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From collision theory, it is found that when the ratedetermining step involves collision of two A molecules, the rate will be proportional to the square of the A concentration. Explain why.

Short Answer

Expert verified
According to collision theory, the rate of a reaction depends on the frequency and effectiveness of collisions between reactant molecules. In a reaction where the rate-determining step involves the collision of two A molecules, the probability of effective collisions is directly proportional to the concentration of A times the concentration of A again. Therefore, the rate is proportional to the square of the A concentration, represented mathematically as: Rate \( \propto \) [A]^2. This relationship highlights the increased probability of effective collisions between A molecules leading to a higher reaction rate as their concentration increases.

Step by step solution

01

1. Understanding Collision Theory

Collision theory is a model used to explain the rates of chemical reactions. It states that for a reaction to occur, the reactant molecules must collide with each other with sufficient energy (known as the activation energy) and the correct orientation. The rate of a reaction depends on the frequency and effectiveness of these collisions.
02

2. Relationship between concentration and rate of reaction

In a given reaction, the concentration of reactants has a direct effect on the rate of reaction. This is because the more molecules present, the higher the probability of effective collisions occurring per unit time. As a result, increasing the concentration of a reactant will lead to an increase in the rate of the reaction, while decreasing the concentration will lead to a decrease in the rate.
03

3. Applying Collision Theory to the given reaction

In this exercise, we are dealing with a reaction where the rate-determining step involves the collision of two A molecules. Since each collision involves two A molecules, the probability of an effective collision occurring is directly proportional to the concentration of A times the concentration of A again. Mathematically, this can be represented as: Rate \( \propto \) [A] * [A] Combining the concentrations gives us: Rate \( \propto \) [A]^2
04

4. Conclusion

According to collision theory, when the rate-determining step involves the collision of two A molecules, the rate will be proportional to the square of the concentration of A. This is due to the increased probability of effective collisions between A molecules as their concentration increases, resulting in a higher reaction rate.

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Key Concepts

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

Rate of Reaction
In chemical reactions, the **rate of reaction** refers to how quickly reactants are converted into products. This rate is influenced by several factors, including temperature, surface area, and most importantly for this concept, the frequency and conditions of collisions between molecules. Collision theory posits that reactant molecules must collide with sufficient energy and proper orientation for a reaction to take place. If these conditions are met more frequently, the reaction rate will naturally increase.

When we consider a situation where two molecules of reactant A need to collide, the reaction rate becomes dependent on the frequency of these specific collisions. Every time two A molecules collide effectively, they can move one step closer to forming the desired products, hence defining the rate at which the reaction proceeds under given conditions. This intrinsic linkage between collision occurrence and reaction speed is why increased collision chances directly translate to a higher rate of reaction.
Concentration
**Concentration** pertains to the amount of a reactant present in a unit volume of the reaction mixture. In the context of reaction rates, increasing the concentration of the reactant molecules means a higher number of those molecules is available to participate in collisions. More molecules result in a greater probability of effective collisions.

For instance, if you have a higher concentration of molecule A, not only are there more individual A molecules floating around waiting to collide, but the chance that any two of them encounter each other head-on is also heightened. This idea is critical in understanding why, when the rate-determining step involves two A molecules colliding, the reaction rate becomes dependent on the square of the concentration of A. Each A effectively has many more opportunities to collide, and additional collisions effectively scale with the square of the concentration.
Reaction Mechanism
A **reaction mechanism** details the step-by-step sequence of elementary reactions by which a chemical change occurs. For many reactions, the mechanism includes a distinct step called the rate-determining step. This step usually represents the slowest and most energy-demanding part of the process, controlling the overall rate of the entire reaction.

In our specific context, the rate-determining step hinges on the collision of two identical molecules, A. This dependence on a bimolecular event (two-parallel molecules colliding) significantly impacts how we assess the reaction's kinetics. By relying on collision theory, we recognize the relationship between the number of these collisions and the reactant concentration is pivotal.
  • Consider a four-step mechanism where the second step involves A+A colliding. If the second step is slowest, it bottlenecks the entire process.
  • The collision needs both energy and correct orientation for the reaction to progress, making it a prime candidate for rate-determined kinetic influence.
Understanding this helps in constructing reaction rate laws like the one given, where rate \( \propto [A]^2 \). This gives a clear mathematical picture of how reactants are transformed and why concentration impacts are squared for particular steps.

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Most popular questions from this chapter

The reaction \(A+B \rightarrow C\) was studied kinetically and the following data was obtained.$$ \begin{aligned} &\begin{array}{|c|c|c|c|} \hline \text { Experiment } & \text { A } & \text { B } & \text { Rate(mole/litter - min) } \\ \hline 1 & 1.0 \mathrm{M} & 1.0 \mathrm{M} & 0.15 \\ \hline 2 & 2.0 \mathrm{M} & 1.0 \mathrm{M} & 0.30 \\ \hline 3 & 1.0 \mathrm{M} & 2.0 \mathrm{M} & 0.15 \\ \hline \end{array}\\\ &\text { Determine the rate expression. } \end{aligned} $$

lwo ditterent molecules, \(\mathrm{A}\) and \(\mathrm{B}\), were reacted to give products according to the equation \(\mathrm{A}+\mathrm{B} \rightarrow\) products The rate of reaction was measured for fixed concentrations of \(\mathrm{A}\) and \(\mathrm{B}\) and the following data was obtained:$$ \begin{aligned} &\begin{array}{|l|l|l|l|} \hline \text { Experiment } & \begin{array}{l} {[\mathrm{A}]} \\ \text { (moles /iter) } \end{array} & \begin{array}{l} \text { [B] } \\ \text { (moles/iter) } \end{array} & \begin{array}{l} \text { Rate } \\ \text { (moles } \left.^{2} \text { /liter }^{2}-\text { sec }\right) \end{array} \\ \hline 1 & 1.0 & 1.0 & 0.05 \\ \hline 2 & 1.0 & 2.0 & 0.10 \\ \hline 3 & 3.0 & 1.0 & 0.45 \\ \hline \end{array}\\\ &\text { Write an expression for the rate of reaction. } \end{aligned} $$

You have the general reaction: \(\mathrm{A}+\mathrm{B} \rightarrow\) products. Consider the following reaction rate data and then determine the rate law for each case: (a) When \([\mathrm{A}]\) is doubled, the initial rates doubles; when [B] is doubled, the initial rate doubles, (b) When [A] doubles, initial rate doubles; when [B] doubles, the (c) Doubling [A], doubles initial rate; but initial rate is halved, doubling [B] leaves the rate unchanged.

In the reaction \(\mathrm{N}_{2} \mathrm{O}_{5} \rightarrow \mathrm{N}_{2} \mathrm{O}_{4}+(1 / 2) \mathrm{O}_{2}\), the \(\mathrm{N}_{2} \mathrm{O}_{5}\) decomposes by a first-order mechanism. At \(298^{\circ} \mathrm{K}\), the half- life is 340 minutes. Find the value of the reaction rate constant. Calculate the number of minutes required for the reaction to proceed 70 percent towards completion.

The reaction \(\mathrm{A}+2 \mathrm{~B} \rightarrow\) products was studied kinetically and the following data was obtained.$$ \begin{aligned} &\begin{array}{|l|l|l|l|} \hline \text { Experiment } & \begin{array}{l} {[\mathrm{A}]} \\ \text { (moles/liter) } \end{array} & \begin{array}{l} {[\mathrm{B}]} \\ \text { (moles/liter) } \end{array} & \text { Rate(moleliter-sec) } \\ \hline 1 & 2.0 \times 10^{-3} & 4.0 \times 10^{-3} & 5.7 \times 10^{-7} \\ \hline 2 & 2.0 \times 10^{-3} & 8.0 \times 10^{-3} & 11.4 \times 10^{-7} \\ \hline 3 & 4.0 \times 10^{-3} & 4.0 \times 10^{-3} & 22.8 \times 10^{-7} \\ \hline \end{array}\\\ &\text { Determine the rate law for this reaction. } \end{aligned} $$

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