Ozone in the upper atmosphere can be destroyed by the following two-step mechanism:

Cl(g) + O₃(g) → ClO(g) + O₂(g)

ClO(g) + O(g) → Cl(g) + O₂(g) 

(b) What is the catalyst in the reaction?

The gas-phase decomposition of ozone is thought to occur by the following two-step mechanism.

Step 1: O₃(g) ⇌ O₂(g) + O(g) (fast)

Step 2: O(g) + O₃(g) → 2 O₂ (slow)

(a) Write the balanced equation for the overall reaction.

The gas-phase decomposition of ozone is thought to occur by the following two-step mechanism.

Step 1: O₃(g) ⇌ O₂(g) + O(g) (fast)

Step 2: O(g) + O₃(g) → 2 O₂ (slow)

(b) Derive the rate law that is consistent with this mechanism. (Hint: The product appears in the rate law.)

The gas-phase decomposition of ozone is thought to occur by the following two-step mechanism.

Step 1: O₃(g) ⇌ O₂(g) + O(g) (fast)

Step 2: O(g) + O₃(g) → 2 O₂ (slow)

(d) If instead the reaction occurred in a single step, would the rate law change? If so, what would it be?

The following mechanism has been proposed for the gasphase reaction of chloroform 1CHCl32 and chlorine:

Step 1: Cl₂(g) k₁⇌ k_-1 2 Cl(g) (fast)

Step 2: Cl(g) + CHCl₃(g) k₂→ HCl(g) + CCl₃(g) (slow)

Step 3: Cl(g0 + CCl3(g) k₃→ CCl₄ (fast)

(e) What is the rate law predicted by this mechanism? (Hint: The overall reaction order is not an integer.)

In a hydrocarbon solution, the gold compound (CH₃)₃AuPH₃ decomposes into ethane (C₂H₆) and a different gold compound, (CH₃)AuPH₃. The following mechanism has been proposed for the decomposition of (CH₃)₃AuPH₃:

Step 1: (CH₃)₃AuPH₃ k₁⇌k_-1 (CH₃)₃Au + PH₃ (fast)

Step 2: (CH₃)₃Au k₂→ C₂H₆ + (CH₃)Au (slow)

Step 3: (CH₃)Au + PH₃ k₃→ (CH₃)AuPH₃ (fast)

(a) What is the overall reaction?

(b) What are the intermediates in the mechanism?

In a hydrocarbon solution, the gold compound (CH₃)₃AuPH₃ decomposes into ethane (C₂H₆) and a different gold compound, (CH₃)AuPH₃. The following mechanism has been proposed for the decomposition of (CH₃)₃AuPH₃:

Step 1: (CH₃)₃AuPH₃ k₁⇌k_-1 (CH₃)₃Au + PH₃ (fast)

Step 2: (CH₃)₃Au k₂→ C₂H₆ + (CH₃)Au (slow)

Step 3: (CH₃)Au + PH₃ k₃→ (CH₃)AuPH₃ (fast)

(c) What is the molecularity of each of the elementary steps?

In a hydrocarbon solution, the gold compound (CH₃)₃AuPH₃ decomposes into ethane (C₂H₆) and a different gold compound, (CH₃)AuPH₃. The following mechanism has been proposed for the decomposition of (CH₃)₃AuPH₃:

Step 1: (CH₃)₃AuPH₃ k₁⇌k_-1 (CH₃)₃Au + PH₃ (fast)

Step 2: (CH₃)₃Au k₂→ C₂H₆ + (CH₃)Au (slow)

Step 3: (CH₃)Au + PH₃ k₃→ (CH₃)AuPH₃ (fast)

(e) What is the rate law predicted by this mechanism?

In a hydrocarbon solution, the gold compound (CH₃)₃AuPH₃ decomposes into ethane (C₂H₆) and a different gold compound, (CH₃)AuPH₃. The following mechanism has been proposed for the decomposition of (CH₃)₃AuPH₃:

Step 1: (CH₃)₃AuPH₃ k₁⇌k_-1 (CH₃)₃Au + PH₃ (fast)

Step 2: (CH₃)₃Au k₂→ C₂H₆ + (CH₃)Au (slow)

Step 3: (CH₃)Au + PH₃ k₃→ (CH₃)AuPH₃ (fast)

(f) What would be the effect on the reaction rate of adding PH₃ to the solution of (CH₃)₃AuPH₃?

Platinum nanoparticles of diameter 2 nm are important catalysts in carbon monoxide oxidation to carbon dioxide. Platinum crystallizes in a face-centered cubic arrangement with an edge length of 3.924 Å. (b) Estimate how many platinum atoms are on the surface of a 2.0-nm Pt sphere, using the surface area of a sphere (4πr²) and assuming that the 'footprint' of one Pt atom can be estimated from its atomic diameter of 2.8 A .

Platinum nanoparticles of diameter 2 nm are important catalysts in carbon monoxide oxidation to carbon dioxide. Platinum crystallizes in a face-centered cubic arrangement with an edge length of 3.924 Å. (c) Using your results from (a) and (b), calculate the percentage of Pt atoms that are on the surface of a 2.0-nm nanoparticle. (d) Repeat these calculations for a 5.0-nm platinum nanoparticle.

One of the many remarkable enzymes in the human body is carbonic anhydrase, which catalyzes the interconversion of carbon dioxide and water with bicarbonate ion and protons. If it were not for this enzyme, the body could not rid itself rapidly enough of the CO2 accumulated by cell metabolism. The enzyme catalyzes the dehydration (release to air) of up to 107 CO2 molecules per second. Which components of this description correspond to the terms enzyme, substrate, and turnover number?

Enzymes are often described as following the two-step mechanism:

E + S ⇌ ES (fast)

ES → E + P (slow)

where E = enzyme, S = substrate, ES = enzyme9substrate complex, and P = product.

(a) If an enzyme follows this mechanism, what rate law is expected for the reaction?

Enzymes are often described as following the two-step mechanism:

E + S  ⇌ ES (fast)

ES → E + P (slow)

where E = enzyme, S = substrate, ES = enzyme9substrate complex, and P = product.

(b) Molecules that can bind to the active site of an enzyme but are not converted into product are called enzyme inhibitors. Write an additional elementary step to add into the preceding mechanism to account for the reaction of E with I, an inhibitor.

The reaction between ethyl iodide and hydroxide ion in ethanol (C₂H₅OH) solution, C₂H₅I(alc) + OH^-(alc) → C₂H₅OH(l) + I^-(alc), has an activation energy of 86.8 kJ/mol and a frequency factor of 2.10 × 10¹¹ M^-1 s^-1. (c) Which reagent in the reaction is limiting, assuming the reaction proceeds to completion?

The reaction between ethyl iodide and hydroxide ion in ethanol (C₂H₅OH) solution, C₂H₅I(alc) + OH^-(alc) → C₂H₅OH(l) + I^-(alc), has an activation energy of 86.8 kJ/mol and a frequency factor of 2.10 × 10¹¹ M^-1 s^-1. (d) Assuming the frequency factor and activation energy do not change as a function of temperature, calculate the rate constant for the reaction at 50 C.

The gas-phase reaction of NO with F2 to form NOF and F has an activation energy of Ea = 6.3 kJ>mol. and a frequency factor of A = 6.0 * 108 M-1 s-1. The reaction is believed to be bimolecular: NO1g2 + F21g2 ¡ NOF1g2 + F1g2 (e) Suggest a reason for the low activation energy for the reaction.

The mechanism for the oxidation of HBr by O₂ to form 2 H₂O and Br₂ is shown in Exercise 14.74. (a) Calculate the overall standard enthalpy change for the reaction process.

The mechanism for the oxidation of HBr by O₂ to form 2 H₂O and Br₂ is shown in Exercise 14.74. (c) Draw a plausible Lewis structure for the intermediate HOOBr. To what familiar compound of hydrogen and oxygen does it appear similar?

The rates of many atmospheric reactions are accelerated by the absorption of light by one of the reactants. For example, consider the reaction between methane and chlorine to produce methyl chloride and hydrogen chloride:

Reaction 1: CH₄(g) + Cl₂(g) → CH₃Cl(g) + HCl(g)

This reaction is very slow in the absence of light. However, Cl₂(g) can absorb light to form Cl atoms:

Reaction 2: Cl₂(g) + hv → 2 Cl(g)

Once the Cl atoms are generated, they can catalyze the reaction of CH₄ and Cl₂, according to the following proposed mechanism:

Reaction 3: CH₄(g) + Cl(g) → CH₃(g) + HCl(g)

Reaction 4: CH₃(g) + Cl₂(g) → CH₃Cl(g) + Cl(g)

The enthalpy changes and activation energies for these two reactions are tabulated as follows:

Reaction ΔH° (kJ/mol) E_a (kJ/mol)

3 +4 17

4 -109 4 

(b) By using the data tabulated here, sketch a quantitative energy profile for the catalyzed reaction represented by reactions 3 and 4.

Many primary amines, RNH₂, where R is a carboncontaining fragment such as CH₃, CH₃CH₂, and so on, undergo reactions where the transition state is tetrahedral. (a) Draw a hybrid orbital picture to visualize the bonding at the nitrogen in a primary amine (just use a C atom for 'R').