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For a first order reaction, which is/are correct here? a. The time taken for the completion of \(75 \%\) reaction is twice the \(t_{1 / 2}\) of the reaction b. The degree of dissociation is equal to \(1-\mathrm{e}^{-k t}\). c. A plot of reciprocal concentration of the reactant versus time gives a straight line d. The pre-exponential factor in the Arrhenius equation has the dimension of time, \(\mathrm{T}^{-1}\).

Short Answer

Expert verified
Correct options: a and b.

Step by step solution

01

Evaluating Option a

For a first-order reaction, the time to complete 75% of the reaction \(t_{75}\%\) is related to the half-life \(t_{1/2}\) by the equation \(t_{75\%} = 2 \times t_{1/2}\). Thus, statement a is correct for a first-order reaction.
02

Evaluating Option b

The degree of dissociation for a first-order reaction can be expressed as \(\alpha = 1 - e^{-kt}\). This statement matches with the given statement, thus it is correct.
03

Evaluating Option c

For a first-order reaction, the concentration of the reactant versus time \(t\) follows the equation \[ [A]_t = [A]_0 e^{-kt} \] and plotting the natural logarithm of the concentration vs. time gives a straight line, \( \ln [A]_t = \ln [A]_0 - kt \). Therefore, statement c is incorrect because a plot of reciprocal concentration does not give a straight line.
04

Evaluating Option d

The pre-exponential factor in the Arrhenius equation \is usually denoted by \((A)\), and it typically has the dimension of frequency \( \text{T}^{-1} \ imes \text{concentration}^{1-n}\). In this statement, the context is not complete, but it is not exclusively the dimension of time, thus statement d is not correct.

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Key Concepts

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

Half-life (t1/2)
In the context of first-order reactions, the concept of half-life (\( t_{1/2} \)) is fundamental. The half-life is defined as the time required for half of the initial concentration of a reactant to be consumed. One interesting property of first-order reactions is that the half-life remains constant, regardless of the starting concentration. This makes it unique compared to other reaction orders. If you know the rate constant (\( k \)) of a first-order reaction, you can calculate the half-life using the equation: \( t_{1/2} = \frac{0.693}{k} \). Thus, for a reaction with a given rate constant, the time for half of the substance to react will always be the same.
Additionally, the relationship between half-life and the time taken to complete any percentage of the reaction is simple for first-order kinetics. For example, the time required to complete 75% of the reaction (\( t_{75\%} \)) is exactly twice the half-life. This is derived by recognizing that reaching 75% completion involves effectively two half-lives: the first half-life takes the concentration from 100% to 50%, and the second takes it from 50% to 25% remaining (or 75% reacted).
  • Half-life remains constant for first-order reactions.
  • Can be calculated via \( t_{1/2} = \frac{0.693}{k} \).
  • \( t_{75\%} \) = 2 \times \( t_{1/2} \).
Understanding these characteristics can help clarify many problems involving time calculations in first-order reactions.
Degree of dissociation
The degree of dissociation is crucial for understanding how much a reactant has transformed into products in a reaction. For first-order reactions, it is determined by the expression \( \alpha = 1 - e^{-kt} \), where \( \alpha \) represents the fraction of the original reactant that has dissociated.
The formula indicates that as time (\( t \)) increases, the factor \( e^{-kt} \) becomes smaller, leading to a larger \( \alpha \). This shows that over time, more and more reactant molecules are breaking down into products.
It is particularly insightful because it links the concept of reaction kinetics with real-time changes in the chemical system. The degree of dissociation can help predict how complete a reaction is at any given moment, based on the known rate constant (\( k \)) and time.
  • Represents how much reactant has turned into product.
  • Given by formula \( \alpha = 1 - e^{-kt} \).
  • Increases over time as the reaction proceeds.
This allows chemists to predict the extent of reactions, which is essential for both analytical and synthetic purposes.
Arrhenius equation
The Arrhenius equation is fundamental in understanding the effect of temperature on reaction rates. It is represented by \( k = A e^{-Ea/RT} \), where \( k \) is the rate constant, \( A \) is the pre-exponential factor, \( Ea \) is the activation energy, \( R \) is the gas constant, and \( T \) is the temperature in Kelvin.
The pre-exponential factor, \( A \), sometimes called the frequency factor, indicates the number of times that reactants approach the activation barrier per unit time. While it is often expressed in units of frequency or concentration, it embodies the idea of the likelihood of the reactants successfully colliding in the correct orientation.
In the equation, as the temperature (\( T \)) increases, the factor \( e^{-Ea/RT} \) increases, resulting in a higher rate constant (\( k \)). This means reactions proceed faster at higher temperatures. The Arrhenius equation helps chemists to understand and predict how reaction rates change with temperature, which is crucial for processes where control over speed is important.
  • Describes temperature dependence of reaction rates.
  • \( A \) suggests successful collision frequency.
  • Higher temperature leads to increased reaction speed.
Having the capability to assess reaction behavior with temperature variations is key in both laboratory and industrial chemistry.

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Most popular questions from this chapter

The rate law for the reaction \(\mathrm{RCl}+\mathrm{NaOH}\) (aq) \(\rightarrow \mathrm{ROH}+\mathrm{NaCl}\) is given by Rate \(=\mathrm{k}[\mathrm{RCl}] .\) The rate of the reaction will be a. Doubled on doubling the concentration of sodium hydroxide b. Halved on reducing the concentration of alkyl halide to one half c. Decreased on increasing the temperature of reaction d. Unaffected by increasing the temperature of the reaction.

For a reaction \(\mathrm{X} \rightarrow \mathrm{Y}\), the rate law is \(\mathrm{R}=\mathrm{k}[\mathrm{X}]^{2}\), which of the following statements are correct? a. \(\mathrm{K}\) remains constant in the reaction at all temperatures. b. The unit of rate constant Litre \(\mathrm{mole}^{-1} \mathrm{~s}^{-1}\). c. The reaction follows first order kinetics. d. The \(t_{1 / 2}\) of the reaction depends upon initial concentration of reactant.

The rate equation for a chemical reaction is Rate of reaction \(=\mathrm{k}[\mathrm{X}][\mathrm{Y}]\) Consider the following statements in this regard (1) The order of reaction is one (2) The molecularity of reaction is two (3) The rate constant depends upon temperature Of these statements: a. 1 and 3 are correct b. 1 and 2 are correct c. 2 and 3 are correct d. 1,2 and 3 are correct

From the following data for the reaction between \(\mathrm{A}\) and \(\mathrm{B}\) \(\begin{array}{llll}{[\mathrm{A}]} & {[\mathrm{B}]} & \text { initial rate } & \left.(\mathrm{mol}]^{-1} \mathrm{~s}^{-1}\right) \\ \mathrm{mol} 1^{-1} & \mathrm{~mol} & \mathrm{l}^{-1} 300 \mathrm{~K} & 320 \mathrm{~K} \\ 2.5 \times 10^{-4} & 3.0 \times 10^{-5} & 5.0 \times 10^{-4} & 2.0 \times 10^{-3} \\\ 5.0 \times 10^{-4} & 6.0 \times 10^{-5} & 4.0 \times 10^{-3} & \- \\ 1.0 \times 10^{-3} & 6.0 \times 10^{-5} & 1.6 \times 10^{-2} & -\end{array}\) Calculate the rate of the equation. a. \(\mathrm{r}=\mathrm{k}[\mathrm{B}]^{1}\) b. \(\mathrm{r}=\mathrm{k}[\mathrm{A}]^{2}\) c. \(r=k[A]^{2}[B]^{1}\) d. \(\mathrm{r}=\mathrm{k}[\mathrm{A}][\mathrm{B}]\)

Which of the following is/are experimentally determined? a. Rate law b. Order c. Molecularity d. Rate constant

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