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Chapter 10: Problem 92

The rate constant of a first-order reaction, \(\mathrm{A} \longrightarrow\) products, is \(60 \times 10^{-4} \mathrm{~s}^{-1} .\) Its rate at \([\mathrm{A}]=\) \(0.01 \mathrm{~mol} \mathrm{~L}^{-1}\) would be (a) \(60 \times 10^{-6} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~min}^{-1}\) (b) \(36 \times 10^{-4} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~min}^{-1}\) (c) \(60 \times 10^{-2} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~min}^{-1}\) (d) \(36 \times 10^{-1} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~min}^{-1}\)

Short Answer

Expert verified
The correct answer is (b) \(36 \times 10^{-4} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~min}^{-1}\).

Step by step solution

01

Understand the Rate Law for First-Order Reactions

For a first-order reaction, the rate of the reaction is given by the expression \( \text{rate} = k[A] \), where \( k \) is the rate constant and \( [A] \) is the concentration of the reactant A.
02

Convert Rate Constant to the Same Unit of Time

The given rate constant \( k = 60 \times 10^{-4} \ \mathrm{s}^{-1} \) is in seconds. Since all the options are given in \( \mathrm{min}^{-1} \), convert \( \mathrm{s}^{-1} \) to \( \mathrm{min}^{-1} \) by multiplying by 60. Therefore, \( k = 60 \times 10^{-4} \times 60 = 36 \times 10^{-2} \ \mathrm{min}^{-1} \).
03

Calculate the Reaction Rate

The rate of the reaction can now be calculated using the formula: \( \text{rate} = k[A] = (36 \times 10^{-2} \ \mathrm{min}^{-1})(0.01 \ \mathrm{mol} \ \mathrm{L}^{-1}) = 36 \times 10^{-4} \ \mathrm{mol} \ \mathrm{L}^{-1} \ \mathrm{min}^{-1} \).
04

Compare with Given Options

The calculated reaction rate is \( 36 \times 10^{-4} \ \mathrm{mol} \ \mathrm{L}^{-1} \ \mathrm{min}^{-1} \), which matches option (b).

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Key Concepts

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

Rate Constant
In the world of chemical reactions, the **rate constant** is a critical parameter. It gives you an idea of how quickly a reaction proceeds. For a first-order reaction, the rate constant, often symbolized as \( k \), has units of \( ext{time}^{-1} \). This is because it reflects a change over time of the concentration of reactants.
In simple terms, it signifies how fast the reaction is moving and depends on factors like temperature and presence of a catalyst.To better understand this, consider the problem involving a reaction \( \text{A} \rightarrow \text{Products} \) with a rate constant of \( k = 60 \times 10^{-4} \ ext{s}^{-1} \). Since the options given are in \( ext{min}^{-1} \), we converted this rate constant from seconds to minutes by multiplying it by 60, leading us to \( k = 36 \times 10^{-2} \ ext{min}^{-1} \). This conversion is essential to correctly compute further values, ensuring uniformity in units.
Reaction Rate
The **reaction rate** tells us how fast a reactant is used up or a product is formed in a given time. For first-order reactions, this rate is directly proportional to the concentration of the reactant. In our problem, the relationship is expressed with the formula \[ \text{rate} = k[A] \], where \( [A] \) is the concentration of the reactant and \( k \) is the rate constant.With our converted rate constant of \( 36 \times 10^{-2} \ ext{min}^{-1} \) and a reactant concentration \( [A] = 0.01 \ ext{mol} \ ext{L}^{-1} \), we apply these values:- Reaction rate = \( k [A] = (36 \times 10^{-2} \ ext{min}^{-1})(0.01 \ ext{mol} \ ext{L}^{-1}) \)After calculation, this results in: - \( 36 \times 10^{-4} \ ext{mol} \ ext{L}^{-1} \ ext{min}^{-1} \)This value matches option (b), showcasing how using the correct formula and units can precisely determine the reaction rate.
Rate Law
The **rate law** is a fundamental concept in chemical kinetics linking the progression of the reaction to the concentration of reactants. For a first-order reaction, the rate law states:\[ \text{rate} = k[A] \]Here, the reaction rate is directly proportional to the concentration of one reactant raised to the first power. Understanding the rate law is crucial because it provides the mathematical relationship needed to connect the rate constant \( k \) and the concentration \( [A] \).
In this law, the proportionality means that doubling the concentration of \( A \) would double the rate of the reaction if the conditions are optimal and other factors like temperature remain constant.This concept enables us to assess how various reactant concentrations impact the overall reaction pace. Therefore, for a given reaction with a known rate constant, you can forecast the speed of the reaction under different conditions by applying the rate law diligently, as seen in our exercise.

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Most popular questions from this chapter

For a reaction, \(\mathrm{A} \rightarrow \mathrm{B}+\mathrm{C}\), it was found that at the end of 10 min from the start the total optical, rotation of the system was \(50^{\circ} \mathrm{C}\) and when the reaction is complete it was \(100^{\circ}\). Assuming that only \(\mathrm{B}\) and \(\mathrm{C}\) are optically active and dextro rotator, the rate constant of this first order reaction would be (a) \(6.9 \mathrm{~min}^{-1}\) (b) \(0.069 \mathrm{~min}^{-1}\) (c) \(6.9 \times 10^{-2} \mathrm{~min}^{-1}\) (d) \(0.69 \mathrm{~min}^{-1}\)

For the reaction, \(\mathrm{N}_{2} \mathrm{O}_{5} \longrightarrow 2 \mathrm{NO}_{2}+1 / 2 \mathrm{O}_{2} .\) Given values are \(-\frac{\mathrm{d}\left[\mathrm{N}_{2} \mathrm{O}_{5}\right]}{\mathrm{dt}}=k_{1}\left[\mathrm{~N}_{2} \mathrm{O}_{5}\right]\) \(\frac{\mathrm{d}\left[\mathrm{NO}_{2}\right]}{\mathrm{dt}}=k_{2}\left[\mathrm{~N}_{2} \mathrm{O}_{5}\right]\) \(\frac{\mathrm{d}\left[\mathrm{O}_{2}\right]}{\mathrm{dt}}=k_{3}\left[\mathrm{~N}_{2} \mathrm{O}_{5}\right]\) Now the relation in between \(k_{1} k_{2}\) and \(k_{3}\) is (a) \(k_{1}=k_{2}=k_{3}\) (b) \(3 k_{1}=k_{2}=2 k_{3}\) (c) \(2 k_{1}=4 k_{2}=k_{3}\) (d) \(2 k_{1}=k_{2}=4 k_{3}\)

For an endothermic reaction, where \(\Delta \mathrm{H}\) represents the enthalpy of the reaction in \(\mathrm{kJ} / \mathrm{mole}\), the minimum value for the energy of activation will be (a) less than \(\Delta \mathrm{H}\) (b) zero (c) more than \(\Delta \mathrm{H}\) (d) equal to \(\Delta \mathrm{H}\)

The energy of activation and specific rate constant for a first order reaction at \(25^{\circ} \mathrm{C}\) are \(100 \mathrm{~kJ} / \mathrm{mol}\) and \(3.46\) \(\times 10^{-5} \mathrm{sec}^{-1}\) respectively. Determine the temperature at which half life of reaction is 2 hour. \(2 \mathrm{~N}_{2} \mathrm{O}_{5} \rightarrow 2 \mathrm{~N}_{2} \mathrm{O}_{4}+\mathrm{O}_{2}\) (in \(\left.\mathrm{CCl}_{4}\right) \quad\) (in \(\left.\mathrm{CCl}_{4}\right)\) (a) \(300 \mathrm{~K}\) (b) \(302 \mathrm{~K}\) (c) \(304 \mathrm{~K}\) (d) \(306 \mathrm{~K}\)

For the reaction \(\mathrm{A} \longrightarrow\) Products, it is found that the rate of reaction increases by a factor of \(6.25\), when the concentration of \(\mathrm{A}\) is increased by a factor of \(2.5 .\) The order of reaction with respect to \(\mathrm{A}\) is (a) \(0.5\) (b) 1 (c) 2 (d) 3
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