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Ch.14 - Chemical Kinetics
All textbooksBrown 14th EditionCh.14 - Chemical KineticsProblem 117
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Chapter 14, Problem 117

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.)

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Step 1: Determine the initial moles of N2O5 using the concentration and volume of the solution. Use the formula: moles = concentration (M) * volume (L).
Step 2: Use the first-order kinetics equation to find the remaining concentration of N2O5 after 20.0 hours. The equation is: [A] = [A]0 * e^(-kt), where [A]0 is the initial concentration, k is the rate constant, and t is the time in seconds.
Step 3: Calculate the change in moles of N2O5, which is the initial moles minus the moles remaining after 20.0 hours. This change represents the moles of N2O5 that have decomposed.
Step 4: Use the stoichiometry of the reaction to determine the moles of O2 produced. The balanced equation for the decomposition is: 2 N2O5 -> 4 NO2 + O2. From this, determine the moles of O2 produced from the decomposed N2O5.
Step 5: Calculate the partial pressure of O2 using the ideal gas law: PV = nRT. Use the moles of O2, the volume of the container (10.0 L), and the appropriate value for R (0.0821 L atm/mol K) to find the pressure, assuming the temperature is constant at 45°C (convert to Kelvin).
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Textbook Question

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?

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Open Question
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?
Textbook Question

Enzymes are often described as following the two-step mechanism: E + S Δ ES 1fast2 ES ¡ E + P 1slow2 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.

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Textbook Question

The reaction between ethyl iodide and hydroxide ion in ethanol 1C2H5OH2 solution, C2H5I1alc2 + OH- 1alc2 ¡ C2H5OH1l2 + I - 1alc2, 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?

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Textbook Question

The reaction between ethyl iodide and hydroxide ion in ethanol 1C2H5OH2 solution, C2H5I1alc2 + OH- 1alc2 ¡ C2H5OH1l2 + I - 1alc2, 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.

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Textbook Question

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.

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