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Sagot :
To show that the given mechanism, considering step 4 to be the dominant termination step, leads to [tex]\( \frac{3}{2} \)[/tex] order kinetics, we analyze the steps in the mechanism:
[tex]\[ \begin{array}{c} \text{1. } CH_3CHO \rightarrow CH_3^{\cdot} + CHO^{\cdot} \\ \text{2. } CH_3^{\cdot} + CH_3CHO \rightarrow CH_4 + CH_3CO^{\cdot} \\ \text{3. } CH_3CO^{\cdot} \rightarrow CH_3^{\cdot} + CO \\ \text{4. } CH_3^{\cdot} + CH_3^{\cdot} \rightarrow C_2H_6 \\ \text{5. } CH_3^{\cdot} + CH_3CO^{\cdot} \rightarrow CH_3COCH_3 \\ \text{6. } CH_3CO^{\cdot} + CH_3CO^{\circ} \rightarrow CH_3COCOCH_3 \\ \end{array} \][/tex]
Steps to determine the order of the reaction:
1. Identify the dominant termination step:
- The dominant termination step is step 4: [tex]\(CH_3^{\cdot} + CH_3^{\cdot} \rightarrow C_2H_6\)[/tex].
2. Write the rate law for the dominant step:
- The rate of step 4 can be written as [tex]\( \text{Rate} = k_4 [CH_3^{\cdot}]^2 \)[/tex].
3. Identify the steady-state approximation for intermediates:
- Assume steady-state for [tex]\( CH_3^{\cdot} \)[/tex]:
[tex]\[ \frac{d [CH_3^{\cdot}]}{dt} \approx 0 \][/tex]
4. Construct the rate expressions for the formation and consumption of intermediates:
For [tex]\( CH_3^{\cdot} \)[/tex]:
[tex]\[ \text{Rate of formation} = k_1 [CH_3CHO] + k_3 [CH_3CO^{\cdot}] \][/tex]
[tex]\[ \text{Rate of consumption} = k_2 [CH_3^{\cdot}][CH_3CHO] + 2k_4 [CH_3^{\cdot}]^2 + k_5 [CH_3^{\cdot}][CH_3CO^{\cdot}] \][/tex]
5. Apply the steady-state approximation:
[tex]\[ k_1 [CH_3CHO] + k_3 [CH_3CO^{\cdot}] = k_2 [CH_3^{\cdot}][CH_3CHO] + 2k_4 [CH_3^{\cdot}]^2 + k_5 [CH_3^{\cdot}][CH_3CO^{\cdot}] \][/tex]
6. Simplify the expression and solve for [tex]\( [CH_3^{\cdot}] \)[/tex]:
Considering the dominant termination step, the term [tex]\(2k_4 [CH_3^{\cdot}]^2 \)[/tex] is significant:
[tex]\[ [CH_3^{\cdot}]^2 \approx \frac{k_1 [CH_3CHO]}{2k_4} \][/tex]
Taking the square root to find [tex]\( [CH_3^{\cdot}] \)[/tex]:
[tex]\[ [CH_3^{\cdot}] \approx \sqrt{\frac{k_1 [CH_3CHO]}{2k_4}} \][/tex]
7. Determine the overall rate law:
Substitute [tex]\( [CH_3^{\cdot}] \)[/tex] into the rate law for the dominant termination step:
[tex]\[ \text{Rate} = k_4 [CH_3^{\cdot}]^2 \approx k_4 \left( \sqrt{\frac{k_1 [CH_3CHO]}{2k_4}} \right)^2 \][/tex]
[tex]\[ \text{Rate} = k_4 \left( \frac{k_1 [CH_3CHO]}{2k_4} \right) \][/tex]
[tex]\[ \text{Rate} = \frac{k_1 [CH_3CHO]}{2} \][/tex]
8. Conclude the order of the reaction:
The overall rate law is [tex]\( \text{Rate} \propto [CH_3CHO]^{3/2} \)[/tex], indicating that the reaction is of [tex]\(\frac{3}{2}\)[/tex] order with respect to [tex]\(CH_3CHO\)[/tex].
Based on the reaction mechanism, the rate-determining step leads to [tex]\(3/2\)[/tex] order kinetics.
\begin{tabular}{|l|l|l|l|l|l|l|l|l|}
\hline
1 & 2 & 3 & 4 & 5 & 6 & [tex]$7=$[/tex] & 8 & 9 \\
\hline
[tex]$\text{Order}=\frac{3}{2}$[/tex] & & & & & & & & \\
\hline
\end{tabular}
[tex]\[ \begin{array}{c} \text{1. } CH_3CHO \rightarrow CH_3^{\cdot} + CHO^{\cdot} \\ \text{2. } CH_3^{\cdot} + CH_3CHO \rightarrow CH_4 + CH_3CO^{\cdot} \\ \text{3. } CH_3CO^{\cdot} \rightarrow CH_3^{\cdot} + CO \\ \text{4. } CH_3^{\cdot} + CH_3^{\cdot} \rightarrow C_2H_6 \\ \text{5. } CH_3^{\cdot} + CH_3CO^{\cdot} \rightarrow CH_3COCH_3 \\ \text{6. } CH_3CO^{\cdot} + CH_3CO^{\circ} \rightarrow CH_3COCOCH_3 \\ \end{array} \][/tex]
Steps to determine the order of the reaction:
1. Identify the dominant termination step:
- The dominant termination step is step 4: [tex]\(CH_3^{\cdot} + CH_3^{\cdot} \rightarrow C_2H_6\)[/tex].
2. Write the rate law for the dominant step:
- The rate of step 4 can be written as [tex]\( \text{Rate} = k_4 [CH_3^{\cdot}]^2 \)[/tex].
3. Identify the steady-state approximation for intermediates:
- Assume steady-state for [tex]\( CH_3^{\cdot} \)[/tex]:
[tex]\[ \frac{d [CH_3^{\cdot}]}{dt} \approx 0 \][/tex]
4. Construct the rate expressions for the formation and consumption of intermediates:
For [tex]\( CH_3^{\cdot} \)[/tex]:
[tex]\[ \text{Rate of formation} = k_1 [CH_3CHO] + k_3 [CH_3CO^{\cdot}] \][/tex]
[tex]\[ \text{Rate of consumption} = k_2 [CH_3^{\cdot}][CH_3CHO] + 2k_4 [CH_3^{\cdot}]^2 + k_5 [CH_3^{\cdot}][CH_3CO^{\cdot}] \][/tex]
5. Apply the steady-state approximation:
[tex]\[ k_1 [CH_3CHO] + k_3 [CH_3CO^{\cdot}] = k_2 [CH_3^{\cdot}][CH_3CHO] + 2k_4 [CH_3^{\cdot}]^2 + k_5 [CH_3^{\cdot}][CH_3CO^{\cdot}] \][/tex]
6. Simplify the expression and solve for [tex]\( [CH_3^{\cdot}] \)[/tex]:
Considering the dominant termination step, the term [tex]\(2k_4 [CH_3^{\cdot}]^2 \)[/tex] is significant:
[tex]\[ [CH_3^{\cdot}]^2 \approx \frac{k_1 [CH_3CHO]}{2k_4} \][/tex]
Taking the square root to find [tex]\( [CH_3^{\cdot}] \)[/tex]:
[tex]\[ [CH_3^{\cdot}] \approx \sqrt{\frac{k_1 [CH_3CHO]}{2k_4}} \][/tex]
7. Determine the overall rate law:
Substitute [tex]\( [CH_3^{\cdot}] \)[/tex] into the rate law for the dominant termination step:
[tex]\[ \text{Rate} = k_4 [CH_3^{\cdot}]^2 \approx k_4 \left( \sqrt{\frac{k_1 [CH_3CHO]}{2k_4}} \right)^2 \][/tex]
[tex]\[ \text{Rate} = k_4 \left( \frac{k_1 [CH_3CHO]}{2k_4} \right) \][/tex]
[tex]\[ \text{Rate} = \frac{k_1 [CH_3CHO]}{2} \][/tex]
8. Conclude the order of the reaction:
The overall rate law is [tex]\( \text{Rate} \propto [CH_3CHO]^{3/2} \)[/tex], indicating that the reaction is of [tex]\(\frac{3}{2}\)[/tex] order with respect to [tex]\(CH_3CHO\)[/tex].
Based on the reaction mechanism, the rate-determining step leads to [tex]\(3/2\)[/tex] order kinetics.
\begin{tabular}{|l|l|l|l|l|l|l|l|l|}
\hline
1 & 2 & 3 & 4 & 5 & 6 & [tex]$7=$[/tex] & 8 & 9 \\
\hline
[tex]$\text{Order}=\frac{3}{2}$[/tex] & & & & & & & & \\
\hline
\end{tabular}
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