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Chapter 11: Problem 27

For a first-order reaction: \(\mathrm{A} \rightarrow\) Product, the initial concentration of \(\mathrm{A}\) is \(0.1 \mathrm{M}\) and after time \(40 \mathrm{~min}\), it becomes \(0.025 \mathrm{M}\). What is the rate of reaction at reactant concentration \(0.01 \mathrm{M} ?\) (a) \(3.465 \times 10^{-4} \mathrm{~mol} \mathrm{lit}^{-1} \mathrm{~min}^{-1}\) (b) \(3.465 \times 10^{-5} \mathrm{~mol} \mathrm{lit}^{-1} \mathrm{~min}^{-1}\) (c) \(6.93 \times 10^{-4} \mathrm{~mol} \mathrm{lit}^{-1} \mathrm{~min}^{-1}\) (d) \(1.7325 \times 10^{-4} \mathrm{~mol} \mathrm{lit}^{-1} \mathrm{~min}^{-1}\)

Short Answer

Expert verified
The rate of reaction at reactant concentration 0.01 M is (b) \(3.465 \times 10^{-5} \mathrm{~mol} \mathrm{lit}^{-1} \mathrm{~min}^{-1}\).

Step by step solution

01

Understand the First-Order Reaction Kinetics

For a first-order reaction, the rate of reaction can be related to the concentration of the reactant (A) by the rate law, which is rate = k[A], where k is the rate constant, and [A] is the concentration of A. The relation between the time and concentration for a first-order reaction is given by ln([A]0/[A]) = kt, where [A]0 is the initial concentration and [A] is the concentration at time t.
02

Calculate the Rate Constant (k)

Use the given concentration values to calculate the rate constant. Applying the first-order reaction formula: ln([A]0/[A]) = kt, and substituting the initial concentration [A]0 = 0.1 M and concentration at time t, [A] = 0.025 M at t = 40 min, calculate k. First, find ln([A]0/[A]) = ln(0.1/0.025) = ln(4). Then divide by the time t to get k.
03

Determine the Reaction Rate at [A] = 0.01 M

With the rate constant k calculated in Step 2, use the rate law rate = k[A] with [A] = 0.01 M to find the reaction rate at this concentration.

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Key Concepts

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

Reaction Kinetics
Understanding reaction kinetics is integral to grasping how the speed of a chemical reaction changes depending on various factors, including reactant concentrations. Kinetics investigates not just the speed, or rate, of chemical reactions, but it also sheds light on the mechanism by which reactions proceed.

In the example given, we deal with the concept of a first-order reaction. This classification indicates that the rate of the reaction is directly proportional to the concentration of one reactant. Mathematically, this can be expressed as rate = k[A], where [A] represents the concentration of reactant A, and k is a proportionality constant known as the rate constant. In the context of a first-order reaction, if you were to double the concentration of A, you would also double the reaction rate.
Rate Constant
The rate constant, denoted as k, plays a pivotal role in the quantitative analysis of reaction kinetics. It is a unique value for each chemical reaction at a given temperature, and it essentially captures the intrinsic speed of a reaction. For a first-order reaction, k has units of inverse time, like s-1 or min-1.

To calculate the rate constant, we turn to the formula for first-order kinetics: ln([A]鈧�/[A]) = kt. By using known concentrations over time, this equation allows us to deduce the value of k. This calculated k then becomes a powerful tool, allowing us to predict the rate of reaction at any given concentration of reactant.
Chemical Kinetics
Chemical kinetics includes the study of reaction rates and extends far beyond, to understand factors that influence these rates, and to determine reaction mechanisms. The key to mastering chemical kinetics is to connect the theoretical aspects, such as the rate equations, to the practical, observable changes in reactant and product concentrations over time.

This study's distinction is crucial because it transcends knowing the 鈥榟ow fast鈥� and delves into the 鈥榳hy鈥� and 鈥榟ow鈥� of reaction rates. Chemical kinetics provides a microscopic view of what is happening on a molecular level during reactions, which can often inform us about the macroscopic properties we observe.
Concentration-Time Relationship
Analyzing the concentration-time relationship is vital for predicting how concentrations of reactants and products change as a reaction progresses. For first-order reactions, the relationship is logarithmic, as shown by the equation ln([A]鈧�/[A]) = kt. This equation establishes that the natural logarithm of the ratio of initial concentration to concentration at a given time is proportional to that time.

In simpler terms, if you graph the natural log of reactant concentration vs. time for a first-order reaction, you will obtain a straight line. The negative of the slope of this line is equal to the rate constant k. This characteristic provides a practical method for experimentally determining the rate constant. Once the rate constant is known, the concentration of reactants at any time can be calculated, enabling predictions about the reaction's progress.

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Most popular questions from this chapter

For the chemical reaction: \(\mathrm{A} \rightarrow\) products, the rate of disappearance of \(\mathrm{A}\) is a given by $$ r_{A}=-\frac{\mathrm{d} C_{\mathrm{A}}}{\mathrm{d} t}=\frac{K_{1} \cdot C_{\mathrm{A}}}{1+K_{2} \cdot C_{\mathrm{A}}} $$ At low concentration of A, \(C_{A}\), the reaction is of the first order with the rate constant (a) \(\frac{K_{1}}{K_{2}}\) (b) \(K_{1}\) (c) \(\frac{K_{2}}{K_{1}}\) (d) \(\frac{K_{1}}{K_{1}+K_{2}}\)

For the following first-order competing reaction: $$ \begin{array}{l} \text { A + Reagent } \rightarrow \text { Product } \\ \text { B + Reagent } \rightarrow \text { Product } \end{array} $$ the ratio of \(K_{1} / K_{2}\), if only \(50 \%\) of ' \(\mathrm{B}\) ' will have been reacted when \(94 \%\) of ' \(\mathrm{A}\) ' has been reacted is \((\log 2=0.3, \log 3=0.48)\) (a) \(4.06\) (b) \(0.246\) (c) \(8.33\) (d) \(0.12\)

When the concentration of 'A' is 0.1 M, it decomposes to give ' \(\mathrm{X}\) ' by a firstorder process with a rate constant of \(6.93 \times 10^{-2} \mathrm{~min}^{-1}\). The reactant 'A', in the presence of catalyst, gives ' \(\mathrm{Y}\) ' by a secondorder mechanism with the rate constant of \(0.2 \mathrm{~min}^{-1} \mathrm{M}^{-1} .\) In order to make half-life of both the processes, same, one should start the second-order reaction with an initial concentration of 'A' equal to (a) \(0.01 \mathrm{M}\) (b) \(2.0 \mathrm{M}\) (c) \(1.0 \mathrm{M}\) (d) \(0.5 \mathrm{M}\)

A zero-order reaction is one (a) in which reactants do not react. (b) in which one of the reactants is in large excess. (c) whose rate does not change with time. (d) whose rate increases with time.

\(\begin{array}{lll}\text { For the } & \text { sequential } & \text { reactions: }\end{array}\) \(\mathrm{A} \stackrel{K_{1}=0.02 \mathrm{~min}^{-1}}{\longrightarrow} \mathrm{B} \stackrel{K_{2}=0.02 \mathrm{~min}^{-1}}{\longrightarrow} \mathrm{C}\), the initial concentration of 'A' was \(0.2 \mathrm{M}\) and initially 'B' and 'C' were absent. The time at which the concentration of ' \(\mathrm{B}\) ' becomes maximum and the maximum concentration of ' B' are, respectively, (a) \(50 \mathrm{~min},\left(\frac{0.2}{e}\right) \mathrm{M}\) (b) \(50 \mathrm{~min}, 0.2 \mathrm{M}\) (c) infinite, \(0.2 \mathrm{M}\) (d) \(25 \mathrm{~min},\left(\frac{0.2}{e}\right) \mathrm{M}\)
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