Chapter 3, Problem 3.1P

1. (a) Example 3-1. Make a plot of k versus T and ln k versus (1/T) for E = 240 kJ/mol and for E = 60 kJ/mol. (1) Write a couple of sentences describing what you find. (2) Next, write a paragraph describing the activation, how it affects chemical reaction rates, and what its origins are.
2. (b) Collision Theory—Professional Reference Shelf. Make an outline of the steps that were used to derive

   $-r_{\text{A}}=A{e}^{-E/RT}{C}_{\text{A}}{C}_{\text{B}}$

3. (c) The rate law for the reaction (2A + B → C) is $-r_{\text{A}}=k_{\text{A}}{C}_{\text{A}}^2{C}_{\text{B}}$ with k_A = 25(dm³/mol)²/s. What are k_B and k_C?

Interpretation Introduction

Interpretation:

A plot of $k$ versus $T$ and $\ln k$ versus $\left(\frac{1}{T}\right)$ for $\text{E}=\text{240}\text{kJ/mol}$ and for $\text{E}=\text{60kJ/mol}$ has to be given along with some sentences describing about what we found from the graph. A paragraph describing the activation and its affects on chemical reaction rate and its orgin have to be given.

Concept Introduction:

Arrhenius equation is a formula that represents the temperature dependence of reaction rates

  The Arrhenius equation can be represented as,

$k=A{e}^{-E/RT}$

• $E$ represents the activation energy and it’s unit is kJ/mol
• R represents the universal gas constant and it has the value of 8.314 J/K.mol
• T represents the absolute temperature
• A represents the frequency factor or collision frequency

Rate of a reaction: It represents the speed at which a chemical reaction runs.  How much concentration of substrates (reactants) consumed and how much concentration of targets (products) formed in a unit of time is said to be rate of reaction.

Rate of reaction depends on time, temperature, pressure, concentration, and $\text{pH}$ of the reaction.

Explanation of Solution

According to the Example 3.1, value of A for the given reaction is $1.32\times {10}^{16}$.  Temperture and rate constant for the given reaction are mentioned below,

$\begin{array}{cccccc}\text{k}\left({\text{s}}^{\text{-1}}\right)& \text{0}\text{.00043}& \text{0}\text{.00103}& \text{0}\text{.00180}& \text{0}\text{.00355}& \text{0}\text{.00717}\\ \text{T}\left(\text{K}\right)& \text{313}\text{.0}& \text{319}\text{.0}& \text{323}\text{.0}& \text{328}\text{.0}& \text{333}\text{.0}\end{array}$

Now,

A plot of $k$ versus $T$ and $\ln k$ versus $\left(\frac{1}{T}\right)$ for $\text{E}=\text{240}\text{kJ/mol}$ and for $\text{E}=\text{60kJ/mol}$ can be drawn as follows,

                  Figure 1

               Figure 2

The equation of the line from any of the two values of $\ln k$ and $\left(\frac{1}{T}\right)$ is calculated as,

    $y-y_1=\left(\frac{y_2-y_1}{x_2-x_1}\right)x-x_1$

By substituting the known values of $xandy$ slope of the line in the graph can be calculated and that slope will give $-\frac{E}{R}$ from this, the given reaction’s activation energy can be calculated.

A successful reaction occurs only after a successful collision between reactant molecules. Successful collision activates reactant molecules and they become an activation complex and having high amount energy. If the activation energy level decreased then the rate of the reaction will increases. 

Interpretation Introduction

Interpretation:

An outline of the steps that were used to derive the equation, $-r_A=A{e}^{\frac{-E}{RT}}{C}_A{C}_B$ has to be given.

Concept Introduction:

• The collision theory explains about the collisions that takes place between the molecules and results in the formation of a new chemical product.
• The rate of the chemical reaction depends on the number of effective collisions that takes place between the reactant molecules.
• There are many factors included in the collision theory that determines the formation of product.

The rate at which the molecule collides is called collision frequency and it can be represented as Z.

$\begin{array}{l}Collisionfrequency,Z=N_A{N}_B\sqrt{\frac{8K_{B}T}{\pi {\mu }_{AB}}}\\ \mu isthereducedmass=\frac{m_A{m}_B}{m_A+m_B}\end{array}$

There are many requirements for the effective the collisions for the reactions to take place.

• The reactant molecules must collide with each other.
• The reactant molecules must orient in proper orientation.
• The energy of reactant molecules must be greater than the activation energy.

Activation Energy

The minimum energy that is required for the reactant molecules to overcome the energy barrier for completing the reaction.

Steric factor $\left(\rho \right)$ is the probability of reactant molecules colliding with right position and orientation so that the desirable product can be formed.

Equilibrium constant: At equilibrium the ratio of products to reactants has a constant value. And it is represented by the letter K. 

For a general reaction, $\text{aA}\text{+}\text{bB}\rightleftarrows \text{cC}\text{+}\text{dD}$

The equilibrium constant $Kc$ = $\frac{{\text{[C]}}^{\text{c}}{\text{[D]}}^{\text{d}}}{{\text{[A]}}^{\text{a}}{\text{[B]}}^{\text{b}}}$ a, b, c and d are the stoichiometric coefficients of reactant and product in the reaction. Concentration value for solid substance is 1.

Explanation of Solution

The equation $-r_A=A{e}^{\frac{-E}{RT}}{C}_A{C}_B$ can be derived as follows,

Considering a general reaction,

$\text{A+}\text{B}\to \text{Product}\left(C\right)$

According to the collision theory,

$\begin{array}{l}Collisionfrequency,Z=N_A{N}_B\sqrt{\frac{8K_{B}T}{\pi {\mu }_{AB}}}\\ \mu isthereducedmass=\frac{m_A{m}_B}{m_A+m_B}\end{array}$

From the energy barrier consideration,

Fraction of collision which occurs with energy above activation energy is,

               $f=e^{\frac{{}^{-E_{}}}{RT}}$

Including steric factor with the above two terms,

Then,

$\begin{array}{c}\text{k}\text{=}\text{Z}{\text{ρe}}^{\frac{{\text{-E}}_{}}{\text{RT}}}\\ \text{A}\text{=}\text{Pre-exponential}\text{factor}\text{=}\text{Zρ}\\ \text{k}\text{=}\text{Rate}\text{constant}\end{array}$

Therefore,

$\text{k}\text{=}{\text{Ae}}^{\frac{{\text{-E}}_{}}{\text{RT}}}$

Equailbrium constant for the given reaction ($\text{A+}\text{B}\to \text{Product}\left(C\right)$) is

$K=\frac{\left[C\right]}{\left[A\right]\left[B\right]}$

According to transition state model,

$\underset{\text{Reactants}}{\text{A+}\text{B}}\to \underset{\text{Transition}\text{state}}{{\text{AB}}^{\text{*}}}\to \underset{\text{Product}}{\text{C}}$

Rate constant for this step can be written as follows,

$\begin{array}{c}{K}^{*}=\frac{\left[A{B}^{*}\right]}{\left[A\right]\left[B\right]}\\ Rate=k^{\text{'}}\left[A{B}^{*}\right]\\ =k^{\text{'}}.k^{*}\left[A\right]\left[B\right]\\ =k\left[A\right]\left[B\right];k=k^{\text{'}}.k^{*}\\ Rate=A{e}^{\frac{-E}{RT}}.\left[A\right]\left[B\right]\\ =A{e}^{\frac{-E}{RT}}{C}_A{C}_B\end{array}$

Thus,

$-r_A=A{e}^{\frac{-E}{RT}}{C}_A{C}_B$

Interpretation Introduction

Interpretation:

The value for $k_{B}and{k}_C$ has to be given for the reaction $2A+B\to C$

Concept Introduction:

Rate equation for the general reaction $\text{A+B}\to \text{Product}$ is,

$\begin{array}{l}Rate=\underset{\begin{array}{l}rate\\ constat\end{array}}{k}\left[A\right]\left[B\right]\\ or\\ -r_A=\underset{\begin{array}{l}rate\\ constat\end{array}}{k}{C}_A.C_B\end{array}$

Explanation of Solution

The given reaction is $2A+B\to C$.

The reaction can also be written as,

$A+\frac{1}{2}B\to \frac{1}{2}C$

The rate law for is $-r_A=k_A{C}_A^2{C}_B$

${\text{k}}_{\text{A}}=\text{25}{\left({\text{dm}}^{\text{3}}\text{/mol}\right)}^{\text{2}}\text{/s}$

$\begin{array}{c}\frac{{\text{r}}_{\text{A}}}{\text{-1}}\text{=}\frac{{\text{r}}_{\text{B}}}{\text{-1/2}}\text{=}\frac{{\text{r}}_{\text{C}}}{\text{1/2}}\\ \text{=}\text{25}{\text{C}}_{\text{A}}^{\text{2}}{\text{C}}_{\text{B}}\\ \text{=}\frac{{\text{k}}_{\text{B}}{\text{C}}_{\text{A}}^{\text{2}}{\text{C}}_{\text{B}}}{\text{1/2}}\text{=}\frac{{\text{k}}_{\text{c}}{\text{C}}_{\text{A}}^{\text{2}}{\text{C}}_{\text{B}}}{\text{-1/2}}\\ {\text{k}}_{\text{C}}\text{=}{\text{k}}_{\text{B}}\text{=}\text{12}\text{.5}\frac{\text{1}}{\text{s}}{\left(\frac{{\text{dm}}^{\text{3}}}{\text{mol}}\right)}^{\text{2}}\end{array}$