Ch.14 - Chemical Kinetics

McMurry8th EditionChemistryISBN: 9781292336145당신이 사용하는 게 아니라요?교과서 변경

McMurry 8th Edition

Ch.14 - Chemical Kinetics

문제 130a

14장, 문제 130a

Some reactions are so rapid that they are said to be diffusion-controlled; that is, the reactants react as quickly as they can collide. An example is the neutralization of H₃O⁺ by OH^-, which has a second-order rate constant of 1.3⨉10¹¹ M^-1 s^-1 at 25 °C. (a) If equal volumes of 2.0 M HCl and 2.0 M NaOH are mixed instantaneously, how much time is required for 99.999% of the acid to be neutralized?

검증된 단계별 안내

Identify the reaction: HCl + NaOH -> NaCl + H2O. This is a neutralization reaction where H3O+ is neutralized by OH-.

Since the reaction is diffusion-controlled, use the second-order rate equation: \( \frac{1}{[A]} - \frac{1}{[A]_0} = kt \), where \([A]_0\) is the initial concentration and \([A]\) is the concentration at time \(t\).

Calculate the initial concentration of H3O+ and OH- after mixing. Since equal volumes of 2.0 M solutions are mixed, the concentration of each will be halved: \([H3O^+]_0 = 1.0\, M\) and \([OH^-]_0 = 1.0\, M\).

Determine the concentration of H3O+ when 99.999% is neutralized. This means only 0.001% remains: \([H3O^+] = 0.00001 \times [H3O^+]_0 = 0.00001 \times 1.0\, M\).

Substitute \([A]_0\), \([A]\), and \(k = 1.3 \times 10^{11} \text{ M}^{-1} \text{s}^{-1}\) into the rate equation to solve for \(t\).

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주요 개념

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Diffusion-Controlled Reactions

Diffusion-controlled reactions occur when the rate of reaction is limited by the rate at which reactants collide in solution. In these cases, the reaction proceeds as quickly as the molecules can move and encounter each other, leading to very high rate constants. This concept is crucial for understanding how quickly certain reactions can occur, especially in dilute solutions.

Second-Order Kinetics

Second-order reactions depend on the concentration of two reactants or the square of the concentration of one reactant. The rate law for a second-order reaction can be expressed as rate = k[A][B], where k is the rate constant. In the context of the given question, the second-order rate constant indicates how the reaction rate changes with varying concentrations of H3O+ and OH-.

Rate Constant and Reaction Time

The rate constant (k) is a proportionality factor in the rate law that quantifies the speed of a reaction at a given temperature. For the neutralization reaction mentioned, the rate constant is 1.3⨉1011 M-1 s-1. To determine the time required for a specific percentage of reactants to react, one can use integrated rate laws that relate concentration changes over time to the rate constant.

관련 실천

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The half-life for the first-order decomposition of N2O4 is 1.3 * 10-5 s. N2O41g2S 2 NO21g2 If N2O4 is introduced into an evacuated flask at a pressure of 17.0 mm Hg, how many seconds are required for the pressure of NO2 to reach 1.3 mm Hg?

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교과서 질문

The reaction 2 NO1g2 + O21g2S 2 NO21g2 has the thirdorder rate law rate = k3NO423O24, where k = 25 M-2 s-1. Under the condition that 3NO4 = 2 3O24, the integrated rate law is 13O242 = 8 kt +113O24022 What are the concentrations of NO, O2, and NO2 after 100.0 s if the initial concentrations are 3NO4 = 0.0200 M and 3O24 = 0.0100 M?

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교과서 질문

Consider the reversible, first-order interconversion of two molecules A and B: where k_f = 3.0⨉10-3 s^-1 is the rate constant for the forward reaction and kr = 1.0⨉10^-3 s^-1 is the rate constant for the reverse reaction. We'll see in Chapter 15 that a reaction does not go to completion but instead reaches a state of equilibrium with comparable concentrations of reactants and products if the rate constants k_f and k_r have comparable values.

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교과서 질문

Assume that you are studying the first-order conversion of a reactant X to products in a reaction vessel with a constant volume of 1.000 L. At 1 p.m., you start the reaction at 25 °C with 1.000 mol of X. At 2 p.m., you find that 0.600 mol of X remains, and you immediately increase the temperature of the reaction mixture to 35 °C. At 3 p.m., you discover that 0.200 mol of X is still present. You want to finish the reaction by 4 p.m. but need to continue it until only 0.010 mol of X remains, so you decide to increase the temperature once again. What is the minimum temperature required to convert all but 0.010 mol of X to products by 4 p.m.?

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교과서 질문

Some reactions are so rapid that they are said to be diffusion-controlled; that is, the reactants react as quickly as they can collide. An example is the neutralization of H₃O⁺ by OH^-, which has a second-order rate constant of 1.3⨉10¹¹ M^-1 s^-1 at 25 °C. (b) Under normal laboratory conditions, would you expect the rate of the acid–base neutralization to be limited by the rate of the reaction or by the speed of mixing?

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