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\(3 \mathrm{~A} \rightarrow 2 \mathrm{~B}\), rate of reaction \(+\mathrm{d}(\mathrm{B})\) is equal to \(\mathrm{dt}\) a. \(-\frac{3}{2} \frac{\mathrm{d}[\mathrm{A}]}{\mathrm{dt}}\) b. \(-\frac{2}{3} \frac{\mathrm{d}[\mathrm{A}]}{\mathrm{dt}}\) c. \(-\frac{1}{3} \frac{\mathrm{d}[\mathrm{A}]}{\mathrm{dt}}\) d. \(+2 \frac{\mathrm{d}[\mathrm{A}]}{\mathrm{dt}}\)

Short Answer

Expert verified
The correct answer is (b) \(-\frac{2}{3} \frac{d[A]}{dt}\).

Step by step solution

01

Understand the Reaction

We have a reaction: \( 3\,A \rightarrow 2\,B \), where substance \( A \) is consumed and \( B \) is produced.
02

Define the Rate of Reaction for B

The rate of reaction regarding \( B \) is given by the increase in concentration per unit time, \( \frac{d[B]}{dt} \). This represents how fast \( B \) is being produced.
03

Relate the Rate of A to the Rate of B

Based on stoichiometry, for every 3 moles of \( A \) consumed, 2 moles of \( B \) are formed. Therefore, the rate of change of \( A \) is linked to \( B \) by the stoichiometric coefficients.
04

Write the Expression for Rate of A

Using the stoichiometric relationship, the rate of \( A \) can be expressed as \(-\frac{1}{3} \frac{d[A]}{dt} = \frac{1}{2} \frac{d[B]}{dt}\), which also means, \(-\frac{d[A]}{dt} = \frac{3}{2} \frac{d[B]}{dt}\).
05

Select the Correct Option

The correct expression for \( \frac{d[B]}{dt} \) in terms of \( \frac{d[A]}{dt} \) is \(-\frac{2}{3} \frac{d[A]}{dt}\), matching option (b).

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Key Concepts

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

Stoichiometry
Understanding stoichiometry is crucial when analyzing chemical reactions. It involves the quantitative relationship between reactants and products in a chemical reaction. For the reaction \( 3A \rightarrow 2B \), stoichiometry tells us that 3 moles of \( A \) are consumed for every 2 moles of \( B \) produced.
This relationship allows us to determine how the rate of disappearance of \( A \) is related to the rate of appearance of \( B \). The stoichiometric coefficients (3 and 2 in this reaction) give us the conversion factors needed.
In essence, stoichiometry provides the balance needed in a reaction, ensuring that the amount of reactants used and products formed are consistent with the law of conservation of mass. It helps chemists determine how much of each substance is involved in a reaction and predict the quantities of products formed as a result.
Rate of Reaction
The rate of reaction is an essential concept in chemical kinetics, describing how fast or slow a reaction proceeds. It can be measured by observing the change in concentration of reactants or products over time.
In the given reaction, the rate at which \( A \) is consumed is expressed as \( \frac{d[A]}{dt} \) and is usually a negative value since the concentration of \( A \) decreases. Conversely, \( \frac{d[B]}{dt} \) is positive, indicating that the concentration of \( B \) increases.
  • The rate of a reaction is influenced by various factors such as temperature, pressure, concentration of reactants, and presence of catalysts.
  • By understanding the rate of reaction, chemists can adjust these factors to control how quickly a reaction occurs or optimize conditions for maximum yield.
Overall, knowing the rate of reaction helps in predicting how a reaction progresses and in designing chemical processes.
Chemical Kinetics
Chemical kinetics is the branch of chemistry that studies the rates of chemical reactions and the mechanisms by which they occur. It involves detailed analysis of reaction rates and how they change under different conditions.
In a reaction like \( 3A \rightarrow 2B \), kinetics helps to elucidate how exactly the atoms or molecules convert from reactants to products. This includes understanding transitions, energy barriers, and the influence of various factors.
  • Kinetics provides insight into the necessary energy (activation energy) to initiate a reaction.
  • It offers a glimpse into the stepwise changes and intermediate stages between reactants and products.
  • Studying kinetics can also reveal new pathways to achieving desired chemical transformations more efficiently.
With chemical kinetics, researchers can develop better catalysts, improve industrial processes, and innovate new reaction paths, enhancing the functionality and efficiency of chemical production.

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Most popular questions from this chapter

Consider a reaction \(\mathrm{aG}+\mathrm{bH} \rightarrow\) Products. When concentration of both the reactants \(\mathrm{G}\) and \(\mathrm{H}\) is doubled, the rate increases by eight times. However when concentration of \(\mathrm{G}\) is doubled keeping the concentration of \(\mathrm{H}\) fixed, the rate is doubled. The overall order of the reaction is a. 0 b. 1 c. 2 d. 3

A mechanism for a naturally occurring reaction that destroys ozone is: Step I: \(\mathrm{O}_{3}(\mathrm{~g})+\mathrm{HO}(\mathrm{g}) \rightarrow \mathrm{HO}_{2}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g})\) Step II: \(\mathrm{HO}_{2}(\mathrm{~g})+\mathrm{O}(\mathrm{g}) \rightarrow \mathrm{HO}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{~g})\) Which species is a catalyst? a. \(\mathrm{O}\) b. \(\mathrm{O}_{3}\) c. \(\mathrm{HO}_{2}\) d. \(\mathrm{HO}\)

In the following question two statements Assertion (A) and Reason (R) are given Mark. a. If \(\mathrm{A}\) and \(\mathrm{R}\) both are correct and \(\mathrm{R}\) is the correct explanation of \(\mathrm{A}\); b. If \(A\) and \(R\) both are correct but \(R\) is not the correct explanation of \(\mathrm{A}\); c. \(\mathrm{A}\) is true but \(\mathrm{R}\) is false; d. \(\mathrm{A}\) is false but \(\mathrm{R}\) is true, e. \(\mathrm{A}\) and \(\mathrm{R}\) both are false. (A): Order can be different from molecularity of a reaction. (R): Slow step is the rate determining step and may involve lesser number of reactants.

When the concentration of \(\mathrm{A}\) is doubled, the rate for the reaction: \(2 \mathrm{~A}+\mathrm{B} \rightarrow 2 \mathrm{C}\) quadruples. When the concentration of \(\mathrm{B}\) is doubled the rate remains the same. Which mechanism below is consistent with the experimental observations? a. Step I: \(2 \mathrm{~A} \rightleftharpoons \mathrm{D}\) (fast equilibrium) Step II: \(\mathrm{B}+\mathrm{D} \rightarrow \mathrm{E}\) (slow) Step III: \(\mathrm{E} \rightarrow 2 \mathrm{C}\) (fast) b. Step I: \(\mathrm{A}+\mathrm{B} \rightleftharpoons \mathrm{D}\) (fast equilibrium) Step II: \(\mathrm{A}+\mathrm{D} \rightarrow 2 \mathrm{C}\) (slow) c. Step \(\mathrm{I}: \mathrm{A}+\mathrm{B} \rightarrow \mathrm{D}\) (slow) Step II: \(\mathrm{A}+\mathrm{D} \rightleftharpoons 2 \mathrm{C}\) (fast equilibrium) d. Step I: \(2 \mathrm{~A} \rightarrow \mathrm{D}\) (slow) Step II: \(\mathrm{B}+\mathrm{D} \rightarrow \mathrm{E}\) (fast) Step III: \(\mathrm{E} \rightarrow 2 \mathrm{C}\) (fast)

Observe the reaction given below \(\mathrm{A}+2 \mathrm{~B} \rightarrow 3 \mathrm{C}\) \(\begin{array}{lll}\mathrm{g} & \mathrm{g} & \mathrm{g}\end{array}\) If the rate of this reaction \(-\frac{\mathrm{d} \mathrm{A}}{\mathrm{dt}}\) is \(2 \times 10^{-3} \mathrm{~mol} \mathrm{lit}^{-1} \mathrm{~min}^{-1}\) than the value of \(\frac{\mathrm{dB}}{\mathrm{dt}}\) and \(\frac{\mathrm{dC}}{\mathrm{dt}}\) will be respectively a. \(1 \times 10^{-3}, 2 / 3 \times 10^{-3}\) b. \(4 \times 10^{-3}, 6 \times 10^{-3}\) c. \(6 \times 10^{-3}, 4 \times 10^{-3}\) d. \(2 / 3 \times 10^{-3}, 1 \times 10^{-3}\)

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