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Chapter 13: Problem 59

A reaction of the form \(a \mathrm{~A} \longrightarrow\) Products is second- order with a rate constant of \(0.225 \mathrm{~L} /(\mathrm{mol} \cdot \mathrm{s})\). If the initial concentration of \(\mathrm{A}\) is \(0.293 \mathrm{~mol} / \mathrm{L}\), what is the molar concentration of \(\mathrm{A}\) after \(35.4 \mathrm{~s} ?\)

Short Answer

Expert verified
The molar concentration of \(\mathrm{A}\) after 35.4 s is approximately 0.088 mol/L.

Step by step solution

01

Identify the Formula

For a second-order reaction, the integrated rate law is given by \[ \frac{1}{[A]} = \frac{1}{[A]_0} + kt \]where - \( [A] \) is the final concentration,- \( [A]_0 \) is the initial concentration,- \( k \) is the rate constant,- \( t \) is the time.
02

Substitute Values

Using the given values:- \([A]_0 = 0.293 \text{ mol/L} \)- \(k = 0.225 \text{ L/(mol} \cdot \text{s)}\)- \(t = 35.4 \text{ s}\) Substitute these values into the integrated rate law formula: \[ \frac{1}{[A]} = \frac{1}{0.293} + (0.225)(35.4) \]
03

Calculate Initial Part of Equation

First, calculate the term \( \frac{1}{0.293} \):\[ \frac{1}{0.293} \approx 3.4123 \]
04

Calculate Second Part of Equation

Now calculate the product of \(k\) and \(t\):\[ 0.225 \times 35.4 = 7.965 \]
05

Combine Components

Add the results from the previous steps to calculate \( \frac{1}{[A]} \):\[ \frac{1}{[A]} = 3.4123 + 7.965 \approx 11.3773 \]
06

Solve for Final Concentration

Now, solve for \([A]\) by taking the reciprocal of \(11.3773\):\[ [A] = \frac{1}{11.3773} \approx 0.0879 \text{ mol/L} \]

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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 expression that describes the speed of a chemical reaction as a function of the concentration of reactants. For a reaction labeled as second-order, like the one in our exercise example, the rate of reaction is proportional to the square of the concentration of one reactant.
  • This is usually expressed as: \( ext{Rate} = k [A]^2 \)
  • Here, \( k \) is the rate constant, which remains constant at a given temperature, while \( [A] \) is the concentration of reactant A.
Understanding the rate law helps us predict how quickly a reaction will proceed under certain conditions. The knowledge of the rate law is crucial in fields like chemical engineering and pharmaceuticals, where controlling reaction rates is vital.
Reaction Kinetics
Reaction kinetics involves the study of the rate at which chemical reactions occur and the mechanisms through which they proceed. By knowing the order of a reaction, we can understand the relationship between reactant concentration and reaction rate.
  • For second-order reactions, common in many real-world scenarios, understanding rate laws and kinetics help infer how reactants are consumed over time.
  • Having knowledge of kinetic concepts allows chemists to modify conditions to control reaction speed. This is necessary for industrial processes where specific rates are desired.
Deep diving into reaction kinetics gives insight into the time scales of reactions, enabling chemists to predict and manipulate the outcomes effectively.
Integrated Rate Law
The integrated rate law provides a direct link between the concentration of reactants and time. For second-order reactions, we use:\[\frac{1}{[A]} = \frac{1}{[A]_0} + kt\]
  • In this equation, \( [A] \) denotes the concentration at any given time \( t \), \([A]_0\) is the initial concentration, and \( k \) is the rate constant.
  • This formula is vital as it allows for the calculation of the concentration of reactants at any given time, portraying the reaction's progress.

The integrated rate law helps us translate changes in concentration over time, making it integral in designing reactions and understanding the extent of reaction progression.
Chemical Concentration
Chemical concentration refers to the amount of a substance in a given volume, and it鈥檚 a key player in reaction kinetics. In our exercise, concentration is treated both as an initial condition \( [A]_0 \) and as a variable \( [A] \) that evolves with time.
  • Concentration is usually expressed in molarity (moles per liter - mol/L).
  • In second-order reactions, the concentration change is nonlinear, as seen in the rate law where concentration impacts the rate exponentially.
Understanding concentration is essential for evaluating how fast a reaction can proceed and to what extent reactants are converted into products. Accurate concentration measurements are fundamental for reaction analysis and control.
Half-life of Chemical Reactions
The half-life of a reaction is the time taken for half the reactant to be consumed.
  • For second-order reactions, the half-life formula is:\[ t_{1/2} = \frac{1}{k[A]_0} \]
  • Unlike first-order reactions, the half-life for second-order reactions depends on the initial concentration, making it longer as the reaction proceeds.
Understanding half-life is crucial in predicting how long a reactant will last under specific conditions and is widely used in environmental chemistry, pharmacology, and other fields where chemical longevity is vital.

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Most popular questions from this chapter

In experiments on the decomposition of azomethane, the following data were obtained: \(\begin{array}{lll} & \text { Initial } & \\ & \begin{array}{l}\text { Concentration } \\ \text { of Azomethane }\end{array} & \text { Initial Rate } \\ \text { Exp. 1 } & 1.13 \times 10^{-2} \mathrm{M} & 2.8 \times 10^{-6} \mathrm{M} / \mathrm{s} \\ \text { Exp. } 2 & 2.26 \times 10^{-2} \mathrm{M} & 5.6 \times 10^{-6} \mathrm{M} / \mathrm{s}\end{array}\) What is the rate law? What is the value of the rate constant?

Sketch a potential-energy diagram for the exothermic, elementary reaction $$\mathrm{A}+\mathrm{B} \longrightarrow \mathrm{C}+\mathrm{D}$$ and on it denote the activation energies for the forward and reverse reactions. Also indicate the reactants, products, and activated complex.

The rate law for the reaction $$2 \mathrm{NO}_{2} \mathrm{Cl}(g) \longrightarrow 2 \mathrm{NO}_{2}(g)+\mathrm{Cl}_{2}(g) \quad \text { (overall equation) }$$ is first order in nitryl chloride, \(\mathrm{NO}_{2} \mathrm{Cl}\). $$\text { Rate }=k\left[\mathrm{NO}_{2} \mathrm{Cl}\right]$$ Explain why the mechanism for this reaction cannot be the single elementary reaction $$2 \mathrm{NO}_{2} \mathrm{Cl} \longrightarrow 2 \mathrm{NO}_{2}+\mathrm{Cl}_{2} \quad \text { (elementary reaction) }$$

What is meant by the term rate of a chemical reaction? Why does the rate of a reaction normally change with time? When does the rate of a chemical reaction equal the rate constant?

For the reaction of nitrogen monoxide, NO, with chlorine, \(\mathrm{Cl}_{2}\) $$2 \mathrm{NO}(g)+\mathrm{Cl}_{2}(g) \longrightarrow 2 \mathrm{NOCl}(g)$$ the observed rate law is $$\text { Rate }=k[\mathrm{NO}]^{2}\left[\mathrm{Cl}_{2}\right]$$ What is the reaction order with respect to nitrogen monoxide? with respect to \(\mathrm{Cl}_{2}\) ? What is the overall order?
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