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What is the half-life of a compound if 75 percent of a given sample of the compound decomposes in \(60 \mathrm{~min}\) ? Assume first-order kinetics.

Short Answer

Expert verified
The half-life of the compound is approximately 30 minutes.

Step by step solution

01

Understand First-order Kinetics

In first-order kinetics, the rate of reaction depends only on the concentration of one reactant. The rate law is expressed as \( \text{Rate} = k[A] \), where \( k \) is the rate constant and \( [A] \) is the concentration of the reactant.
02

Recognize the Percentage Decomposition

Given that 75% of the compound has decomposed, it means that 25% of the original concentration remains. This is used to determine the rate constant \( k \).
03

Use the Integrated Rate Law

For first-order reactions, the integrated rate law is \( \ln \left(\frac{[A]_t}{[A]_0}\right) = -kt \). Here, \( [A]_t \) is the final concentration (25% of \([A]_0\)), \( [A]_0 \) is the initial concentration, and \( t = 60 \) minutes.
04

Calculate the Rate Constant

Substitute into the equation: \( \ln \left( \frac{0.25[A]_0}{[A]_0} \right) = -k(60) \). Simplify to get \( \ln(0.25) = -k(60) \). Then solve for \( k \) to find \( k = -\frac{\ln(0.25)}{60} \approx 0.0231 \, ext{min}^{-1} \).
05

Find the Half-life Formula

For first-order reactions, the half-life \( t_{1/2} \) is given by \( t_{1/2} = \frac{0.693}{k} \).
06

Calculate the Half-life

Substitute \( k = 0.0231 \, ext{min}^{-1} \) into the half-life formula: \( t_{1/2} = \frac{0.693}{0.0231} \approx 30 \, \text{minutes} \).

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Key Concepts

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

First-order kinetics
First-order kinetics is an essential concept in chemical kinetics, explaining how reaction rates depend on the concentration of a single reactant. In these reactions, the rate is directly proportional to the concentration of the reactant. Mathematically, this is shown as:
  • Rate = k[A]
Here, "Rate" is the speed at which the reaction occurs, "k" is the rate constant, and "[A]" is the concentration of the reactant. An important feature of first-order reactions is that the half-life, which is the time it takes for half of the reactant to decompose, remains constant regardless of the initial concentration. This is different from reactions of other orders where the half-life varies with concentration. Understanding first-order kinetics provides insight into predicting how long it will take for a reactant to decrease to a certain concentration over time.
Rate constant
The rate constant, often symbolized as "k", is a pivotal parameter in the study of reaction kinetics. It provides the proportionality factor between the reaction rate and the concentration of reactants, especially in first-order reactions where the relationship is simple and direct. A higher value of "k" indicates a faster reaction. The units of the rate constant depend on the order of reaction.
  • For first-order reactions, the unit is usually min-1 or s-1, as it involves time after simplifying the rate expression.
Determining "k" is crucial for finding out other kinetic properties, such as the half-life of a reaction. Experimental values of "k" can sometimes vary with temperature, so it must be considered if the reaction is being compared under differing conditions.
Integrated rate law
The integrated rate law for first-order reactions is a powerful tool enabling the transformation of rates into concentrations over time. This helps in calculating the remaining concentration of a reactant after a given period. For a first-order reaction, it takes the form:
  • \( \ln \left(\frac{[A]_t}{[A]_0}\right) = -kt \)
Here, \([A]_t\) is the concentration of the reactant at time \( t \), \([A]_0\) is the initial concentration, and \( k \) is the rate constant. By rearranging this formula, you can find any missing parameter. For example, by inserting the percentage left after a specific time or experimenting with arbitrary concentrations, it allows the calculation of "k". The integrated rate law helps us understand how reactors behave over time and is frequently employed in laboratory settings to gather kinetic data.
Percentage decomposition
Percentage decomposition measures how much of the original material has reacted over a given period, usually shown as a percentage. It tells you how far a reaction has progressed and is particularly useful in kinetics to interpret the extent of reactions. For instance, saying that 75% has decomposed means only 25% of the original reactant remains. This information can then be used in the integrated rate law.In first-order kinetics:
  • Calculate the remaining concentration, knowing that \(%\text{decomposed} = 100 - \%[A]_t\).
This inference enables the calculation of other parameters like the rate constant or time, given the equation:
  • \( \ln \left(\frac{[A]_t}{[A]_0}\right) = -kt \)
Understanding percentage decomposition and applying it helps streamline the process of kinetic data analysis, making it easier to solve problems or predict reaction behavior.

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Most popular questions from this chapter

The following scheme in which \(\mathrm{A}\) is converted to \(\mathrm{B}\), which is then converted to \(\mathrm{C}\), is known as a consecutive reaction: \(\mathrm{A} \longrightarrow \mathrm{B} \longrightarrow \mathrm{C}\) Assuming that both steps are first order, sketch on the same graph the variations of \([\mathrm{A}],[\mathrm{B}],\) and \([\mathrm{C}]\) with time.

On which of the following properties does the rate constant of a reaction depend: (a) reactant concentrations, (b) nature of reactants, (c) temperature?

Radioactive plutonium- \(239\left(t_{1 / 2}=2.44 \times 10^{5} \mathrm{yr}\right)\) is used in nuclear reactors and atomic bombs. If there are \(5.0 \times 10^{2} \mathrm{~g}\) of the isotope in a small atomic bomb, how long will it take for the substance to decay to \(1.0 \times 10^{2} \mathrm{~g}\), too small an amount for an effective bomb?

To carry out metabolism, oxygen is taken up by hemoglobin \((\mathrm{Hb})\) to form oxyhemoglobin \(\left(\mathrm{Hb} \mathrm{O}_{2}\right)\) according to the simplified equation: $$ \mathrm{Hb}(a q)+\mathrm{O}_{2}(a q) \stackrel{k}{\longrightarrow} \mathrm{HbO}_{2}(a q) $$ where the second-order rate constant is \(2.1 \times 10^{6} / M \cdot \mathrm{s}\) at \(37^{\circ} \mathrm{C}\). For an average adult, the concentrations of \(\mathrm{Hb}\) and \(\mathrm{O}_{2}\) in the blood at the lungs are \(8.0 \times 10^{-6} \mathrm{M}\) and \(1.5 \times 10^{-6} M,\) respectively. (a) Calculate the rate of formation of \(\mathrm{HbO}_{2}\). (b) Calculate the rate of consumption of \(\mathrm{O}_{2}\). (c) The rate of formation of \(\mathrm{HbO}_{2}\) increases to \(1.4 \times 10^{-4} M / \mathrm{s}\) during exercise to meet the demand of the increased metabolism rate. Assuming the \(\mathrm{Hb}\) concentration to remain the same, what must the oxygen concentration be to sustain this rate of \(\mathrm{HbO}_{2}\) formation?

The rate constant for the second-order reaction: $$ 2 \mathrm{NO}_{2}(g) \longrightarrow 2 \mathrm{NO}(g)+\mathrm{O}_{2}(g) $$ is \(0.54 / M \cdot \mathrm{s}\) at \(300^{\circ} \mathrm{C}\). How long (in seconds) would it take for the concentration of \(\mathrm{NO}_{2}\) to decrease from \(0.65 M\) to \(0.18 M ?\)

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