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In a first order reaction \(\mathrm{A} \rightarrow \mathrm{B}\) if \(\mathrm{k}\) is rate constant and initial concentration of the reactant \(\mathrm{A}\) is \(0.5 \mathrm{M}\) then the half life is a. \(0.693 / 0.5 \mathrm{k}\) b. \(\log 2 / \mathrm{k}\) c. \(\log 2 / \mathrm{k} \sqrt{0} .5\) d. In \(2 / \mathrm{k}\)

Short Answer

Expert verified
The correct answer is d: \( \ln 2 / k \).

Step by step solution

01

Understand the Half-Life Formula for First Order Reaction

For a first-order reaction, the half-life \( t_{1/2} \) is given by:\[t_{1/2} = \frac{0.693}{k}\]where \( k \) is the rate constant.
02

Identify the Given Values

We are given that the initial concentration of \( \text{A} \) is \( 0.5 \, \text{M} \). However, for a first-order reaction, the half-life does not depend on the initial concentration, so this value is not needed for calculating \( t_{1/2} \).
03

Match with the Options

Using the formula \( t_{1/2} = \frac{0.693}{k} \), we can match the correct expression to the given options:- Option a: \( \frac{0.693}{0.5k} \) (Incorrect, as it involves initial concentration which is not needed for first order)- Option b: \( \frac{\log 2}{k} \) (Incorrect format, not \( 0.693 \))- Option c: \( \frac{\log 2}{k \sqrt{0.5}} \) (Incorrect, involves \( \sqrt{0.5} \))- Option d: \( \frac{\ln 2}{k} \) (Correct, as \( \ln 2 = 0.693 \), and matches the formula)

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Key Concepts

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

Half-Life Formula
The concept of half-life is a crucial element in understanding first-order reactions. The half-life of a reaction, commonly denoted as \( t_{1/2} \), is the time it takes for half of the reactant to convert into the product. For first-order reactions, this period remains constant regardless of the initial concentration of the reactant.
Using a specific formula for first-order reactions, you can determine the half-life with ease:
  • \[ t_{1/2} = \frac{0.693}{k} \]
In this equation:
  • \( t_{1/2} \) is the half-life of the reaction.
  • \( k \) is the rate constant, which quantifies the speed of the reaction.
  • \( 0.693 \) is a natural logarithm of 2, stemming from the nature of exponential decay.
An important aspect of the half-life formula for first-order reactions is its independence from initial concentration. This means that no matter how much reactant you start with, the time taken to reduce to half always stays the same during a first-order process.
Rate Constant
The rate constant, often represented by \( k \), is a crucial parameter in chemical kinetics, particularly in describing the speed of a chemical reaction. For a first-order reaction, the rate constant is instrumental in determining not just how fast a reaction proceeds but also in calculating the half-life.
The rate constant relates to how reactions progress over time. It is used in the half-life formula and is influential in many other aspects of kinetic calculations too. Here are some essential qualities of the rate constant in first-order reactions:
  • The unit of \( k \) for first-order reactions is \( \text{s}^{-1} \), indicating how the reaction's speed is measured in per second intervals.
  • Higher values of \( k \) signify that the reaction is comparatively faster.
To solve problems involving first-order reactions, understanding how the rate constant factors into equations like the half-life formula is vital, as it shows how time and reaction speed interlink. Thus, by combining the values of \( k \) and half-life, one can understand the temporal dynamics of a given chemical reaction.
Reaction Order
The concept of reaction order is fundamental in chemical kinetics to describe how the concentration of reactants affects the rate of a reaction. In the simplest terms, the reaction order is an exponent in the rate equation that shows the dependence of the reaction rate on the concentration of reactants.
For a first-order reaction, the rate is directly proportional to the concentration of a single reactant. The general rate law for a first-order reaction can be expressed as:
  • Rate = \( k[A] \)
Here, \([A]\) is the concentration of the reactant A, and \( k \) is the rate constant. This linear relationship implies:
  • If you double the concentration of A, the rate of the reaction also doubles.
  • The half-life remains constant as it does not depend on the concentration of reactant, which is a unique aspect of first-order kinetics.
Understanding the role of reaction order is key to predicting how changes in concentration affect the speed of a reaction. It offers insight into the molecular mechanisms behind the transformation of reactants into products, providing a window into the reaction's dynamic nature.

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Most popular questions from this chapter

In the following question two statements Assertion (A) and Reason (R) are given Mark. a. If \(\mathrm{A}\) and \(\mathrm{R}\) both are correct and \(\mathrm{R}\) is the correct explanation of \(\mathrm{A}\); b. If \(A\) and \(R\) both are correct but \(R\) is not the correct explanation of \(\mathrm{A}\); c. \(\mathrm{A}\) is true but \(\mathrm{R}\) is false; d. \(\mathrm{A}\) is false but \(\mathrm{R}\) is true, e. \(\mathrm{A}\) and \(\mathrm{R}\) both are false. (A): A catalyst enhances the rate of reaction. ( \(\mathbf{R}\) ): The energy of activation of the reaction is lowered in presence of a catalyst.

In aqueous solution, hypobromite ion \(\left(\mathrm{BrO}^{-}\right)\), reacts to produce bromate ion \(\left(\mathrm{BrO}_{3}^{-}\right)\), and bromide ion (Br), according to the following chemical equation. \(3 \mathrm{BrO}^{-}\)(aq) \(\rightarrow \mathrm{BrO}_{3}^{-}(\mathrm{aq})+2 \mathrm{Br}\) (aq) A plot of \(1 /\left[\mathrm{BrO}^{-}\right] \mathrm{vs}\). time is linear and the slope is equal to \(0.056 \mathrm{M}^{-1} \mathrm{~s}^{-1} .\) If the initial concentration of \(\mathrm{BrO}^{-}\)is \(0.80 \mathrm{M}\), how long will it take one-half of the \(\mathrm{BrO}^{-}\)ion to react? a. \(2.12 \mathrm{~s}\) b. \(22 \mathrm{~s}\) c. \(12 \mathrm{~s}\) d. \(3.22 \mathrm{~s}\)

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 rate constant for the reaction, \(2 \mathrm{~N}_{2} \mathrm{O}_{5} \rightarrow 4 \mathrm{NO}_{2}+\mathrm{O}_{2}\) is \(3.0 \times 10^{-4} \mathrm{~s}^{-1}\). If start made with \(1.0 \mathrm{~mol} \mathrm{~L}^{-1}\) of \(\mathrm{N}_{2} \mathrm{O}_{5}\), calculate the rate of formation of \(\mathrm{NO}_{2}\) at the moment of the reaction when concentration of \(\mathrm{O}_{2}\) is \(0.1 \mathrm{~mol} \mathrm{~L}^{-1}\). a. \(1.2 \times 10^{-4} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1}\) b. \(3.6 \times 10^{-4} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1}\) c. \(9.6 \times 10^{-4} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1}\) d. \(4.8 \times 10^{-4} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1}\)

For this reaction \(\mathrm{X}^{-}+\mathrm{OH}^{-} \rightarrow \mathrm{X}^{-}+\mathrm{XO}^{-}\)in an aque- ons medium, the rate of the reaction is given as \(\frac{\left(\mathrm{d}\left(\mathrm{XO}^{-}\right)\right.}{\mathrm{dt}}=\mathrm{K} \frac{\left[\mathrm{X}^{-}\right]\left[\mathrm{XO}^{-}\right]}{\left[\mathrm{OH}^{-}\right]}\) The overall order for this reaction is a. Zero b. 1 c. \(-1\) d. \(1 / 2\)

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