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For a first order reaction, a. The degree of dissociation is equal to \(\left(1-\mathrm{e}^{-\mathrm{k}} \mathrm{t}\right)\) b. The pre-exponential factor in the Arrhenius equation has the dimensions of time \(\mathrm{T}^{-1}\). c. The time taken for the completion of \(75 \%\) reaction is thrice the \(t 1 / 2\) of the reaction. d. both (a) and (b)

Short Answer

Expert verified
The correct answer is (d).

Step by step solution

01

Identify characteristics of a first order reaction

A first order reaction has a rate that is directly proportional to the concentration of one reactant. The rate equation can be expressed as \(-\frac{d[A]}{dt} = k[A]\), where \(k\) is the rate constant.
02

Consider option (a)

For a first order reaction, the concentration of a reactant over time is given by \([A] = [A]_0\mathrm{e}^{-kt}\). The degree of dissociation at any time \(t\) is \(1 - \frac{[A]}{[A]_0} = 1 - \mathrm{e}^{-kt}\), which matches the expression given in option (a). So, option (a) is correct.
03

Check option (b)

The Arrhenius equation \(k = Ae^{-Ea/RT}\) involves the pre-exponential factor \(A\) that dictates the rate constant's units. For a first-order reaction, \(k\) has units \(T^{-1}\), and since \(e^{-Ea/RT}\) is dimensionless, \(A\) also has units of \(T^{-1}\). Hence, option (b) is correct.
04

Evaluate option (c)

In a first-order reaction, the half-life \(t_{1/2} = \frac{0.693}{k}\). If it takes thrice the half-life for 75% completion, that time would be \(t = 3t_{1/2}\), so \(t = 3 \times \frac{0.693}{k} = \frac{2.079}{k}\). Solving \(1 - 0.25 = \mathrm{e}^{-kt}\) gives \(t = \frac{1.386}{k}\), meaning \(t > 3t_{1/2}\) doesn't align with option (c). Thus, option (c) is incorrect.
05

Determine the correct answer

Both options (a) and (b) are true as evaluated in the steps above, and option (c) was shown incorrect. Hence, option (d) suggesting both (a) and (b) is correct is the right choice.

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Key Concepts

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

Rate Equation
The rate equation is a mathematical representation of how the rate of a chemical reaction depends on the concentration of the reactants. For a first order reaction, the rate is directly proportional to the concentration of one reactant. This can be expressed with the equation: \(-\frac{d[A]}{dt} = k[A]\). Here, \([A]\) represents the concentration of the reactant at time \(t\), and \(k\) is the rate constant.
This equation shows that as the concentration of \([A]\) decreases over time, the rate of the reaction also decreases.
  • The negative sign indicates that the concentration of \([A]\) is decreasing.
  • The rate constant \(k\) is specific to a reaction at a given temperature.
Understanding the rate equation helps in predicting how fast a reaction will proceed under certain conditions.
Arrhenius Equation
The Arrhenius equation describes how the rate constant \(k\) of a reaction changes with temperature. It is given by \(k = Ae^{-Ea/RT}\). This equation highlights several key points:
  • \(A\) is known as the pre-exponential factor, which we will discuss in detail later.
  • \(E_a\) is the activation energy required for the reaction to occur.
  • \(R\) is the universal gas constant.
  • \(T\) is the temperature in Kelvin.
The Arrhenius equation shows that as temperature increases, the exponential term becomes larger, increasing \(k\) and thus, the reaction rate. A lower activation energy \(E_a\) will also increase the reaction rate as it requires less energy for the reactants to be converted into products.
Degree of Dissociation
The degree of dissociation is a measure of how much of a reactant has been converted into products over time. For a first order reaction, this can be expressed as \(1-\mathrm{e}^{-kt}\), where \(\mathrm{e}^{-kt}\) represents the fraction of the original concentration \([A]_0\) that remains unreacted at time \(t\).
This formula shows that as time progresses, \(\mathrm{e}^{-kt}\) becomes smaller, meaning more reactant has dissociated.
  • At \(t = 0\), no dissociation has occurred, so the degree of dissociation is 0.
  • As \(t \to \infty\), the entire reactant would ideally dissociate, reaching a degree of close to 1.
Understanding the degree of dissociation allows chemists to predict how long it will take for a certain amount of reactant to be consumed in a reaction.
Pre-exponential Factor
The pre-exponential factor, denoted as \(A\) in the Arrhenius equation, is a constant that represents the frequency of collisions and the orientation factors necessary for effective collisions at any given temperature.
In first order reactions, \(A\) shares the same units as the rate constant \(k\) which is \(T^{-1}\).
  • This supports the collision theory, where more frequent and appropriately oriented collisions increase the likelihood of a reaction occurring.
  • The pre-exponential factor can vary widely between different reactions based on the complexity and nature of the molecules involved.
By combining \(A\) with the exponential term, the Arrhenius equation offers a comprehensive view of how factors such as temperature and molecular orientation affect reaction rates.

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Most popular questions from this chapter

Consider the following statements: (1) Rate of a process is directly proportional to its free energy change. (2) The order of an elementary reaction step can be determined by examining the stoichiometry. (3) The first order reaction describe exponential time coarse. Of the statements a. 1 and 2 are correct b. 1 and 3 are correct c. 2 and 3 are correct d. 1,2 and 3 are correct

The following set of data was obtained by the method of initial rates for the reaction: $$ \begin{array}{r} 2 \mathrm{HgCl}_{2}(\mathrm{aq})+\mathrm{C}_{2} \mathrm{O}_{4}^{2-}(\mathrm{aq}) \rightarrow \\ 2 \mathrm{Cl}^{-}(\mathrm{aq})+2 \mathrm{CO}_{2}(\mathrm{~g})+\mathrm{Hg}_{2} \mathrm{Cl}_{2} \end{array} $$ $$ \begin{array}{lll} \hline\left[\mathrm{HgCl}_{2}\right], \mathrm{M} & {\left[\mathrm{C}_{2} \mathrm{O}_{4}^{2-}\right], \mathrm{M}} & \text { Rate, } \mathrm{M} / \mathrm{s} \\ \hline 0.10 & 0.10 & 1.3 \times 10^{-7} \\ 0.10 & 0.20 & 5.2 \times 10^{-7} \\ 0.20 & 0.20 & 1.0 \times 10^{-6} \\ \hline \end{array} $$ What is the value of the rate constant, \(\mathrm{k}\) ? a. \(1.6 \times 10^{-4} 1 / \mathrm{M}^{2} \mathrm{~s}\) b. \(1.3 \times 10^{-4} 1 / \mathrm{M}^{2} . \mathrm{s}\) c. \(1.4 \times 10^{-7} 1 / \mathrm{M}^{2} . \mathrm{s}\) d. \(1.3 \times 10^{-6} 1 / \mathrm{M}^{2} \cdot \mathrm{s}\)

The following set of data was obtained by the method of initial rates for the reaction: $$ \begin{array}{r} \mathrm{BrO}_{3}^{-}(\mathrm{aq})+5 \mathrm{Br}(\mathrm{aq})+6 \mathrm{H}^{+}(\mathrm{aq}) \rightarrow \\ 3 \mathrm{Br}_{2}(\mathrm{aq})+3 \mathrm{H}_{2} \mathrm{O} \text { (1) } \end{array} $$ Calculate the initial rate when \(\mathrm{BrO}_{3}^{-}\)is \(0.30 \mathrm{M}\), \(\mathrm{Br}\) is \(0.050 \mathrm{M}\) and \(\mathrm{H}^{+}\)is \(0.15 \mathrm{M}\). $$ \begin{array}{llll} \hline\left[\mathrm{BrO}_{3}^{-}\right], \mathrm{M} & {[\mathrm{Br}], \mathrm{M}} & {\left[\mathrm{H}^{+}\right], \mathrm{M}} & \text { Rate, } \mathrm{M} / \mathrm{s} \\ \hline 0.10 & 0.10 & 0.10 & 8.0 \times 10^{-4} \\ 0.20 & 0.10 & 0.10 & 1.6 \times 10^{-3} \\ 0.20 & 0.15 & 0.10 & 2.4 \times 10^{-3} \\ 0.10 & 0.10 & 0.25 & 5.0 \times 10^{-3} \\ \hline \end{array} $$ a. \(3.17 \times 10^{-4} \mathrm{M} / \mathrm{s}\) b. \(6.7 \times 10^{-3} \mathrm{M} / \mathrm{s}\) c. \(2.7 \times 10^{-3} \mathrm{M} / \mathrm{s}\) d. \(1.71 \times 10^{-3} \mathrm{M} / \mathrm{s}\)

The first order isomerization reaction: Cyclopropane \(\rightarrow\) Propene, has a rate constant of \(1.10 \times 10^{-4} \mathrm{~s}^{-1}\) at \(470^{\circ} \mathrm{C}\) and \(5.70 \times 10^{-4} \mathrm{~s}^{-1}\) at \(500^{\circ} \mathrm{C}\). What is the activation energy (Ea) for the reaction? a. \(340 \mathrm{~kJ} / \mathrm{mol}\) b. \(260 \mathrm{~kJ} / \mathrm{mol}\) c. \(160 \mathrm{~kJ} / \mathrm{mol}\) d. \(620 \mathrm{~kJ} / \mathrm{mol}\)

The following set of data was obtained by the method of initial rates for the reaction: $$ \begin{aligned} &\mathrm{S}_{2} \mathrm{O}_{8}^{2-}(\mathrm{aq})+3 \mathrm{I}^{-}(\mathrm{aq}) \rightarrow \\ &2 \mathrm{SO}_{4}^{2-}(\mathrm{aq})+\mathrm{I}_{3}-(\mathrm{aq}) \end{aligned} $$ What is the rate law for the reaction? $$ \begin{array}{lll} \hline\left[\mathrm{S}_{2} \mathrm{O}_{8}^{2-}\right], \mathrm{M} & {[\mathrm{I}-], \mathrm{M}} & \text { Initial rate, } \mathrm{M} \mathrm{s}^{-1} \\ \hline 0.25 & 0.10 & 9.00 \times 10^{-3} \\ 0.10 & 0.10 & 3.60 \times 10^{-3} \\ 0.20 & 0.30 & 2.16 \times 10^{-2} \\ \hline \end{array} $$ a. Rate \(=\mathrm{k}\left[\mathrm{S}_{2} \mathrm{O}_{8}^{2-}\right]\left[\mathrm{I}^{-}\right]^{2}\) b. Rate \(=\mathrm{k}\left[\mathrm{S}_{2} \mathrm{O}_{8}^{2-}\right]^{2}\left[\mathrm{I}^{-}\right]\) c. Rate \(=\mathrm{k}\left[\mathrm{S}_{2} \mathrm{O}_{8}^{2-}\right]\left[\mathrm{I}^{-}\right]\) d. Rate \(=\mathrm{k}\left[\mathrm{S}_{2} \mathrm{O}_{8}^{2-}\right]\left[\mathrm{I}^{-}\right]^{5}\)

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