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Chapter 10: Problem 113

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

Short Answer

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

Step by step solution

01

Understanding the Reaction Relationship

Given the reaction, \(3\, \text{A} \rightarrow 2\, \text{B}\), means that 3 moles of A are transformed into 2 moles of B. We can write the rate of disappearance of A as \(-\frac{1}{3} \frac{\mathrm{d}[\mathrm{A}]}{\mathrm{dt}}\) and the rate of formation of B as \(+\frac{1}{2} \frac{\mathrm{d}[\mathrm{B}]}{\mathrm{dt}}\).
02

Relate the Rates of A and B

From the stoichiometry of the reaction, the relationship between the rate of disappearance of A and the formation of B is \(3 \times \frac{\Delta [B]}{2} = - \Delta [A]\). Therefore, \(\frac{1}{2} \frac{\mathrm{d}[\mathrm{B}]}{\mathrm{dt}} = -\frac{1}{3} \frac{\mathrm{d}[\mathrm{A}]}{\mathrm{dt}}\).
03

Solving for the Rate of Formation of B

Solving for \(+\frac{\mathrm{d}[\mathrm{B}]}{\mathrm{dt}}\): Multiply both sides by 2 to solve for \(\frac{\mathrm{d}[\mathrm{B}]}{\mathrm{dt}}\). Thus, \(+\frac{\mathrm{d}[\mathrm{B}]}{\mathrm{dt}} = -\frac{2}{3} \frac{\mathrm{d}[\mathrm{A}]}{\mathrm{dt}}\).

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Key Concepts

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

Reaction Stoichiometry
When we talk about reaction stoichiometry, we are referring to the quantitative relationships between the reactants and products in a chemical reaction. In simple terms, it tells us how much of one substance will react with a given amount of another substance to form a product.

For example, in the reaction \(3 \, \text{A} \rightarrow 2 \, \text{B}\), 3 moles of substance A will produce 2 moles of substance B. This relationship is crucial because it helps us understand how the reactants transform into products.

Stoichiometry allows us to predict how much product will form in a reaction, given a certain amount of reactant. It's like a recipe in cooking: if you know the recipe, you can figure out how much of each ingredient you need.
  • Helps in predicting the amounts in chemical reactions.
  • Balances chemical equations to conserve mass.
  • Critical for understanding the yield and efficiency of reactions.
Rate of Reaction
The rate of reaction is a measure of how quickly a reaction takes place. It is determined by looking at the change in concentration of either the reactants or products over time.

For the reaction \(3 \, \text{A} \rightarrow 2 \, \text{B}\), the rate at which A disappears is connected to the rate at which B appears. The rate is expressed as \(-\frac{\text{d}[\text{A}]}{\text{dt}}\) for the disappearance of A and \(+\frac{\text{d}[\text{B}]}{\text{dt}}\) for the formation of B.
  • Reaction rate is influenced by concentration, temperature, and pressure.
  • Helps in understanding how fast a reaction can be completed.
  • Allows chemists to control processes and optimize conditions.
Rate Laws
Rate laws are equations that relate the rate of reaction to the concentration of reactants. They tell us how changing the concentration of one or more reactants will affect the speed of the reaction.

In our reaction \(3 \, \text{A} \rightarrow 2 \, \text{B}\), the rate law can be used to calculate how the concentration of A or B will change over time. Typically, rate laws are expressed in terms of derivatives, such as \[- \frac{\text{d}[ ext{A}]}{\text{dt}} = k [ ext{A}]^m \], where \(k\) is the rate constant, and \(m\) is the order of the reaction with respect to A.
  • Allows prediction of how changes in conditions affect the reaction rate.
  • Helps to determine kinetic parameters like rate constant and reaction order.
  • Essential for scaling up reactions from lab to industrial scale.

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Most popular questions from this chapter

A substance undergoes first-order decomposition, it follows two parallel reactions \(k_{1}=1.26 \times 10^{-4} \mathrm{~s}^{-1}\) and \(k_{2}=3.8 \times 10^{-5} \mathrm{~s}^{-1}\) The percentage distribution of \(\mathrm{Y}\) and \(\mathrm{Z}\) are (a) \(80 \% \mathrm{Y}\) and \(20 \% \mathrm{Z}\) (b) \(72.83 \% \mathrm{Y}\) and \(32.71 \% \mathrm{Z}\) (c) \(76.83 \% \mathrm{Y}\) and \(23.17 \% \mathrm{Z}\) (d) \(62.4 \% \mathrm{Y}\) and \(90.5 \% \mathrm{Z}\)

At \(380^{\circ} \mathrm{C}\), half-life period for the first-order decomposition of \(\mathrm{H}_{2} \mathrm{O}_{2}\) is \(360 \mathrm{~min}\). The energy of activation of the reaction is \(200 \mathrm{~kJ} \mathrm{~mol}^{1} .\) Calculate the time required for \(75 \%\) decomposition at \(450^{\circ} \mathrm{C}\) if half-life for decomposition of \(\mathrm{H}_{2} \mathrm{O}_{2}\) is \(10.17 \mathrm{~min}\) at \(450^{\circ} \mathrm{C}\). (a) \(20.4 \mathrm{~min}\) (b) \(408 \mathrm{~min}\) (c) \(10.2 \mathrm{~min}\) (d) none of these

For a first-order reaction, \(t_{0.9}\) is \(138.6\) seconds. Its specific reaction rate constant (in \(\left.\sec ^{-1}\right)\) is (a) \(10^{-2}\) (b) \(10^{-4}\) (c) \(10^{-5}\) (d) \(10^{-6}\)

The following data are obtained from the decomposition of a gaseous compound Initial pressure in arm \(\quad 1.6 \quad 0.8 \quad 0.4\) Time for \(50 \%\) reaction in min \(80 \quad 113 \quad 160\) The order of the reaction is (a) \(0.5\) (b) \(1.0\) (c) \(1.5\) (d) \(2.0\)

A catalyst is a substance which (a) supplies energy to the reaction (b) increases the equilibrium concentration of the product (c) changes the equilibrium constant of the reaction (d) shortens the time to each equilibrium
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