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(a) Most commercial heterogeneous catalysts are extremely finely divided solid materials. Why is particle size important? (b) What role does adsorption play in the action of a heterogeneous catalyst?

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(a) Particle size is important in commercial heterogeneous catalysts because smaller particles provide a larger surface area per unit mass, leading to a greater number of active sites for catalytic reactions. Consequently, this results in higher catalytic activity and efficiency. (b) Adsorption plays a crucial role in heterogeneous catalysis by increasing reactant concentrations on the catalyst surface and promoting more effective collisions between reactants, thus enhancing the overall reaction rate. It consists of three steps: adsorption of reactants, reaction on the catalyst surface, and desorption of products.

Step by step solution

01

(a) Importance of particle size in heterogeneous catalysts

Particle size plays a major role in heterogeneous catalysts because it directly affects the surface area. When the particle size is smaller, the surface area of the catalyst per unit mass increases, leading to a larger number of active sites available for catalytic reactions. This increased surface area results in a higher catalytic activity for reactions occurring on the surface of the catalyst. Therefore, commercial heterogeneous catalysts are designed to have small, finely divided particles to maximize their activity, effectiveness, and overall efficiency.
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(b) Role of adsorption in heterogeneous catalysis

Adsorption is the process where a substance, in this case, the reactants, form a thin layer on the surface of a solid or a liquid, such as the heterogeneous catalyst. In heterogeneous catalysis, the catalyst is typically a solid, while the reactants are in a gas or liquid phase. The role of adsorption in heterogeneous catalysis can be divided into the following steps: 1. Adsorption of reactants: Reactant molecules are adsorbed onto the surface of the catalyst, which increases their concentration on the catalyst's surface. This causes the probability of effective collisions between the reactants to increase, thus increasing the reaction rate. 2. Reaction on the catalyst surface: Within the adsorbed layer, the reactant molecules are brought closer together, allowing them to form bonds more easily. The energy required for the reaction to proceed is often lower due to the orientation and proximity of the reactants, which enhances the reaction rate. 3. Desorption of products: After the reaction occurs, the product molecules no longer need to remain on the catalyst's surface, so they desorb and are released into the surrounding medium. Overall, adsorption plays a vital role in the action of a heterogeneous catalyst by increasing reactant concentrations on the catalyst surface and promoting more effective collisions between the reactants, which in turn enhances the overall reaction rate.

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Most popular questions from this chapter

For the elementary process \(\mathrm{N}_{2} \mathrm{O}_{5}(g) \longrightarrow \mathrm{NO}_{2}(g)+\mathrm{NO}_{3}(g)\) the activation energy \(\left(E_{a}\right)\) and overall \(\Delta E\) are 154 \(\mathrm{kJ} / \mathrm{mol}\) and 136 \(\mathrm{kJ} / \mathrm{mol}\) , respectively. (a) Sketch the energy profile for this reaction, and label \(E_{a}\) and \(\Delta E\) . (b) What is the activation energy for the reverse reaction?

Consider the hypothetical reaction \(2 \mathrm{A}+\mathrm{B} \longrightarrow 2 \mathrm{C}+\mathrm{D}\) . The following two-step mechanism is proposed for the reaction: $$ \begin{array}{l}{\text { Step } 1 : \mathrm{A}+\mathrm{B} \longrightarrow \mathrm{C}+\mathrm{X}} \\ {\text { Step } 2 : \mathrm{A}+\mathrm{X} \longrightarrow \mathrm{C}+\mathrm{D}}\end{array}$$ \(X\) is an unstable intermediate. (a) What is the predicted rate law expression if Step 1 is rate determining? (b) What is the predicted rate law expression if Step 2 is rate determining? (c) Your result for part (b) might be considered surprising for which of the following reasons: (i) The concentration of a product is in the rate law. (ii) There is a negative reaction order in the rate law. (ii) Both reasons (i) and (ii). (iv) Neither reasons (i) nor (ii).

The reaction \(2 \mathrm{NO}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{NO}_{2}(g)\) is second order in \(\mathrm{NO}\) and first order in \(\mathrm{O}_{2}\) . When \([\mathrm{NO}]=0.040 \mathrm{M}\) and \(\left[\mathrm{O}_{2}\right]=0.035 \mathrm{M},\) the observed rate of disappearance of \(\mathrm{NO}\) is \(9.3 \times 10^{-5} \mathrm{M} / \mathrm{s}\) . (a) What is the rate of disappearance of \(\mathrm{O}_{2}\) at this moment? (b) What is the value of the rate constant? (c) What are the units of the rate constant? (d) What would happen to the rate if the concentration of NO were increased by a factor of 1.8\(?\)

The gas-phase reaction of NO with \(\mathrm{F}_{2}\) to form \(\mathrm{NOF}\) and \(\mathrm{F}\) has an activation energy of \(E_{a}=6.3 \mathrm{kJ} / \mathrm{mol} .\) and a frequency factor of \(A=6.0 \times 10^{8} M^{-1} \mathrm{s}^{-1} .\) The reaction is believed to be bimolecular: $$ \mathrm{NO}(g)+\mathrm{F}_{2}(g) \longrightarrow \mathrm{NOF}(g)+\mathrm{F}(g)$$ (a) Calculate the rate constant at \(100^{\circ} \mathrm{C}\) . (b) Draw the Lewis structures for the NO and the NOF molecules, given that the chemical formula for NOF is misleading because the nitrogen atom is actually the central atom in the molecule, (c) Predict the shape for the NOF molecule.Draw a possible transition state for the formation of NOF, using dashed lines to indicate the weak bonds that are beginning to form. (e) Suggest a reason for the low activation energy for the reaction.

The following mechanism has been proposed for the gasphase reaction of chloroform (CHCl_ ) and chlorine:$$\begin{array}{l}{\text { Step } 1 : \mathrm{Cl}_{2}(g) \frac{k_{1}}{k_{-1}} 2 \mathrm{Cl}(g) \text { (fast) }} \\\ {\text { Step } 2 : \mathrm{Cl}(g)+\mathrm{CHCl}_{3}(g) \stackrel{k_{2}}{\longrightarrow} \mathrm{HCl}(g)+\mathrm{CCl}_{3}(g) \text { (slow) }}\end{array}$$ $$ { Step } \quad3 : \quad \mathrm{Cl}(g)+\mathrm{CCl}_{3}(g) \stackrel{k_{3}}{\longrightarrow} \mathrm{CCl}_{4} \quad(\text { fast })$$ (a) What is the overall reaction? (b) What are the intermedi- ates in the mechanism? (c) What is the molecularity of each of the elementary reactions? (d) What is the rate-determining step? (e) What is the rate law predicted by this mechanism? (Hint: The overall reaction order is not an integer.)

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