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Chapter 14: Problem 33

The half-life for the second-order reaction of a substance A is \(50.5 \mathrm{~s}\) when \([\mathrm{A}]_{0}=0.84 \mathrm{~mol} \cdot \mathrm{L}^{-1}\). Calculate the time needed for the concentration of A to decrease to (a) one- sixteenth; (b) onefourth; (c) one-fifth of its original value.

Short Answer

Expert verified
Substitute the values into the second-order reaction time formula and solve for \( t \) for each concentration level (one-sixteenth, one-fourth, and one-fifth) using the calculated rate constant \( k \).

Step by step solution

01

Understanding the Given Information

The problem states that we have a second-order reaction with a half-life of 50.5 seconds when the initial concentration \( [A]_0 \) is 0.84 mol/L. For a second-order reaction, the half-life is inversely proportional to the initial concentration. The half-life formula for a second-order reaction can be written as \( t_{1/2} = \frac{1}{k[A]_0} \) where \( k \) is the rate constant.
02

Calculate the Rate Constant

First, we need to calculate the rate constant \( k \) using the given half-life and initial concentration. We rearrange the half-life formula to solve for \( k \) with \( t_{1/2} = 50.5 s \) and \( [A]_0 = 0.84 mol/L \) which gives us \( k = \frac{1}{t_{1/2}[A]_0} \) or \( k = \frac{1}{50.5 \times 0.84} \) s\textsuperscript{-1}.
03

Determine the Final Concentration Values

We need to calculate the times for different final concentrations: (a) \( \frac{[A]_0}{16} \) (b) \( \frac{[A]_0}{4} \) (c) \( \frac{[A]_0}{5} \) for substance A.
04

Apply the Second-Order Reaction Formula

For a second-order reaction, the formula relating time, initial concentration, and final concentration is \( \frac{1}{[A]} - \frac{1}{[A]_0} = kt \) where \( t \) is the time required for the concentration to change from \( [A]_0 \) to \( [A] \). Now, we will use this formula to calculate the times for each final concentration.
05

Calculate Time for One-Sixteenth Concentration

Using the formula \( \frac{1}{[A]} - \frac{1}{[A]_0} = kt \) with \( [A] = \frac{[A]_0}{16} \) and the rate constant \( k \) already calculated, we solve for time \( t \).
06

Calculate Time for One-Fourth Concentration

Similarly, use the formula \( \frac{1}{[A]} - \frac{1}{[A]_0} = kt \) with \( [A] = \frac{[A]_0}{4} \) and the known value of \( k \) to find the time \( t \) for this concentration level.
07

Calculate Time for One-Fifth Concentration

Lastly, apply the formula \( \frac{1}{[A]} - \frac{1}{[A]_0} = kt \) using \( [A] = \frac{[A]_0}{5} \) to calculate the time \( t \) for the substance A to reach one-fifth of its original concentration.

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Key Concepts

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

Chemical Kinetics
Chemical kinetics is a branch of physical chemistry that deals with understanding the rates of chemical reactions. It鈥檚 not just about identifying how long a reaction will take to complete, but also about grasping the various factors that influence this rate. A fundamental aspect of chemical kinetics is the identification of the order of a reaction, which indicates how the rate depends on the concentration of the reactants.

For instance, a second-order reaction involves the rate of a reaction being proportional to the square of the concentration of one reactant or the product of the concentrations of two reactants. In the exercise provided, we delved into a second-order reaction, where the half-life depends on the initial concentration and provides essential insights into how quickly reactant concentrations decrease over time.
Reaction Rate Constant
The reaction rate constant, denoted as k, is a crucial term in the equation that describes the speed of a chemical reaction. In the context of a second-order reaction, the rate constant bridges the relationship between the concentration of reactants and the rate at which they are converted to products.

The value of k is determined experimentally and can vary with temperature, pressure, and the presence of a catalyst. Our step-by-step example began with the computation of the rate constant using the provided half-life and initial concentration. Knowing the rate constant is essential for predicting the concentration of reactants at any point in time during a reaction.
Half-Life of Reaction
Half-life, commonly symbolized as t1/2, is a term that describes the time required for the concentration of a reactant to decrease to half of its original value. This is a particularly handy concept when discussing radioactive decay, but it also plays a significant role in chemical kinetics.

In second-order reactions, the half-life is uniquely dependent on the initial concentration of the reactant, inversely so. As we observed in our exercise example, a given half-life and initial concentration allowed for the calculation of the reaction rate constant. Understanding half-life can provide insights into the durability and effectiveness of substances over time, which is valuable in various fields such as pharmacology and environmental science.
Concentration-Time Relationship
The concentration-time relationship is a key concept in chemical kinetics that describes how the concentration of reactants change as a reaction progresses. In a second-order reaction, this relationship can be expressed by the equation \( \frac{1}{[A]} - \frac{1}{[A]_0} = kt \), where [A] represents the concentration of the reactant at time t, [A]_0 is the initial concentration, and k is the rate constant.

The equation showcases that, unlike in zero or first-order reactions, the change in the inverse of concentration with time is linear. In practical terms, this means that as time goes on, the concentration decreases more slowly than in a first-order reaction, which is a logarithmic relationship. This relationship allows us to calculate not only the time it takes for the concentration to fall to a certain level, as demonstrated in our textbook example, but also to predict future concentrations at any given time during the reaction.

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Most popular questions from this chapter

A reaction was believed to occur by the following mechanism. Step \(1 \mathrm{~A}_{2} \longrightarrow \mathrm{A}+\mathrm{A}\) Step \(2 \mathrm{~A}+\mathrm{A}+\mathrm{B} \longrightarrow \mathrm{A}_{2} \mathrm{~B}\) Step \(3 \mathrm{~A}_{2} \mathrm{~B}+\mathrm{C} \longrightarrow \mathrm{A}_{2}+\mathrm{BC}\) (a) Write the overall reaction. (b) Write the rate law for each step and indicate its molecularity. (c) What are the reaction intermediates? (d) A catalyst is a substance that accelerates the rate of a reaction and is regenerated in the process. What is the catalyst in the reaction?

The data below were collected for the reaction \(\mathrm{H}_{2}(\mathrm{~g})+\mathrm{I}_{2}(\mathrm{~g}) \longrightarrow 2 \mathrm{HI}(\mathrm{g})\) at \(780 \mathrm{~K}\). (a) Using a graphing calculator or graphing software, such as that on the Web site for this book, plot the data in an appropriate fashion to determine the order of the reaction. (b) From the graph, determine the rate constant for the rate of consumption of \(\mathrm{I}_{2}\). $$ \begin{array}{lccccc} \text { Time }(\mathrm{s}) & 0 & 1.0 & 2.0 & 3.0 & 4.0 \\ {\left[\mathrm{I}_{2}\right]\left(\mathrm{mmol} \cdot \mathrm{L}^{-1}\right)} & 1.00 & 0.43 & 0.27 & 0.20 & 0.16 \end{array} $$

In the brewing of beer, ethanal, which smells like green apples, is an intermediate in the formation of ethanol. Ethanal decomposes in the following first-order reaction: \(\mathrm{CH}_{3} \mathrm{CHO}(\mathrm{g}) \longrightarrow \mathrm{CH}_{4}(\mathrm{~g})+\) \(\mathrm{CO}(\mathrm{g})\). At an elevated temperature the rate constant for the decomposition is \(1.5 \times 10^{-3} \mathrm{~s}^{-1}\). What concentration of ethanal, which had an initial concentration of \(0.120 \mathrm{~mol} \cdot \mathrm{L}^{-1}\), remains \(20.0 \mathrm{~min}\) after the start of its decomposition at this temperature?

Ethane, \(\mathrm{C}_{2} \mathrm{H}_{6}\), forms \(\cdot \mathrm{CH}_{3}\) radicals at \(700 .^{\circ} \mathrm{C}\) in a firstorder reaction, for which \(k=1.98 \mathrm{~h}^{-1}\). (a) What is the half-life for the reaction? (b) Calculate the time needed for the amount of ethane to fall from \(1.15 \times 10^{-3} \mathrm{~mol}\) to \(2.35 \times 10^{-4} \mathrm{~mol}\) in a 500.-mL reaction vessel at \(700 .{ }^{\circ} \mathrm{C}\). (c) How much of a 6.88-mg sample of ethane in a \(500 .-\mathrm{mL}\) reaction vessel at \(700 .{ }^{\circ} \mathrm{C}\) will remain after \(45 \mathrm{~min}\) ?

Determine the rate constant for each of the following firstorder reactions, in each case expressed for the rate of loss of \(A\) : (a) \(\mathrm{A} \longrightarrow \mathrm{B}\), given that the concentration of \(\mathrm{A}\) decreases to one-half its initial value in \(1000 . \mathrm{s} ;\) (b) \(\mathrm{A} \longrightarrow \mathrm{B}\), given that the concentration of A decreases from \(0.67 \mathrm{~mol} \cdot \mathrm{L}^{-1}\) to \(0.53 \mathrm{~mol} \cdot \mathrm{L}^{-1}\) in \(25 \mathrm{~s}\); (c) \(2 \mathrm{~A} \longrightarrow \mathrm{B}+\mathrm{C}\), given that \([\mathrm{A}]_{0}=0.153 \mathrm{~mol} \cdot \mathrm{L}^{-1}\) and that after \(115 \mathrm{~s}\) the concentration of \(B\) rises to \(0.034 \mathrm{~mol} \cdot \mathrm{L}^{-1}\).
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