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For a first order reaction, which of the following are not correct? a. \(t_{3 / 8}=2 t_{3 / 4}\) b. \(t_{3 / 4}=2 t_{1 / 2}\) c. \(t_{15 / 6}=4 t_{1 / 2}\) d. \(t_{15 / 16}=3 t_{3 / 4}\)

Short Answer

Expert verified
Options (a), (c), and (d) are not correct for a first-order reaction.

Step by step solution

01

Understand First-Order Reaction Kinetics

For a first-order reaction, the rate of reaction is proportional to the concentration of a single reactant. The time to reach a certain fraction of completion is related to the half-life of the reaction, which is constant.
02

Analyze Option (a)

Option (a) claims \(t_{3 / 8}=2 t_{3 / 4}\). For a first-order reaction, the time required to reduce concentration from 3/8 to zero would not be twice the time to reduce from 3/4 because the time relationships depend on logarithmic decay. This statement is incorrect.
03

Analyze Option (b)

Option (b) is \(t_{3 / 4}=2 t_{1 / 2}\). For first-order reactions, \(t_{3/4}\) should indeed be twice \(t_{1/2}\) because reducing concentration by a further half from 1/2 to 1/4 requires the same time as the first half-life. This statement is correct.
04

Analyze Option (c)

Option (c) is \(t_{15 / 6}=4 t_{1 / 2}\). The notation seems incorrect; let's correct it to \(t_{5/6}\). It is incorrect for a first-order reaction, as \(t_{5/6}\) does not equal four half-lives. The accurate conversion involves using the equation for decay time based on natural logarithms.
05

Analyze Option (d)

Option (d) states \(t_{15 / 16}=3 t_{3 / 4}\). Reducing the concentration from \(3/4\) to \(15/16\) does not follow a straightforward multiple of \(t_{3/4}\), and the exact reduction would require applying the integrated rate law. This statement is incorrect.

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Key Concepts

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

Reaction Kinetics
Understanding reaction kinetics is essential when studying chemical reactions. Reaction kinetics describes how the rate of a chemical reaction depends on the concentration of reactants and the specific rate constant. For a first-order reaction, the rate is directly proportional to the concentration of a single reactant. This means if the concentration of the reactant is doubled, the rate of the reaction will also double. By understanding reaction kinetics, you can predict how fast a reaction will proceed under certain conditions.
This concept helps in designing chemical processes and in understanding various natural phenomena.
  • The rate of reaction is typically measured in moles per liter per second (M/L/s).
  • Reaction mechanisms provide insight into the steps involved in transforming reactants into products.
Grasping these basics can demystify why some reactions happen quickly while others take longer, paving the way for more advanced studies in chemistry.
Half-Life
Half-life is a crucial concept in first-order reactions. It is the time required for half of the reactant to be consumed in a chemical reaction. In other words, it is the time taken for the concentration of a reactant to decrease by half. A special feature of first-order reactions is that the half-life is constant, meaning it doesn't change as the concentration decreases.
Understanding half-life is particularly useful in fields such as pharmacology, where you need to know how quickly a drug is metabolized. It is also important in radioactive decay calculations.
  • The formula for half-life ( t_{1/2} ) in a first-order reaction is given as t_{1/2} = rac{0.693}{k} , where k is the rate constant.
  • This constant time period creates straightforward experimental and theoretical analysis of decay processes.
By familiarizing oneself with half-life, one can predict how much time is required for a given fraction of a reactive species to break down, which is invaluable in many scientific applications.
Logarithmic Decay
Logarithmic decay is an intriguing phenomenon observed in first-order reactions. It describes how the concentration of reactants decreases over time in a manner that's exponential and not linear. In these reactions, as time progresses, the concentration decreases similarly to an exponential decay curve. This type of decay follows the equation [A] = [A]_0e^{-kt} , where [A] is the concentration at time t , [A]_0 is the initial concentration, and k is the rate constant.
This logarithmic relationship is important when determining how long a reaction will take to reach a certain completion stage.
  • The rate of decay indicates how quickly the concentration of reactants changes.
  • This decay pattern allows chemists to predict the concentrations of reactants at any given time during the reaction.
Understanding logarithmic decay can help you grasp why reactions slow down as they near completion and how to quantify these dynamics in chemical processes.
Integrated Rate Law
The integrated rate law is a mathematical expression that provides a relationship between the concentration of reactants and time for first-order reactions. This is particularly useful for calculating concentrations at various time points during a reaction. The integrated rate law for first-order kinetics can be expressed as ext{ln} [A] = ext{ln} [A]_0 - kt .
This equation shows a linear relationship between the natural log of concentration and time, making it straightforward to plot and analyze experimental data.
  • The slope of a plot of ext{ln} [A] versus time gives the rate constant k .
  • The intercept corresponds to ext{ln} [A]_0 , the natural log of the initial concentration.
Using the integrated rate law allows scientists to gain accurate predictions about the behavior of a reaction under varying conditions, essential for both academic studies and practical applications like engineering and pharmacology.

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Most popular questions from this chapter

The rate law for the reaction \(\mathrm{RCl}+\mathrm{NaOH} \rightarrow \mathrm{ROH}+\mathrm{NaCl}\) is given by Rate \(=\mathrm{k}(\mathrm{RCl})\). The rate of the reaction is a. Halved by reducing the concentration of \(\mathrm{RCl}\) by one half. b. Increased by increasing the temperature of the reaction. c. Remains same by change in temperature. d. Doubled by doubling the concentration of \(\mathrm{NaOH}\).

Hydrogen peroxide decomposes to water and oxygen according to the reaction below: $$ 2 \mathrm{H}_{2} \mathrm{O}_{2}(\mathrm{aq}) \rightarrow 2 \mathrm{H}_{2} \mathrm{O}(\mathrm{l})+\mathrm{O}_{2}(\mathrm{~g}) $$ In the presence of large excesses of \(\mathrm{I}^{-}\)ion, the following set of data is obtained. What is the average rate of disappearance of \(\mathrm{H}_{2} \mathrm{O}_{2}\) (aq) in \(\mathrm{M} / \mathrm{s}\) in the first \(45.0\) seconds of the reaction if \(1.00\) litre of \(\mathrm{H}_{2} \mathrm{O}_{2}\) reacts at \(25^{\circ} \mathrm{C}\) and \(1.00 \mathrm{~atm}\) pressure? \begin{tabular}{ll} Time, \(s\) & \(\mathrm{O}_{2}(\mathrm{~g})\) collected, \(\mathrm{ml}\) \\ \(0.0\) & \(0.0\) \\ \(45.0\) & \(2.00\) \\ \(90.0\) & \(4.00\) \\ \(135.0\) & \(6.00\) \\ \hline \end{tabular} 26 \(2.63 \times 10^{-4} \mathrm{M} / \mathrm{s}\) a. \(.2 .63 \times 10^{-4} \mathrm{M} / \mathrm{s}\) \(6.33 \times 10^{-6} \mathrm{M} / \mathrm{s}\) b. \(6.33 \times 10^{-6} \mathrm{M} / \mathrm{s}\) \(3.63 \times 10^{-6} \mathrm{M} / \mathrm{s}\) c. d. \(1.36 \times 10^{-3} \mathrm{M} / \mathrm{s}\)

\(2 \mathrm{P}+3 \mathrm{Q}+\mathrm{R} \rightarrow\) product If for this reaction Rate \((\mathrm{R})=\mathrm{K}[\mathrm{P}]^{-1 / 2}[\mathrm{Q}]^{1}[\mathrm{R}]^{1 / 2}\) The order of this reaction is a. Zero b. Ist c. IInd d. \(3 / 2\)

In hypothetical reaction \(\mathrm{X}_{2}+\mathrm{Y}_{2} \rightarrow 2 \mathrm{XY}\) Follows the mechanism as given below \(\mathrm{X}_{2}=\mathrm{X}+\mathrm{X}\) (fast reaction) \(\mathrm{X}+\mathrm{Y}_{2} \rightarrow \mathrm{XY}+\mathrm{Y}\) (slow reaction) \(\mathrm{X}+\mathrm{Y} \rightarrow \mathrm{XY}\) (fast reaction) Here the correct statement is/are a. Order of reaction is \(3 / 2\). b. Molecularity is 2 . c. \(\mathrm{R}=\mathrm{k}[\mathrm{X}]\left[\mathrm{Y}_{2}\right]\) d. Both molecularity and order \(=3\)

The following data pertains to the reaction between A and B $$ \begin{array}{llll} \hline \text { S. } & {[\mathrm{A}]} & {[\mathrm{B}]} & \text { Rate } \\ \text { No. } & \mathrm{mol} \mathrm{L}^{-1} & \mathrm{~mol} \mathrm{~L}^{-1} & \mathrm{Mol} \mathrm{L}^{-1} \mathrm{t}^{-1} \\ \hline 1 & 1 \times 10^{-2} & 2 \times 10^{-2} & 2 \times 10^{-4} \\ 2 & 2 \times 10^{-2} & 2 \times 10^{-2} & 4 \times 10^{-4} \\ 3 & 2 \times 10^{-2} & 4 \times 10^{-2} & 8 \times 10^{-4} \\ \hline \end{array} $$ Which of the following inferences are drawn from the above data? (1) Rate constant of the reaction is \(10^{-4}\) (2) Rate law of the reaction is k [A][B] (3) Rate of reaction increases four times by doubling the concentration of each reactant. Select the correct answer the codes given below: a. 1 and 3 b. 2 and 3 c. 1 and 2 d. 1,2 and 3

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