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A compound decomposes by a first-order reaction. If the concentration of the compound is \(0.0569 M\) after 40 . s when the initial concentration was \(0.0700 \mathrm{M},\) what is the concentration of the compound after \(91 \mathrm{~s}\) ?

Short Answer

Expert verified
The concentration after 91 seconds is approximately 0.0439 M.

Step by step solution

01

Understanding the First-Order Reaction Equation

A first-order reaction has the rate law: \( \ln \left(\frac{[A]_t}{[A]_0}\right) = -kt \), 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

Using the given data, \([A]_0 = 0.0700\, M\) and \([A]_{40} = 0.0569\, M\). We need to find \(k\). Rearrange the formula to find \(k\): \( k = -\frac{\ln ([A]_{40}/[A]_0)}{t} = -\frac{\ln (0.0569/0.0700)}{40} \).
03

Solve for the Rate Constant

Substitute the values in and solve for \(k\): \( k \approx -\frac{\ln (0.81286)}{40} = 0.004965\, \text{s}^{-1} \).
04

Find the Concentration After 91 Seconds

Apply the reaction equation again: \( \ln \left(\frac{[A]_{91}}{[A]_0}\right) = -kt \). Use \([A]_0 = 0.0700\, M\), \(t = 91\), and \(k = 0.004965\, \text{s}^{-1}\).
05

Rearrange and Solve for New Concentration

Rearrange to solve for \([A]_{91}\): \( [A]_{91} = [A]_0 \times e^{-kt} = 0.0700 \times e^{-0.004965 \times 91} \).
06

Calculate the New Concentration

Compute the value: \( [A]_{91} = 0.0700 \times e^{-0.451815} \approx 0.0439\, M \).

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Key Concepts

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

Rate Law
In a chemical reaction, the rate law is a mathematical equation that describes the relationship between the rate of a reaction and the concentration of its reactants. For a first-order reaction, the rate law is significant because it reveals how the concentration of the reactant decreases over time. The typical rate law for a first-order reaction is given by:\[\ln \left(\frac{[A]_t}{[A]_0}\right) = -kt\]where:
  • \([A]_t\) is the concentration of the reactant at time \(t\).
  • \([A]_0\) is the initial concentration.
  • \(k\) is the rate constant.
This equation helps us understand how quickly a reaction is progressing. In the given exercise, this equation was the foundation to determine how much of the compound remains after a certain time. Simply put, it is the mathematical sentence that gives insights into the change in concentration over time.
Rate Constant
The rate constant \(k\) is a crucial part of the rate law. It is a proportionality factor that gives us an idea of the speed of the reaction. For first-order reactions, the units of \(k\) are typically \ \( \text{s}^{-1} \ \). Understanding \(k\) is key to predicting how the concentration changes over time.To find \(k\) in the exercise, use the rearranged rate law formula:\[k = -\frac{\ln \left(\frac{[A]_t}{[A]_0}\right)}{t}\]Inserting the given concentrations, \([A]_0 = 0.0700\, M\) and \([A]_{40} = 0.0569\, M\), for \(t = 40\) seconds, we calculated \(k\) as approximately 0.004965 \(\text{s}^{-1}\). This value of \(k\) tells us how quickly the concentration is decreasing per second. A larger \(k\) would indicate a faster reaction, whereas a smaller \(k\) would correspond to a slower one.
Concentration
Concentration is the amount of a substance in a given volume, often measured in molarity (M). In chemical kinetics, monitoring how the concentration of a reactant changes over time is essential for understanding the reaction's progress. For a first-order reaction, the concentration decreases exponentially. Using the rate law and knowing the \(k\), you can calculate the concentration at any time \(t\). In the exercise, after determining the rate constant \(k\), we found the concentration after 91 seconds using the formula:\[[A]_{91} = [A]_0 \times e^{-kt}\]Given \([A]_0 = 0.0700\, M\) and \(k = 0.004965\, \text{s}^{-1}\), we calculated \([A]_{91} = 0.0439\, M\). This shows how concentration is predicted to diminish with time, illustrating exponential decay typical of first-order reactions. Understanding these changes helps in controlling and predicting the behavior of chemical reactions.

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Most popular questions from this chapter

The decomposition of hydrogen peroxide is a firstorder reaction: $$ \mathrm{H}_{2} \mathrm{O}_{2}(a q) \longrightarrow \mathrm{H}_{2} \mathrm{O}(l)+\frac{1}{2} \mathrm{O}_{2}(g) $$ The half-life of the reaction is 17.0 minutes. a. What is the rate constant of the reaction? b. If you had a bottle of \(\mathrm{H}_{2} \mathrm{O}_{2},\) how long would it take for \(86.0 \%\) to decompose? c. If you started the reaction with \(\left[\mathrm{H}_{2} \mathrm{O}_{2}\right]=0.100 M,\) what would be the hydrogen peroxide concentration after 15.0 minutes?

A plot of \(1 /[\mathrm{A}]\) versus time for the hypothetical reaction \(\mathrm{A} \longrightarrow \mathrm{B}+\mathrm{C}\) yields a straight line. a) What is the order of the reaction? b) If it took \(66 \mathrm{~s}\) for the concentration to drop 40 percent from its initial value of \(0.50 \mathrm{M},\) what is the rate constant for the reaction?

The rate of a particular reaction quadruples when the temperature is increased from \(25^{\circ} \mathrm{C}\) to \(42^{\circ} \mathrm{C}\). Calculate the activation energy for this reaction.

Relate the rate of decomposition of \(\mathrm{NH}_{4} \mathrm{NO}_{2}\) to the rate of formation of \(\mathrm{N}_{2}\) for the following reaction: $$ \mathrm{NH}_{4} \mathrm{NO}_{2}(a q) \longrightarrow \mathrm{N}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(l) $$

Draw and label the potential-energy curve for the reaction $$ \mathrm{N}_{2} \mathrm{O}_{4}(g) \rightleftharpoons 2 \mathrm{NO}_{2}(g) ; \Delta H=57 \mathrm{~kJ} $$ The activation energy for the reverse reaction is \(23 \mathrm{~kJ}\). Note \(\Delta H\) and \(E_{a}\) on the diagram. What is the activation energy for the forward reaction? For which reaction (forward or reverse) will the reaction rate be most sensitive to a temperature increase? Explain.

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