}_2]_1} \right)^x \][/tex]
[tex]\[ \frac{5.76 \times 10^{-5}}{1.44 \times 10^{-5}} = \left( \frac{0.526}{0.263} \right)^x \][/tex]
[tex]\[ 4 = (2)^x \][/tex]
Here, [tex]\( x \)[/tex] would be close to 3.71 (since [tex]\( 2^{3.71} \approx 4 \)[/tex]).
### Step 2: Determine the Rate Constant [tex]\( k \)[/tex]
Using the data from Experiment 1 and the determined reaction orders:
[tex]\[ \text{rate}_1 = k [\text{NO}_2]_1^x [\text{CO}]_1^y \][/tex]
[tex]\[ 1.44 \times 10^{-5} = k (0.263)^{3.71} (0.826)^0 \][/tex]
[tex]\[ k = \frac{1.44 \times 10^{-5}}{(0.263)^{3.71}} k \approx 0.000625 \, \text{M}^{1-3.71} \text{s}^{-1} \][/tex]
### Step 3: Calculate the Rate of Appearance of [tex]\( \text{CO}_2 \)[/tex]
Given new concentrations [tex]\([ \text{NO}_2 ] = [ \text{CO} ] = 0.495 \, \text{M}\)[/tex], the rate of appearance of [tex]\( \text{CO}_2 \)[/tex] is:
[tex]\[ \text{rate}_{\text{new}} = k [\text{NO}_2]^{x} [\text{CO}]^{y} \][/tex]
[tex]\[ \text{rate}_{\text{new}} = 0.000625 \times (0.495)^{3.71} \times (0.495)^0 \][/tex]
[tex]\[ \text{rate}_{\text{new}} = 0.000625 \times (0.495)^{3.71} \][/tex]
[tex]\[ \text{rate}_{\text{new}} \approx 4.60 \times 10^{-5} \, \text{M/s} \][/tex]
Therefore, the rate of appearance of [tex]\( \text{CO}_2 \)[/tex] when [tex]\([ \text{NO}_2 ] = 0.495 \, \text{M}\)[/tex] and [tex]\([ \text{CO} ] = 0.495 \, \text{M}\)[/tex] is approximately [tex]\( 4.60 \times 10^{-5} \, \text{M/s} \)[/tex].
### Step 1: Determine the Reaction Orders
Given the rate law in the form:
[tex]\[ \text{rate} = k [\text{NO}_2]^x [\text{CO}]^y \][/tex]
We have the experimental data from 3 experiments:
1. Experiment 1:
- [tex]\([ \text{NO}_2 ]_1 = 0.263 \, \text{M}\)[/tex]
- [tex]\([ \text{CO} ]_1 = 0.826 \, \text{M}\)[/tex]
- [tex]\(\text{rate}_1 = 1.44 \times 10^{-5} \, \text{M/s}\)[/tex]
2. Experiment 2:
- [tex]\([ \text{NO}_2 ]_2 = 0.263 \, \text{M}\)[/tex]
- [tex]\([ \text{CO} ]_2 = 0.413 \, \text{M}\)[/tex]
- [tex]\(\text{rate}_2 = 1.44 \times 10^{-5} \, \text{M/s}\)[/tex]
3. Experiment 3:
- [tex]\([ \text{NO}_2 ]_3 = 0.526 \, \text{M}\)[/tex]
- [tex]\([ \text{CO} ]_3 = 0.413 \, \text{M}\)[/tex]
- [tex]\(\text{rate}_3 = 5.76 \times 10^{-5} \, \text{M/s}\)[/tex]
#### Finding the order with respect to [tex]\( \text{CO} \)[/tex] [tex]\((y)\)[/tex]:
Comparing Experiment 1 and Experiment 2 where [tex]\([ \text{NO}_2 ]\)[/tex] is constant:
[tex]\[ \frac{\text{rate}_2}{\text{rate}_1} = \left( \frac{[\text{CO}]_2}{[\text{CO}]_1} \right)^y \][/tex]
[tex]\[ \frac{1.44 \times 10^{-5}}{1.44 \times 10^{-5}} = \left( \frac{0.413}{0.826} \right)^y \][/tex]
[tex]\[ 1 = \left( 0.5 \right)^y \][/tex]
Here, [tex]\( y \)[/tex] would be 0 (since the ratio results in 1, indicating a zero-order dependence on [tex]\( [\text{CO}] \)[/tex]).
#### Finding the order with respect to [tex]\( \text{NO}_2 \)[/tex] [tex]\((x)\)[/tex]:
Comparing Experiment 1 and Experiment 3 where [tex]\([ \text{CO} ]\)[/tex] is constant:
[tex]\[ \frac{\text{rate}_3}{\text{rate}_1} = \left( \frac{[\text{NO}_2]_3}{[\text{NO}_2]_1} \right)^x \][/tex]
[tex]\[ \frac{5.76 \times 10^{-5}}{1.44 \times 10^{-5}} = \left( \frac{0.526}{0.263} \right)^x \][/tex]
[tex]\[ 4 = (2)^x \][/tex]
Here, [tex]\( x \)[/tex] would be close to 3.71 (since [tex]\( 2^{3.71} \approx 4 \)[/tex]).
### Step 2: Determine the Rate Constant [tex]\( k \)[/tex]
Using the data from Experiment 1 and the determined reaction orders:
[tex]\[ \text{rate}_1 = k [\text{NO}_2]_1^x [\text{CO}]_1^y \][/tex]
[tex]\[ 1.44 \times 10^{-5} = k (0.263)^{3.71} (0.826)^0 \][/tex]
[tex]\[ k = \frac{1.44 \times 10^{-5}}{(0.263)^{3.71}} k \approx 0.000625 \, \text{M}^{1-3.71} \text{s}^{-1} \][/tex]
### Step 3: Calculate the Rate of Appearance of [tex]\( \text{CO}_2 \)[/tex]
Given new concentrations [tex]\([ \text{NO}_2 ] = [ \text{CO} ] = 0.495 \, \text{M}\)[/tex], the rate of appearance of [tex]\( \text{CO}_2 \)[/tex] is:
[tex]\[ \text{rate}_{\text{new}} = k [\text{NO}_2]^{x} [\text{CO}]^{y} \][/tex]
[tex]\[ \text{rate}_{\text{new}} = 0.000625 \times (0.495)^{3.71} \times (0.495)^0 \][/tex]
[tex]\[ \text{rate}_{\text{new}} = 0.000625 \times (0.495)^{3.71} \][/tex]
[tex]\[ \text{rate}_{\text{new}} \approx 4.60 \times 10^{-5} \, \text{M/s} \][/tex]
Therefore, the rate of appearance of [tex]\( \text{CO}_2 \)[/tex] when [tex]\([ \text{NO}_2 ] = 0.495 \, \text{M}\)[/tex] and [tex]\([ \text{CO} ] = 0.495 \, \text{M}\)[/tex] is approximately [tex]\( 4.60 \times 10^{-5} \, \text{M/s} \)[/tex].
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