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Chapter 18: Problem 21

The half-life period of a first order reaction is 15 minutes. The amount of substance left after one hour will be: [Main Online April 9, 2014] (a) \(\frac{1}{4}\) of the original amount (b) \(\frac{1}{8}\) of the original amount (c) \(\frac{1}{16}\) of the original amount (d) \(\frac{1}{32}\) of the original amount

Short Answer

Expert verified
The amount left after one hour is \( \frac{1}{16} \), option (c).

Step by step solution

01

Understanding Half-Life for First Order Reactions

For a first-order reaction, the half-life period \( t_{1/2} \) is the time required for the concentration of the substance to decrease to half of its initial value. The half-life \( t_{1/2} \) is defined as \( t_{1/2} = \frac{0.693}{k} \), where \( k \) is the rate constant.
02

Calculate Rate Constant

Given the half-life period \( t_{1/2} = 15 \) minutes, we can solve for the rate constant \( k \):\[ k = \frac{0.693}{t_{1/2}} = \frac{0.693}{15} \space \text{min}^{-1}\]
03

Determine Number of Half-Lives in One Hour

In one hour (60 minutes), we need to determine how many half-lives have passed:\[ \text{Number of half-lives} = \frac{60}{15} = 4\]
04

Calculate Remaining Amount of Substance

For a first-order reaction, after \( n \) half-lives, the amount of substance left is \( \frac{1}{2^n} \) of the original amount. Here, \( n = 4 \):\[ \text{Remaining amount} = \frac{1}{2^4} = \frac{1}{16}\]
05

Identify the Correct Answer Option

The amount of substance left after one hour is \( \frac{1}{16} \) of the original amount, which corresponds to option (c).

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Key Concepts

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

First Order Reaction
In chemical kinetics, reactions are often classified into different types based on their reaction order. A first order reaction is a type where the reaction rate is directly proportional to the concentration of one reactant. This means that as the concentration of the reactant decreases, the rate at which the reaction occurs also slows down.

First order reactions are quite common in nature and laboratory settings. The rate law for a first order reaction can be expressed as:

\( ext{Rate} = k[A] \)

Where:
  • \( ext{Rate} \) is the speed at which the reaction occurs.
  • \( k \) is the reaction rate constant, which is a fixed value specific to each reaction.
  • \( [A] \) is the concentration of the reactant.
Understanding how first order reactions work helps predict how long it will take for a reaction to reach a certain point, such as when a reactant is fully consumed.
Half-Life
The concept of half-life is crucial when discussing reactions, particularly first order reactions. Half-life, denoted as \( t_{1/2} \), is defined as the time required for the concentration of a reactant to reduce to half its initial amount. It is an indication of how quickly a reaction proceeds.

For first order reactions, the half-life is independent of the initial concentration. This means that regardless of how much reactant you start with, the time it takes for half of it to react will be the same. The formula for calculating half-life in first order reactions is:

\( t_{1/2} = \frac{0.693}{k} \)

Here, \( k \) is the reaction rate constant.

The half-life can be used to quickly determine how much of a substance remains after various periods of time by calculating how many half-lives have elapsed.
Reaction Rate Constant
The reaction rate constant, symbolized as \( k \), is a fundamental factor in chemical kinetics. It provides an indication of the speed of a reaction under specific conditions. For first order reactions, the rate constant is determined from the half-life using the equation:

\( k = \frac{0.693}{t_{1/2}} \)

The rate constant \( k \) has units of time\(^{-1}\), often noted as \( ext{min}^{-1} \) in cases where time is measured in minutes.

A larger value of \( k \) indicates a faster reaction, meaning the reactant is consumed more quickly. Conversely, a smaller \( k \) suggests a slower reaction.

Understanding how \( k \) interacts with the half-life and concentration allows researchers to manipulate reaction conditions to achieve desired results, such as increasing the speed of a reaction or prolonging a reactant's availability.

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Most popular questions from this chapter

If uranium (mass number 238 and atomic number 92) emits an \(\alpha\) particle, the product has mass no. and atomic no. (a) 236 and 92 (b) 234 and 90 (c) 238 and 90 (d) 236 and 90

For the non-stoichiometric reaction \(2 \mathrm{~A}+\mathrm{B} \rightarrow \mathrm{C}+\mathrm{D}\), the following kinetic data were obtained in three separate experiments, all at \(298 \mathrm{~K}\). \begin{tabular}{|c|c|c|} \hline Initial Concentration (A) & Initial Concentration (B) & Initial rate of formation of C \(\left(\mathbf{m o l} \mathbf{L}^{-1} \mathbf{s}^{-1} \mathbf{)}\right.\) \\ \hline \(0.1 \mathrm{M}\) & \(0.1 \mathrm{M}\) & \(1.2 \times 10^{-3}\) \\ \hline \(0.1 \mathrm{M}\) & \(0.2 \mathrm{M}\) & \(1.2 \times 10^{-3}\) \\ \hline \(0.2 \mathrm{M}\) & \(0.1 \mathrm{M}\) & \(2.4 \times 10^{-3}\) \\ \hline \end{tabular}The rate law for the formation of \(\mathrm{C}\) is: [Main 2014] (a) \(\frac{d \mathrm{c}}{d t}=k[\mathrm{~A}][\mathrm{B}]\) (b) \(\frac{d \mathrm{c}}{d t}=k[\mathrm{~A}]^{2}[\mathrm{~B}]\)(c) \(\frac{d \mathrm{c}}{d t}=k[\mathrm{~A}][\mathrm{B}]^{2}\) (d) \(\frac{d \mathrm{c}}{d t}=k[\mathrm{~A}]\)

The half-life period of a radioactive element is 140 days. After 560 days, one gram of the element will reduced to : (a) \(\frac{1}{2} \mathrm{~g}\) (b) \(\frac{1}{4} \mathrm{~g}\) (c) \(\frac{1}{8} \mathrm{~g}\) (d) \(\frac{1}{16} \mathrm{~g}\)

\({ }^{64} \mathrm{Cu}\) (half-life \(=12.8 \mathrm{~h}\) ) decays by \(\beta\) - emission \((38 \%), \beta^{+}\)emission \((19 \%)\) and electron capture (43\%). Write the decay products and calculate partial half-lives for each of the decay processes.

During the nuclear explosion, one of the products is \({ }^{90} \mathrm{Sr}\) with half life of \(6.93\) years. If \(1 \square \mathrm{g}\) of \({ }^{90} \mathrm{Sr}\) was absorbed in the bones of a newly born baby in place of \(\mathrm{Ca}\), how much time, in years, is required to reduce it by \(90 \%\) if it is not lost metabolically. [Main Jan. 07, 2020 (I)]
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