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The rate law has the form; rate \(=k[\mathrm{~A}][\mathrm{B}]^{32}\), can the reaction be an elementary process? (a) yes (b) no (c) may be yes or no (d) cannot be predicted

Short Answer

Expert verified
(b) no

Step by step solution

01

Understanding Elementary Reactions

An elementary reaction is a reaction in which reactants convert to products in a single step with no intermediates. For an elementary reaction, the rate law corresponds directly to the stoichiometry of the balanced equation.
02

Interpreting the Given Rate Law

The rate law provided is: \(\text{rate} = k[\text{A}][\text{B}]^{32}\). This implies that the reaction rate depends on the concentration of A raised to the first power and the concentration of B raised to the 32nd power.
03

Checking Stoichiometric Coefficients for Elementary Processes

In an elementary reaction, the coefficients in the rate law must match the stoichiometric coefficients in the balanced chemical equation. The exponent of 32 on [B] suggests that the stoichiometric coefficient for B would need to be 32.
04

Conclusion on Feasibility

An elementary reaction with a stoichiometric coefficient of 32 is highly unlikely, as such large coefficients are not typical for direct, single-step reactions, making it improbable for this to be an elementary process.

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Key Concepts

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

Rate Law
The rate law is a mathematical equation that represents the relationship between the concentration of reactants and the speed or rate of a chemical reaction. In the context of a chemical process, the rate law helps us predict how changes in concentration affect the rate at which products are formed.
The general form of a rate law is given by: \[\text{rate} = k[ ext{A}]^m[ ext{B}]^n\]where:
  • \(k\) is the rate constant, a value that depends on the temperature and nature of the reaction.
  • \([ ext{A}]\) and \([ ext{B}]\) represent the concentrations of reactants A and B.
  • \(m\) and \(n\) are the reaction orders with respect to reactants A and B, respectively.
The reaction order indicates the power to which the concentration of a reactant is raised. This is crucial because it tells us how sensitive the reaction rate is to the concentration changes of these reactants. Unlike stoichiometric coefficients, reaction orders are not always whole numbers and are determined experimentally.
Understanding rate laws is key to mastering reaction kinetics, as it provides insights into how reactions can be controlled and optimized.
Stoichiometry
Stoichiometry involves the quantitative relationships between the amounts of reactants and products in a chemical reaction. For a reaction to proceed with the right yield and efficiency, it is critical to understand the reactant ratios based on a balanced chemical equation.
In an elementary reaction, the rate law is directly determined by the stoichiometry of the reaction. This means the exponents in the rate law should match the coefficients in the balanced equation. For example, if a balanced equation shows that reactant A has a coefficient of 1 and reactant B has a coefficient of 2, this implies in an elementary reaction, the rate law would include these values as orders, such as:\[\text{rate} = k[ ext{A}]^1[ ext{B}]^2\]This close alignment can make it straightforward to infer rate laws from balanced reactions, but only if the reaction is simple and happens in a single step.
Understanding stoichiometry allows chemists to predict the number of products generated from given reactants, ensuring efficient and economical chemical processes.
Reaction Kinetics
Reaction kinetics studies the rates of chemical reactions and the factors affecting these rates. It encompasses not only the formation of products from reactants but also the identification of intermediates and transition states.
In the context of kinetics, elementary reactions are fundamental because they occur in one step and provide direct insight into reaction mechanisms. The kinetics of a proposed elementary reaction can often unravel if it truly occurs in that manner since each step should correspond to a direct transformation, leaving no room for complex intermediates or rearrangements.
  • Reaction kinetics help explain why reactions occur at different speeds and under varying conditions such as temperature or pressure.
  • Understanding reaction kinetics enables chemists to alter the conditions to either speed up or slow down reactions as desired.
  • It also provides an understanding of catalysts and how they lower activation energy to increase reaction rates.
In practice, developing a comprehension of reaction kinetics enhances the ability to create efficient reactions and optimize conditions tailored to industrial and research objectives.

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Most popular questions from this chapter

The rate constant of first-order reaction is \(10^{-2} \mathrm{~min}^{-1}\). The half-life period of reaction is (a) \(693 \mathrm{~min}\) (b) \(69.3 \mathrm{~min}\) (c) \(6.93 \mathrm{~min}\) (d) \(0.693 \mathrm{~min}\)

For a zero-order reaction, the plot of concentration vs time is linear with (a) +ve slope and zero intercept (b) -ve slope and zero intercept (c) tve slope and non-zero intercept (d) -ve slope and non-zero intercept

Reaction \(\mathrm{A}_{2}+\mathrm{B}_{2} \rightarrow 2 \mathrm{AB}\) is completed according to the following mechanism. \(A_{2}=2 A\) \(\mathrm{A}+\mathrm{B}_{2} \rightarrow \mathrm{AB}+\mathrm{B} \quad\) (slow step) \(\mathrm{A}+\mathrm{B} \rightarrow \mathrm{AB}\) The order of reaction is (a) 1 (b) \(3 / 2\) (c) \(1 / 2\) (d) 2

The half-life of a substance in a first-order reaction is 15 minutes. The rate constant is (a) \(2.46 \times 10^{2} \mathrm{~min}^{-1}\) (b) \(4.62 \times 10^{-2} \mathrm{~min}^{-1}\) (c) \(3 \times 10^{-5} \mathrm{~min}^{-1}\) (d) \(3 \times 10^{-4} \mathrm{~min}^{-1}\)

A follows parallel path Ist order reactions giving B and C as shown: If initial concentration of \(\mathrm{A}\) is \(0.25 \mathrm{M}\), calculate the concentration of \(\mathrm{C}\) after 5 hour of reaction. Given, \(\lambda_{1}=1.5 \times 10^{-5} \mathrm{~s}^{-1}, \lambda_{2}=5 \times 10^{-6} \mathrm{~s}^{-1}\) (a) \(7.55 \times 10^{-3} \mathrm{M}\) (b) \(1.89 \times 10^{-2} \mathrm{M}\) (c) \(5.53 \times 10^{-3} \mathrm{M}\) (d) \(3.51 \times 10^{-3} \mathrm{M}\)

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