/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 23 What is the half-life of a compo... [FREE SOLUTION] | 91Ó°ÊÓ

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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 min? Assume first-order kinetics.

Short Answer

Expert verified
The half-life of the compound, to the nearest minute, is approximately 21 minutes.

Step by step solution

01

Understand the principles of first-order kinetics

The general formula for first-order decay is given by: \( N = N_0 e^{-kt} \), where \(N\) is the amount remaining, \(N_0\) is the initial amount, \(k\) is the rate constant, and \(t\) is time. For this problem, we need to rearrange this equation to solve for \(k\), which we can then use to find the half-life
02

Setting up the equation

Given that 75% of the sample decomposes in 60 minutes, meaning that 25% remains. We can substitute this information into the decay formula: \(0.25N_0 = N_0 e^{-k(60)}\)
03

Solve for the rate constant (k)

After getting rid of the \(N_0\) on both sides and taking the natural logarithm (ln) we obtain: \(-ln(4) = -60k\), rearranging this to solve for \(k\) gives \(k = ln(4) / 60\)
04

Derive the half-life from the decay constant

The formula connecting the half-life (t1/2) with the rate constant in a first-order reaction is \(t1/2 = ln(2) / k\). Substituting our calculated \(k\) value into this formula will yield the half-life.

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Key Concepts

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

Half-life calculation
Understanding the half-life of a compound is crucial in different scientific areas, like chemistry and pharmacology. Half-life is the time it takes for a substance to reduce to half of its initial amount. For first-order reactions, calculating the half-life is straightforward due to the relationship between the decay constant and the natural logarithm. The formula to calculate the half-life is: \[ t_{1/2} = \frac{\ln(2)}{k} \] Here, \( t_{1/2} \) represents the half-life, and \( k \) is the rate constant of the reaction. In the example provided, after calculating \( k \) from the decomposition data, you can substitute \( k \) back into this formula to find how long it will take for the compound to decompose to half of its original amount in a first-order kinetic reaction. It's important to remember that this calculation is specific to first-order reactions.
Rate constant
The rate constant, often symbolized as \( k \), is a critical parameter in kinetics that quantifies the speed of a chemical reaction. In the context of first-order reactions, the rate constant determines how quickly the concentration of a reactant decreases over time.The general formula for a first-order reaction is: \[ N = N_0 e^{-kt} \] Where:
  • \( N \) is the amount of substance remaining.
  • \( N_0 \) is the initial amount.
  • \( t \) is the time elapsed.
  • \( k \) is the rate constant.
To find \( k \), you rearrange this formula based on the known quantities such as the time at which a certain percentage of the substance remains. In the exercise, 25% of the compound remains after 60 minutes. By substituting the known values and solving for \( k \), one can understand how swiftly the reaction proceeds.
Exponential decay
Exponential decay describes the process wherein the amount of a substance decreases at a rate proportional to its current value. This is a common characteristic of first-order reactions. The mathematical representation of exponential decay is given by: \[ N = N_0 e^{-kt} \] This formula highlights how the concentration of a reactant diminishes exponentially over time.Key features of exponential decay include:
  • The decay rate, which is proportional to the amount of substance remaining.
  • The quantity decreases rapidly initially and slows down over time.
  • The existence of a constant rate, denoted as \( k \), which dictates the decay speed.
In the decomposition problem, recognizing the exponential nature of decay facilitates the understanding of how 75% of the substance decomposes over 60 minutes. By understanding exponential decay, students can predict future amounts of reactants in similar scenarios.

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Most popular questions from this chapter

Many reactions involving heterogeneous catalysts are zero order; that is, rate \(=k\). An example is the decomposition of phosphine \(\left(\mathrm{PH}_{3}\right)\) over tungsten (W): $$ 4 \mathrm{PH}_{3}(g) \longrightarrow \mathrm{P}_{4}(g)+6 \mathrm{H}_{2}(g) $$ It is found that the reaction is independent of \(\left[\mathrm{PH}_{3}\right]\) as long as phosphine's pressure is sufficiently high \((\geq 1 \mathrm{~atm}) .\) Explain.

Define activation energy. What role does activation energy play in chemical kinetics?

What are the advantages of measuring the initial rate of a reaction?

Polyethylene is used in many items such as water pipes, bottles, electrical insulation, toys, and mailer envelopes. It is a polymer, a molecule with a very high molar mass made by joining many ethylene molecules (the basic unit is called a monomer) together (see p. 369 ). The initiation step is $$ \mathrm{R}_{2} \stackrel{k_{1}}{\longrightarrow} 2 \mathrm{R} \cdot \quad \text { initiation } $$ The \(\mathrm{R}\) - species (called a radical) reacts with an ethylene molecule \((\mathrm{M})\) to generate another radical $$ \mathrm{R} \cdot+\mathrm{M} \longrightarrow \mathrm{M}_{1} \cdot $$ Reaction of \(\mathrm{M}_{1}\). with another monomer leads to the growth or propagation of the polymer chain: $$ \mathrm{M}_{1} \cdot+\mathrm{M} \stackrel{k_{\mathrm{p}}}{\longrightarrow} \mathrm{M}_{2} \cdot \quad \text { propagation } $$ This step can be repeated with hundreds of monomer units. The propagation terminates when two radicals combine $$ \mathrm{M}^{\prime} \cdot+\mathrm{M}^{\prime \prime} \cdot \stackrel{k_{t}}{\longrightarrow} \mathrm{M}^{\prime}-\mathrm{M}^{\prime \prime} \quad \text { termination } $$ (a) The initiator used in the polymerization of ethylene is benzoyl peroxide \(\left[\left(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COO}\right)_{2}\right]:\) $$ \left(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COO}\right)_{2} \longrightarrow 2 \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COO} \cdot $$ This is a first-order reaction. The half-life of benzoyl peroxide at \(100^{\circ} \mathrm{C}\) is \(19.8 \mathrm{~min} .\) (a) Calculate the rate constant (in \(\min ^{-1}\) ) of the reaction. (b) If the half-life of benzoyl peroxide is \(7.30 \mathrm{~h}\) or \(438 \mathrm{~min},\) at \(70^{\circ} \mathrm{C},\) what is the activation energy (in \(\mathrm{kJ} / \mathrm{mol}\) ) for the decomposition of benzoyl peroxide? (c) Write the rate laws for the elementary steps in the above polymerization process and identify the reactant, product, and intermediates. (d) What condition would favor the growth of long high-molar-mass polyethylenes?

What are the units of the rate constant for a thirdorder reaction?

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