Chapter 14: Problem 76
(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?
Chapter 14: Problem 76
(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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Get started for freeFor 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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