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Problem 101a
The rate of a first-order reaction is followed by spectroscopy, monitoring the absorbance of a colored reactant at 520 nm. The reaction occurs in a 1.00-cm sample cell, and the only colored species in the reaction has an extinction coefficient of 5.60 × 103 M-1 cm-1 at 520 nm.
(a) Calculate the initial concentration of the colored reactant if the absorbance is 0.605 at the beginning of the reaction.
Problem 101d
The rate of a first-order reaction is followed by spectroscopy, monitoring the absorbance of a colored reactant at 520 nm. The reaction occurs in a 1.00-cm sample cell, and the only colored species in the reaction has an extinction coefficient of 5.60 × 103 M-1 cm-1 at 520 nm.
(d) How long does it take for the absorbance to fall to 0.100?
- A colored dye compound decomposes to give a colorless product. The original dye absorbs at 608 nm and has an extinction coefficient of 4.7 * 10^4 M^-1 cm^-1 at that wavelength. You perform the decomposition reaction in a 1-cm cuvette in a spectrometer and obtain the following data: Time (min) Absorbance at 608 nm 0 1.254 30 0.941 60 0.752 90 0.672 120 0.545. From these data, determine the rate law for the reaction 'dye → product' and determine the rate constant.
Problem 102
- Cyclopentadiene (C5H6) reacts with itself to form dicyclopentadiene (C10H12). A 0.0400 M solution of C5H6 was monitored as a function of time as the reaction 2 C5H6 → C10H12 proceeded. The following data were collected: Time (s) | [C5H6] (M) 0.0 | 0.0400 50.0 | 0.0300 100.0 | 0.0240 150.0 | 0.0200 200.0 | 0.0174 Plot [C5H6] versus time, ln[C5H6] versus time, and 1/[C5H6] versus time. (b) What is the value of the rate constant?
Problem 103
- The first-order rate constant for the reaction of a particular organic compound with water varies with temperature as follows: Temperature (K) Rate Constant (s⁻¹) 300 3.2 × 10⁻¹¹, 320 1.0 × 10⁻⁹, 340 3.0 × 10⁻⁸, 355 2.4 × 10⁻⁷. From these data, calculate the activation energy in units of kJ/mol.
Problem 104
- The following is a quote from an article in the August 18, 1998, issue of The New York Times about the breakdown of cellulose and starch: “A drop of 18 degrees Fahrenheit [from 77 _x001E_F to 59 _x001E_F] lowers the reaction rate six times; a 36-degree drop [from 77 _x001E_F to 41 _x001E_F] produces a fortyfold decrease in the rate.” (b) Assuming the value of Ea calculated from the 36 _x001E_ drop and that the rate of breakdown is first order with a half-life at 25 _x001E_C of 2.7 yr, calculate the half-life for breakdown at a temperature of -15 _x001E_C.
Problem 106
Problem 107a
The following mechanism has been proposed for the reaction of NO with H2 to form N2O and H2O:
NO(g) + NO(g) → N2O2(g)
N2O2(g) + H2(g) → N2O(g) + H2O(g)
(a) Show that the elementary reactions of the proposed mechanism add to provide a balanced equation for the reaction.
Problem 107d
The following mechanism has been proposed for the reaction of NO with H2 to form N2O and H2O:
NO(g) + NO(g) → N2O2(g)
N2O2(g) + H2(g) → N2O(g) + H2O(g)
(d) The observed rate law is rate = k[NO]2[H2]. If the proposed mechanism is correct, what can we conclude about the relative speeds of the first and second reactions?
Problem 108a
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)
(a) What is the overall equation for this process?
Problem 108b
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?
Problem 109a
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.
Problem 109b
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.)
Problem 109d
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?
Problem 110e
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.)
- Consider the hypothetical reaction 2 A + B → 2 C + D. The following two-step mechanism is proposed for the reaction: Step 1: A + B → C + X Step 2: A + X → C + D. X is an unstable intermediate. (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. (iii) Both reasons (i) and (ii). (iv) Neither reasons (i) nor (ii).
Problem 111
Problem 112a,b
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?
Problem 112c
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?
Problem 112e
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?
Problem 112f
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?
Problem 113b
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 .
Problem 113c,d
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.
Problem 114
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?
- Suppose that, in the absence of a catalyst, a certain biochemical reaction occurs x times per second at normal body temperature 37 °C. In order to be physiologically useful, the reaction needs to occur 5000 times faster than when it is uncatalyzed. By how many kJ/mol must an enzyme lower the activation energy of the reaction to make it useful?
Problem 115
Problem 116a
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?
Problem 116b
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.
- Dinitrogen pentoxide (N2O5) decomposes in chloroform as a solvent to yield NO2 and O2. The decomposition is first order with a rate constant at 45 _x001E_C of 1.0 * 10^-5 s^-1. Calculate the partial pressure of O2 produced from 1.00 L of 0.600 M N2O5 solution at 45 _x001E_C over a period of 20.0 h if the gas is collected in a 10.0-L container. (Assume that the products do not dissolve in chloroform.)
Problem 117
Problem 118c
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?
Problem 118d
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.
Problem 120e
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.
Problem 121a
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.