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Sagot :
To determine the rate law for the given reaction [tex]\( A + B \rightarrow C \)[/tex], we examine how changes in the concentrations of reactants [tex]\( A \)[/tex] and [tex]\( B \)[/tex] affect the initial reaction rate.
First, let's analyze the experimental data for each experiment:
[tex]\[ \begin{tabular}{c|l|l|r} Experiment & $[ A ]( M )$ & $[ B ]( M )$ & Initial Rate (M/s) \\ \hline 1 & 0.451 & 0.885 & 1.13 \\ 2 & 0.451 & 1.77 & 1.13 \\ 3 & 1.35 & 0.885 & 10.17 \\ \end{tabular} \][/tex]
Step-by-Step Solution:
1. Determine the order of reaction with respect to [tex]\( [B] \)[/tex]:
- Compare Experiment 1 and Experiment 2, where [tex]\( [A] \)[/tex] is constant:
[tex]\[ \frac{[A]_{\text{exp2}}}{[A]_{\text{exp1}}} = \frac{0.451}{0.451} = 1 \][/tex]
[tex]\( [B] \)[/tex] is doubled:
[tex]\[ \frac{[B]_{\text{exp2}}}{[B]_{\text{exp1}}} = \frac{1.77}{0.885} = 2 \][/tex]
The initial rate remains the same:
[tex]\[ \frac{\text{Rate}_{\text{exp2}}}{\text{Rate}_{\text{exp1}}} = \frac{1.13}{1.13} = 1 \][/tex]
Since the rate does not change when [tex]\( [B] \)[/tex] is doubled, the reaction is zero-order with respect to [tex]\( [B] \)[/tex]. This means the concentration of [tex]\( [B] \)[/tex] has no effect on the rate of the reaction.
2. Determine the order of reaction with respect to [tex]\( [A] \)[/tex]:
- Compare Experiment 1 and Experiment 3, where [tex]\( [B] \)[/tex] is constant:
[tex]\[ \frac{[B]_{\text{exp3}}}{[B]_{\text{exp1}}} = \frac{0.885}{0.885} = 1 \][/tex]
[tex]\( [A] \)[/tex] is tripled:
[tex]\[ \frac{[A]_{\text{exp3}}}{[A]_{\text{exp1}}} = \frac{1.35}{0.451} \approx 3 \][/tex]
The initial rate increases by approximately 9 times:
[tex]\[ \frac{\text{Rate}_{\text{exp3}}}{\text{Rate}_{\text{exp1}}} = \frac{10.17}{1.13} \approx 9 \][/tex]
Since the rate increases by a factor of 9 when [tex]\( [A] \)[/tex] is tripled, the reaction is second-order with respect to [tex]\( [A] \)[/tex]. The rate of reaction is proportional to the square of the concentration of [tex]\( [A] \)[/tex].
Putting these findings together, the rate law for the reaction is:
[tex]\[ \text{rate} = k[A]^2 \][/tex]
Here, [tex]\( k \)[/tex] is the rate constant.
Therefore, the rate law for the reaction [tex]\( A + B \rightarrow C \)[/tex] is:
[tex]\[ \boxed{k[A]^2} \][/tex]
First, let's analyze the experimental data for each experiment:
[tex]\[ \begin{tabular}{c|l|l|r} Experiment & $[ A ]( M )$ & $[ B ]( M )$ & Initial Rate (M/s) \\ \hline 1 & 0.451 & 0.885 & 1.13 \\ 2 & 0.451 & 1.77 & 1.13 \\ 3 & 1.35 & 0.885 & 10.17 \\ \end{tabular} \][/tex]
Step-by-Step Solution:
1. Determine the order of reaction with respect to [tex]\( [B] \)[/tex]:
- Compare Experiment 1 and Experiment 2, where [tex]\( [A] \)[/tex] is constant:
[tex]\[ \frac{[A]_{\text{exp2}}}{[A]_{\text{exp1}}} = \frac{0.451}{0.451} = 1 \][/tex]
[tex]\( [B] \)[/tex] is doubled:
[tex]\[ \frac{[B]_{\text{exp2}}}{[B]_{\text{exp1}}} = \frac{1.77}{0.885} = 2 \][/tex]
The initial rate remains the same:
[tex]\[ \frac{\text{Rate}_{\text{exp2}}}{\text{Rate}_{\text{exp1}}} = \frac{1.13}{1.13} = 1 \][/tex]
Since the rate does not change when [tex]\( [B] \)[/tex] is doubled, the reaction is zero-order with respect to [tex]\( [B] \)[/tex]. This means the concentration of [tex]\( [B] \)[/tex] has no effect on the rate of the reaction.
2. Determine the order of reaction with respect to [tex]\( [A] \)[/tex]:
- Compare Experiment 1 and Experiment 3, where [tex]\( [B] \)[/tex] is constant:
[tex]\[ \frac{[B]_{\text{exp3}}}{[B]_{\text{exp1}}} = \frac{0.885}{0.885} = 1 \][/tex]
[tex]\( [A] \)[/tex] is tripled:
[tex]\[ \frac{[A]_{\text{exp3}}}{[A]_{\text{exp1}}} = \frac{1.35}{0.451} \approx 3 \][/tex]
The initial rate increases by approximately 9 times:
[tex]\[ \frac{\text{Rate}_{\text{exp3}}}{\text{Rate}_{\text{exp1}}} = \frac{10.17}{1.13} \approx 9 \][/tex]
Since the rate increases by a factor of 9 when [tex]\( [A] \)[/tex] is tripled, the reaction is second-order with respect to [tex]\( [A] \)[/tex]. The rate of reaction is proportional to the square of the concentration of [tex]\( [A] \)[/tex].
Putting these findings together, the rate law for the reaction is:
[tex]\[ \text{rate} = k[A]^2 \][/tex]
Here, [tex]\( k \)[/tex] is the rate constant.
Therefore, the rate law for the reaction [tex]\( A + B \rightarrow C \)[/tex] is:
[tex]\[ \boxed{k[A]^2} \][/tex]
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