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(a) Most heterogeneous catalysts of importance 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 heterogeneous catalysts because smaller particles have a larger surface area to volume ratio, increasing the number of reactive sites available for reactants to interact with the catalyst. This leads to more efficient and faster reactions. (b) Adsorption plays a crucial role in heterogeneous catalysis, as it allows reactants to concentrate at the catalyst's surface, enhancing the likelihood of successful collisions leading to product formation. The catalyst lowers the activation energy barriers, enabling the reaction to occur faster or under milder conditions.

Step by step solution

01

Part (a) Importance of Particle Size in Heterogeneous Catalysts

A heterogeneous catalyst is a catalyst that is in a different phase (solid, liquid, or gas) than the reactants it is working on. The fact that these catalysts are often finely divided solid materials helps to optimize their performance in chemical reactions. The reason for this is that the particle size has a direct impact on the surface area of the catalyst. Smaller particles have a larger surface area to volume ratio, which increases the number of reactive sites available for the reactants to interact with the catalyst. By having a larger surface area, the catalyst can interact with more reactant molecules at the same time, allowing the reaction to occur more efficiently and at a faster rate. This is crucial, as it allows the catalyst to promote the reaction without getting consumed or altering the final product.
02

Part (b) Role of Adsorption in Heterogeneous Catalysts

Adsorption is the process by which molecules or atoms bind to the surface of a solid or liquid (in this case, the surface of the heterogeneous catalyst). This process plays a crucial role in the action of heterogeneous catalysts because it allows the reactants to concentrate at the surface of the catalyst, enhancing the chances of successful collisions that lead to the formation of products. In heterogeneous catalytic reactions, the following process usually occurs: 1. Adsorption: Reactant molecules adsorb to the surface of the catalyst, forming a bond between the reactant and the surface of the catalyst. 2. Reaction: Once the reactants are adsorbed on the surface, they react with one another to form products. 3. Desorption: The products are released from the surface of the catalyst. The main function of the catalyst is to lower the activation energy barriers for reactants to transform into products. Adsorption helps in this process by holding the reactants in close proximity, allowing them to form the transition state more easily. The catalyst provides an alternative pathway for the reaction with lower activation energy, enabling the reaction to occur faster or under milder conditions. In summary, adsorption is an essential step in heterogeneous catalysis, as it allows reactants to interact with the catalyst more effectively, leading to a faster and more efficient reaction.

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Most popular questions from this chapter

Consider the following reaction between mercury(II) chloride and oxalate ion: \(2 \mathrm{HgCl}_{2}(a q)+\mathrm{C}_{2} \mathrm{O}_{4}{ }^{2-}(a q)\) \(2 \mathrm{Cl}^{-}(a q)+2 \mathrm{CO}_{2}(g)+\mathrm{Hg}_{2} \mathrm{Cl}_{2}(s)\) The initial rate of this reaction was determined for several concentrations of \(\mathrm{HgCl}_{2}\) and \(\mathrm{C}_{2} \mathrm{O}_{4}{ }^{2-}\), and the rate constant in units of \(\mathrm{s}^{-1}\). (c) Calculate the half-life of the reaction. (d) How long does it take for the absorbance to fall to \(0.100\) ? $$ \begin{array}{llll} \text { Experiment } & {\left[\mathrm{HgCl}_{2} \mathrm{~J}(\mathrm{M})\right.} & {\left[\mathrm{C}_{2} \mathrm{O}_{4}{ }^{2-}\right](M)} & \text { Rate }(\mathrm{M} / \mathrm{s}) \\ \hline 1 & 0.164 & 0.15 & 3.2 \times 10^{-5} \\ 2 & 0.164 & 0.45 & 2.9 \times 10^{-4} \\ 3 & 0.082 & 0.45 & 1.4 \times 10^{-4} \\ 4 & 0.246 & 0.15 & 4.8 \times 10^{-5} \\ \hline \end{array} $$ (a) What is the rate law for this reaction? (b) What is the value of the rate constant? (c) What is the reaction rate when the concentration of \(\mathrm{HgCl}_{2}\) is \(0.100 \mathrm{M}\) and that of \(\left(\mathrm{C}_{2} \mathrm{O}_{4}{ }^{2}\right)\) is \(0.25 \mathrm{M}\), if the temperature is the same as that used to obtain the data shown?

You have studied the gas-phase oxidation of \(\mathrm{HBr}\) by \(\mathrm{O}_{2}\) : $$ 4 \mathrm{HBr}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{H}_{2} \mathrm{O}(g)+2 \mathrm{Br}_{2}(g) $$ You find the reaction to be first order with respect to \(\mathrm{HBr}\) and first order with respect to \(\mathrm{O}_{2}\). You propose the following mechanism: $$ \begin{aligned} \mathrm{HBr}(g)+\mathrm{O}_{2}(g) & \rightarrow \mathrm{HOOBr}(g) \\ \mathrm{HOOBr}(g)+\mathrm{HBr}(g) & \longrightarrow 2 \mathrm{HOBr}(g) \\ \mathrm{HOBr}(g)+\mathrm{HBr}(g) \longrightarrow & \mathrm{H}_{2} \mathrm{O}(g)+\mathrm{Br}_{2}(g) \end{aligned} $$ (a) Indicate how the elementary reactions add to give the overall reaction. (Hint: You will need to multiply the coefficients of one of the equations by 2.) (b) Based on the rate law, which step is rate determining? (c) What are the intermediates in this mechanism? (d) If you are unable to detect HOBr or HOOBr among the products, does this disprove your mechanism?

NO catalyzes the decomposition of \(\mathrm{N}_{2} \mathrm{O}\), possibly by the following mechanism: $$ \begin{array}{r} \mathrm{NO}(\mathrm{g})+\mathrm{N}_{2} \mathrm{O}(\mathrm{g}) \longrightarrow \mathrm{N}_{2}(g)+\mathrm{NO}_{2}(g) \\ 2 \mathrm{NO}_{2}(g) \longrightarrow 2 \mathrm{NO}(g)+\mathrm{O}_{2}(g) \end{array} $$ (a) What is the chemical equation for the overall reaction? Show how the two steps can be added to give the overall equation. (b) Why is NO considered a catalyst and not an intermediate? (c) If experiments show that during the decomposition of \(\mathrm{N}_{2} \mathrm{O}, \mathrm{NO}_{2}\) does not accumulate in measurable quantities, does this rule out the proposed mechanism? If you think not, suggest what might be going on.

Explain why rate laws generally cannot be written from balanced equations. Under what circumstance is the rate law related directly to the balanced equation for a reaction?

Sucrose \(\left(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\right)\), which is commonly known as table sugar, reacts in dilute acid solutions to form two simpler sugars, glucose and fructose, both of which have the formula \(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\) : At \(23^{\circ} \mathrm{C}\) and in \(0.5 \mathrm{M} \mathrm{HCl}\), the following data were obtained for the disappearance of sucrose: $$ \begin{array}{cl} \hline \text { Time }(\mathrm{min}) & \left.\mathrm{IC}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\right](M) \\ \hline 0 & 0.316 \\ 39 & 0.274 \\ 80 & 0.238 \\ 140 & 0.190 \\ 210 & 0.146 \end{array} $$ (a) Is the reaction first order or second order with respect to \(\left[\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\right] ?\) (b) What is the value of the rate constant?

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