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Chapter 14: Problem 97

A compound decomposes by a first-order reaction. If the concentration of the compound is \(0.0250 M\) after \(65 \mathrm{~s}\) when the initial concentration was \(0.0350 M\), what is the concentration of the compound after 88 s?

Short Answer

Expert verified
The concentration after 88 seconds is approximately 0.0237 M.

Step by step solution

01

Understand the First-Order Reaction Formula

In a first-order reaction, the rate of decomposition can be described by the equation \( \ln \left( \frac{[A]_t}{[A]_0} \right) = -kt \), where \([A]_t\) is the concentration at time \(t\), \([A]_0\) is the initial concentration, and \(k\) is the rate constant.
02

Calculate the Rate Constant (k)

Given the initial concentration \([A]_0 = 0.0350 \, M\) and the concentration \([A]_{65} = 0.0250 \, M\) after 65 seconds, we can rearrange the first-order formula to solve for \(k\): \( k = - \frac{1}{t} \ln \left( \frac{[A]_t}{[A]_0} \right) \). Substitute the values: \( k = - \frac{1}{65} \ln \left( \frac{0.0250}{0.0350} \right) \). Calculate \(k\) to find its value.
03

Use the Rate Constant to Find Future Concentration

With \(k\) known, use the first-order formula to find the concentration after 88 seconds, \( [A]_{88} \): \( \ln \left( \frac{[A]_{88}}{[A]_0} \right) = -k \times 88 \). Rearrange to solve for \([A]_{88}\): \([A]_{88} = [A]_0 \times e^{-k \times 88} \). Substitute \([A]_0 = 0.0350\) and the calculated \(k\) into the equation to find \([A]_{88}\).
04

Calculate to Complete the Solution

Substitute \(k\) and other values into \([A]_{88} = 0.0350 \times e^{-k \times 88}\). Perform the calculations to get the concentration \([A]_{88}\).

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Key Concepts

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

Rate Constant
The rate constant, denoted as \(k\), is an essential parameter in understanding the speed of a chemical reaction, particularly for first-order reactions. In these types of reactions, the rate at which the reaction proceeds is directly proportional to the concentration of one reactant. The rate constant \(k\) is a measure that helps determine how fast this occurs.

To find \(k\) in a first-order reaction, we use the formula \( k = - \frac{1}{t} \ln \left( \frac{[A]_t}{[A]_0} \right) \). This requires knowing the initial concentration \([A]_0\), the concentration at a given time \([A]_t\), and the time elapsed \(t\). With these components, we can isolate \(k\) and calculate its value.
  • The units of \(k\) for a first-order reaction are typically \(s^{-1}\), underscoring that it is a measure of reaction speed.
  • Finding \(k\) allows you to predict future concentrations and understand the behavior of your reacting system over time.
Reaction Kinetics
Reaction kinetics is the study of how and why reactions take place at certain speeds. It provides insight into the factors that influence the reaction rate, such as concentration, temperature, and the presence of catalysts.

In first-order reactions, the rate is dependent on the concentration of a single reactant. This simplicity allows us to model the reaction with straightforward mathematical equations. As time progresses, the concentration of the reactant decreases in a predictable way, allowing us to characterize the reaction with a single rate constant \(k\).
  • Kinetics helps chemists understand how manipulations to reactants or conditions can alter the speed of reactions.
  • Models like this are not only applicable to academic exercises but also real-world scenarios, such as drug decomposition or biochemical processes.
Exponential Decay
Exponential decay refers to a process where a quantity decreases at a rate proportional to its current value. In the context of first-order reactions, this means that the concentration of a reactant declines over time in an exponential manner.

This concept can be mathematically expressed with the equation \([A]_t = [A]_0 \times e^{-kt}\), where the exponential function \(e\) represents this consistent rate of decay. The greater the rate constant \(k\), the faster the concentration decreases. After a specific duration, such as 88 seconds in the provided problem, the concentration will have gone down significantly as predicted by the exponential model.
  • Exponential decay is a fundamental principle used in various scientific fields, such as radiometric dating or population studies.
  • Understanding it is critical for accurately predicting how concentrations will change over time in a reaction.

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Most popular questions from this chapter

Butadiene can undergo the following reaction to form a dimer (two butadiene molecules hooked together). $$ 2 \mathrm{C}_{4} \mathrm{H}_{8}(g) \longrightarrow \mathrm{C}_{8} \mathrm{H}_{12}(g) $$ The half-life for the reaction at a given temperature is \(5.92 \times\) \(10^{-2} \mathrm{~s}\). The reaction is second-order. a. If the initial concentration of \(\mathrm{C}_{4} \mathrm{H}_{8}\) is \(0.50 \mathrm{M}\), what is the rate constant for the reaction? b. If the initial concentration of \(\mathrm{C}_{4} \mathrm{H}_{8}\) is \(0.010 \mathrm{M}\), what will be the concentration of \(\mathrm{C}_{4} \mathrm{H}_{8}\) after \(3.6 \times 10^{2} \mathrm{~s}\) ?

Dinitrogen pentoxide, \(\mathrm{N}_{2} \mathrm{O}_{5}\), decomposes when heated in carbon tetrachloride solvent. $$ \mathrm{N}_{2} \mathrm{O}_{5} \longrightarrow 2 \mathrm{NO}_{2}+\frac{1}{2} \mathrm{O}_{2}(g) $$ If the rate constant for the decomposition of \(\mathrm{N}_{2} \mathrm{O}_{5}\) is \(6.2 \times\) \(10^{-4} / \mathrm{min}\), what is the half-life? (The rate law is first order in \(\mathrm{N}_{2} \mathrm{O}_{5} .\) ) How long would it take for the concentration of \(\mathrm{N}_{2} \mathrm{O}_{5}\) to decrease to \(25 \%\) of its initial value? to \(12.5 \%\) of its initial value?

At high temperature, the reaction $$ \mathrm{NO}_{2}(g)+\mathrm{CO}(g) \longrightarrow \mathrm{NO}(g)+\mathrm{CO}_{2}(g) $$ is thought to occur in a single step. What should be the rate law in that case?

To obtain the rate of the reaction $$ \begin{array}{r} 3 \mathrm{I}^{-}(a q)+\mathrm{H}_{3} \mathrm{AsO}_{4}(a q)+2 \mathrm{H}^{+}(a q) \longrightarrow \\ \mathrm{I}_{3}^{-}(a q)+\mathrm{H}_{3} \mathrm{AsO}_{3}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \end{array} $$ you might follow the \(\mathrm{I}^{-}\) concentration or the \(\mathrm{I}_{3}^{-}\) concentration. How are the rates in terms of these species related?

Consider the following mechanism for a reaction in aqueous solution and indicate the species acting as a catalyst: $$ \begin{aligned} \mathrm{NH}_{2} \mathrm{NO}_{2}+\mathrm{OH}^{-} & \rightleftharpoons \mathrm{H}_{2} \mathrm{O}+\mathrm{NHNO}_{2}^{-} \\ \mathrm{NHNO}_{2}^{-} & \longrightarrow \mathrm{N}_{2} \mathrm{O}(g)+\mathrm{OH}^{-} \end{aligned} $$ Explain why you believe this species is a catalyst. What is the overall reaction? What substance might be added to the reaction mixture to give the catalytic activity?
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