/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 24 The temperature dependence of ra... [FREE SOLUTION] | 91Ó°ÊÓ

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The temperature dependence of rate constant \((\mathrm{k})\) of a chemical reaction is written in terms of Arrhenius equation, \(\mathrm{k}=\mathrm{A} \cdot \mathrm{e}^{-\mathrm{E}^{*} \mathrm{RT}}\). Activation energy \(\left(\mathrm{E}^{*}\right)\) of the reaction can be calculated by ploting a. \(\log \mathrm{k} \operatorname{vs} \mathrm{T}^{-1}\) b. \(\log \mathrm{k} \mathrm{vs} \frac{1}{\log \mathrm{T}}\) c. \(\mathrm{k}\) vs \(\mathrm{T}\) d. \(\mathrm{k} \operatorname{vs} \frac{1}{\log \mathrm{T}}\)

Short Answer

Expert verified
Plot \( \log k \) vs \( T^{-1} \) (Option a) to find the activation energy.

Step by step solution

01

Understand the Arrhenius Equation

The Arrhenius equation is expressed as \( k = A \cdot e^{-E^*/RT} \), where \( k \) is the rate constant, \( A \) is the pre-exponential factor, \( E^* \) is the activation energy, \( R \) is the universal gas constant, and \( T \) is the temperature in Kelvin. To find the activation energy, we need a linear form of this equation.
02

Linearize the Arrhenius Equation

The linear form of the Arrhenius equation can be derived by taking the natural logarithm on both sides: \( \ln k = \ln A - \frac{E^*}{R} \left( \frac{1}{T} \right) \). This is in the form of a straight line equation \( y = mx + c \), where \( y = \ln k \) and \( x = \frac{1}{T} \).
03

Identify Plot for Activation Energy

Since \( \ln k = \ln A - \frac{E^*}{R} \left( \frac{1}{T} \right) \), the slope of the line when plotting \( \ln k \) versus \( \frac{1}{T} \) gives \( -\frac{E^*}{R} \). Therefore, plotting \( \ln k \) against \( \frac{1}{T} \) (which is equivalent to \( \log k \) vs \( T^{-1} \) using base 10 logarithms) allows us to calculate the activation energy from the slope.
04

Select the Correct Plot Option

Of the given options, the plot of \( \log k \) vs \( T^{-1} \) (Option a) corresponds to the linear relationship derived from the Arrhenius equation. Using this plot, the slope can be used to calculate the activation energy \( E^* \).

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Key Concepts

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

Activation Energy
Activation energy is an essential concept in chemistry that defines the minimum amount of energy required for a chemical reaction to occur. It is often symbolized by \( E^* \). Think of it as a barrier that reactants must overcome to be transformed into products. When reactants have enough energy to surpass this barrier, a reaction happens.

In the Arrhenius equation, activation energy plays a key role. This equation, \( k = A \cdot e^{-E^*/RT} \), includes \( E^* \) in an exponent. If the activation energy is high, fewer molecules have enough energy to react, slowing down the reaction. Conversely, a lower activation energy means that more molecules can potentially react, increasing the reaction rate.

For students analyzing experiments, activation energy is significant because it helps predict how both temperature and catalysts can affect the rate of a chemical reaction. Catalysts, for instance, lower the activation energy, allowing more collisions between molecules to result in a reaction.
Rate Constant
The rate constant, denoted as \( k \), is a factor in the Arrhenius equation that quantifies the speed of a chemical reaction under specific conditions. It is not a constant in the everyday sense, since it changes with temperature.

In the equation \( k = A \cdot e^{-E^*/RT} \), the rate constant is affected by both the activation energy and temperature. Understanding \( k \) provides clues into how fast a reaction will proceed. It is influenced by:
  • The frequency of collisions among reactant molecules (represented by \( A \), the pre-exponential factor)
  • The fraction of collisions that lead to a reaction
  • The activation energy and temperature, which are essential for calculating \( k \)
In practice, the rate constant is derived from experimental data. By measuring \( k \) at different temperatures and using the Arrhenius equation, you can estimate the activation energy of the reaction.
Temperature Dependence
Temperature significantly influences the rate at which chemical reactions occur. This is where the Arrhenius equation becomes invaluable:
\[ k = A \cdot e^{-E^*/RT} \]

In this formula, \( T \) represents the temperature in Kelvin. As temperature rises, the exponential part of the equation increases, leading to a larger value of \( k \). This means more molecules have sufficient energy to overcome the activation energy, accelerating the reaction.

Conversely, if the temperature decreases, \( k \) also becomes smaller, and reactions slow down. This dependence on temperature helps chemists understand why some reactions are spontaneous or very slow at room temperature but can be either sped up or slowed down by changing the temperature.

By conducting experiments at various temperatures, students can collect rate constants and use these to analyze how temperature affects the reaction velocity. This understanding is crucial in fields like chemical engineering, where controlling reaction rates is essential.
Linear Plot
The process of creating a linear plot from a nonlinear equation like the Arrhenius equation helps in analyzing the data more easily. It involves transforming the equation into a linear form:
\[ \ln k = \ln A - \frac{E^*}{R} \left( \frac{1}{T} \right) \]

This equation is of the form \( y = mx + c \), which is a straight line. Here, \( y = \ln k \) and \( x = \frac{1}{T} \). The slope of this line is \(-\frac{E^*}{R}\), allowing us to determine activation energy directly from the slope, making the analysis straightforward.

To construct a linear plot, students must plot \( \ln k \) against \( \frac{1}{T} \). From this line, the activation energy can be calculated using the slope. This method is advantageous because it provides a clear visual representation of the relationship between the rate constant \( k \) and temperature and can help predict reaction behavior under different temperature conditions.

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Most popular questions from this chapter

The basic theory of Arrhenius equation is that (1) Activation energy and pre exponential factors are always temperature independent (2) The number of effective collisions is proportional to the number of molecule above a certain threshold energy. (3) As the temperature increases, the number of molecules with energies exceeding the threshold energy increases. (4) The rate constant in a function of temperature a. 2,3 and 4 b. 1,2 and 3 c. 2 and 3 d. 1 and 3

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 \(=\mathrm{k}\left[\mathrm{S}_{2} \mathrm{O}_{8}^{2-}\right]\left[\mathrm{I}^{-}\right]^{5}\)

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): \(2 \mathrm{FeCl}_{3}+\mathrm{SnCl}_{2} \rightarrow \mathrm{FeCl}_{2}+\mathrm{SnCl}_{4}\) is a \(3^{\text {nd }}\) order reaction ( \(\mathbf{R}\) ): The rate constant for third order reaction has unit \(\mathrm{L}^{2} \mathrm{~mol}^{-2} \mathrm{~s}^{-1}\).

What happens when the temperature of a reaction system is increased by \(10^{\circ} \mathrm{C}\) ? a. The effective number of collisions between the molecules possessing certain threshold energy increases atleast by \(100 \%\). b. The total number of collisions between reacting molecules increases atleast by \(100 \%\) c. The activation energy of the reaction is increased d. The total number of collisions between reacting molecules increases merely by \(1-2 \%\).

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. \(1.71 \times 10^{-3} \mathrm{M} / \mathrm{s}\)

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