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The initial rate of the reaction \(A+B \longrightarrow C+D\) is determined for different initial conditions, with the results listed in the table. (a) What is the order of reaction with respect to A and to B? (b) What is the overall reaction order? (c) What is the value of the rate constant, \(k ?\) $$\begin{array}{llll} \hline \text { Expt } & \text { [A], M } & \text { [B], M } & \text { Initial Rate, M s }^{-1} \\ \hline 1 & 0.185 & 0.133 & 3.35 \times 10^{-4} \\ 2 & 0.185 & 0.266 & 1.35 \times 10^{-3} \\ 3 & 0.370 & 0.133 & 6.75 \times 10^{-4} \\ 4 & 0.370 & 0.266 & 2.70 \times 10^{-3} \\ \hline \end{array}$$

Short Answer

Expert verified
The order of reaction with respect to A is 1, with respect to B is 2. Thus, overall reaction order is 3. The rate constant k can be calculated by substituting in the rate law the values from any experiment.

Step by step solution

01

Determine order of reaction with respect to A

To find the order of reaction with respect to 'A', compare Experiments 1 and 3. The concentration of 'B' is constant and the concentration of 'A' doubles from Experiment 1 to 3 ([A] in Expt. 3 = 2 × [A] in Expt. 1). Meanwhile, the initial rate also doubles (Rate in Expt. 3 = 2 × Rate in Expt. 1). Therefore, the order of reaction with respect to A is 1.
02

Determine order of reaction with respect to B

To find the order of reaction with respect to 'B', compare Experiments 1 and 2. The concentration of 'A' is constant while 'B' doubles from Experiment 1 to 2 ([B] in Expt. 2 = 2 × [B] in Expt. 1). The initial rate shows a four-fold increase (Rate in Expt. 2 = 4 × Rate in Expt. 1). Therefore, the order of reaction with respect to B is 2.
03

Identify the overall reaction order

The overall reaction order is simply the sum of the individual reaction orders. Thus, the overall order of reaction = order wrt A + order wrt B = 1 + 2 = 3.
04

Calculation of the rate constant k

Use the rate law and data from any of the experiments (e.g., Experiment 1) to calculate the rate constant. The rate law is given by rate = \(k[A]^m[B]^n\), where m and n are the orders with respect to A and B. Substituting the values from Experiment 1, \(3.35 × 10^{-4}\) MS^-1 = \(k(0.185)^1(0.133)^2\). Solve this equation for k.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Rate Law
Understanding the rate law is essential to predicting how a reaction's rate depends on the concentration of the reactants. The rate law is a mathematical expression that relates the rate of a chemical reaction to the concentration of the reactants. It usually has the form: \[ \text{rate} = k[A]^m[B]^n\] Here,
  • \(k\) is the rate constant, which is a unique value for each reaction at a given temperature.
  • \([A]\) and \([B]\) are the molar concentrations of reactants A and B.
  • \(m\) and \(n\) are the orders of the reaction with respect to A and B, respectively.
These orders \(m\) and \(n\) are determined experimentally and indicate how the rate depends on each reactant. For example, if \(m=1\), doubling the concentration of A doubles the reaction rate.
If \(m=2\), the rate increases fourfold. Importantly, these values do not necessarily correspond to the stoichiometric coefficients of the balanced chemical equation.
Rate Constant
The rate constant, \(k\), is a crucial factor in the rate law, linking reactant concentrations to the reaction rate. Unlike reaction orders, the rate constant is influenced by external factors like temperature and the presence of a catalyst. In our example, we calculated \(k\) using data from Experiment 1:\[3.35 \times 10^{-4} \, \text{Ms}^{-1} = k(0.185)^1(0.133)^2\]Solving for \(k\), we factor in the specific concentrations used in the experiment. The outcome gives us the rate constant's value in units that might vary based on the reaction order. To accurately solve for \(k\), rearrange the formula as follows:\[k = \frac{\text{rate}}{[A]^m[B]^n}\]By performing this calculation, you gain a clearer understanding of \(k\) and how it allows predictions about reaction behavior under similar conditions.
Reaction Kinetics
Reaction kinetics encompasses the rates of chemical reactions and the mechanisms that control these rates. It is a field that focuses on both theoretical and practical aspects of how reactions proceed over time. In reaction kinetics, several key factors influence the speed and progress:
  • Concentration: Higher concentrations of reactants can increase the likelihood of particle collisions, leading to faster reaction rates.
  • Temperature: Raising temperature generally speeds up reactions due to increased particle motion and collision frequency.
  • Presence of Catalysts: Catalysts lower the activation energy required for a reaction, enabling it to proceed quicker without being consumed.
  • Surface Area: More surface area allows more opportunities for collision between particles, often increasing rate in heterogeneous reactions.
By studying these aspects, scientists and engineers can develop methods to control and optimize reactions in industrial settings, such as pharmaceutical synthesis or energy production tasks. Moreover, reaction kinetics plays a fundamental role in understanding natural processes such as metabolic pathways or atmospheric chemistry.

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Most popular questions from this chapter

The half-lives of both zero-order and second-order reactions depend on the initial concentration, as well as on the rate constant. In one case, the half- life gets longer as the initial concentration increases, and in the other it gets shorter. Which is which, and why isn't the situation the same for both?

In three different experiments, the following results were obtained for the reaction \(A \longrightarrow\) products: \([\mathrm{A}]_{0}=1.00 \mathrm{M}, t_{1 / 2}=50 \mathrm{min} ;[\mathrm{A}]_{0}=200 \mathrm{M}, t_{1 / 2}=\) \(25 \min ;[\mathrm{A}]_{0}=0.50 \mathrm{M}, t_{1 / 2}=100 \mathrm{min} .\) Write the rate equation for this reaction, and indicate the value of \(k.\)

For the reaction \(\mathrm{A}+2 \mathrm{B} \longrightarrow \mathrm{C}+\mathrm{D},\) the rate law is rate of reaction \(=k[\mathrm{A}][\mathrm{B}]\) (a) Show that the following mechanism is consistent with the stoichiometry of the overall reaction and with the rate law. $$\begin{array}{l} \mathrm{A}+\mathrm{B} \longrightarrow \mathrm{I} \quad(\text { slow }) \\ \mathrm{I}+\mathrm{B} \longrightarrow \mathrm{C}+\mathrm{D} \quad(\text { fast }) \end{array}$$ (b) Show that the following mechanism is consistent with the stoichiometry of the overall reaction, but not with the rate law. $$\begin{array}{c} 2 \mathrm{B} \stackrel{k_{1}}{\mathrm{k}_{1}} \mathrm{B}_{2} \text { (fast) } \\\ \mathrm{A}+\mathrm{B}_{2} \stackrel{k_{2}}{\longrightarrow} \mathrm{C}+\mathrm{D} \text { (slow) } \end{array}$$

The reaction \(2 \mathrm{NO}+\mathrm{Cl}_{2} \longrightarrow 2 \mathrm{NOCl}\) has the rate law: rate of reaction \(=k[\mathrm{NO}]^{2}\left[\mathrm{Cl}_{2}\right] .\) Propose a twostep mechanism for this reaction consisting of a fast reversible first step, followed by a slow step.

You want to test the following proposed mechanism for the oxidation of HBr. $$\begin{array}{c} \mathrm{HBr}+\mathrm{O}_{2} \stackrel{k_{1}}{\longrightarrow} \mathrm{HOOBr} \\\ \mathrm{HOOBr}+\mathrm{HBr} \stackrel{k_{2}}{\longrightarrow} 2 \mathrm{HOBr} \\\ \mathrm{HOBr}+\mathrm{HBr} \stackrel{k_{3}}{\longrightarrow} \mathrm{H}_{2} \mathrm{O}+\mathrm{Br}_{2} \end{array}$$ You find that the rate is first order with respect to HBr and to \(\mathrm{O}_{2}\). You cannot detect HOBr among the products. (a) If the proposed mechanism is correct, which must be the rate-determining step? (b) Can you prove the mechanism from these observations? (c) Can you disprove the mechanism from these observations?

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